CH 17 and 19

Based on the following thermodynamic data, calculate the boiling point of ethanol in degrees Celsius. Substance _________ ΔH∘f(kJ/mol) _________ ΔS∘[J/(K⋅mol)] C2H5OH(l) ________ −277.7 ___________________ 160.6 C2H5OH(g) ________ −235.1 ___________________ 282.6

76 ∘C (My answer of 76.18 was accepted) The boiling point represents an equilibrium condition between liquid and gas: C2H5OH(l)⇌C2H5OH(g) (Therefore ΔG = 0.) Calculate the value of ΔHvap from ΔHvap=ΔH∘f(products)−ΔH∘f(reactants) Calculate the value of ΔSvap for the reaction from ΔSvap=S∘(products)−S∘(reactants) Convert the units of ΔSvap to be compatible with those of ΔHvap. Calculate the equilibrium temperature in kelvins using the following equation: T = ΔHvap / ΔSvap Convert the temperature from kelvins to degrees Celsius.

What is the pH of a buffer that was prepared by adding 3.96 g of sodium benzoate, NaC7H5O2, to 1.00 L of 0.0100 M benzoic acid, HC7H5O2? Assume that there is no change in volume. The Ka for benzoic acid is 6.3 × 10^-5.

4.64 The applicable equation is the Henderson-Hasselbalch equation: pH = pKa + log [base] / [acid]. See Section 17.2 (Calculate the concentration of benzoate ions created from the sodium benzoate. Calculate pKa. Use Henderson-Hasselbalch to calculate pH where [base] = concentration of sodium benzoate ion and [acid] = concentration of benzoic acid.)

CaCO3(s)+H+(aq)⇌Ca2+(aq)+HCO3−(aq) Equilibrium constants at 25 ∘C are listed in the table below. CaCO3: Ksp = 4.5×10^−9 H2CO3: Ka1 = 4.3×10^−7, Ka2 = 5.6×10^−11 What is the molar solubility of marble (i.e., [Ca2+] in a saturated solution) in normal rainwater, for which pH=5.60? Express your answer to two significant figures and include the appropriate units.

[Ca2+] = 1.4×10^−2 M Based on the stoichiometry of this reaction if [Ca2+] = s then [HCO3-] is also equal to s. ___ CaCO3(s) _____ H+(aq) _____ Ca2+(aq) _____ +HCO3−(aq) initial: __________________________ 0 _______________ 0 final: ___________________________ s _______________ s Based on known Kc values, the Kc value for the overall reaction: CaCO3(s) + H+(aq) ⇌ Ca2+(aq) + HCO3−(aq) is Kc = 80.4 (Multiply Ksp with the reciporacal of Ka2 as these equations are what make up the overall reaction.) [H+] in a rainwater solution with pH=5.60 is 2.51×10^−6 M

CaCO3(s)+H+(aq)⇌Ca2+(aq)+HCO3−(aq) Equilibrium constants at 25 ∘C are listed in the table below. CaCO3: Ksp = 4.5×10^−9 H2CO3: Ka1 = 4.3×10^−7, Ka2 = 5.6×10^−11 What is the molar solubility of marble (i.e., [Ca2+] in a saturated solution) in acid rainwater, for which pH=4.20? Express your answer to two significant figures and include the appropriate units.

[Ca2+] = 7.1×10^−2 M As you saw in Part A, Kc = 80.4. Furthermore, if [Ca2+]=s, then Kc = [Ca2+][HCO3−] / [H+] = s^2 / [H+] Use the pH to determine [H+], then solve for s.

A buffer contains a weak acid, HA, and its conjugate base. The weak acid has a pKa of 4.5, and the buffer has a pH of 4.7. Without doing a calculation, state which of these possibilities are correct. [HA]=[A−] [HA]<[A−] [HA]>[A−]

[HA]<[A−] (Because pKa is less than pH that means there is more base than acid in the buffer. If you take the Henderson-Hasselbalch equation and solve for what the ratio of [base] / [acid] is you'll get 1.58. Because this value is greater than 1 it shows that [acid] must be less than [base].)

You are asked to prepare a pH=4.00 buffer starting from 1.60 L of 0.0400 M solution of benzoic acid (C6H5COOH) and an excess of sodium benzoate (C6H5COONa). Given that the dissociation constant Ka for benzoic acid is equal to 6.3×10^−5. How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

m = 5.8 g (Use pH = pKa + log(base / acid) to find what the ratio of base to acid is: base / acid = 0.63 By finding the moles of acid we have we can plug it into the above ratio to find how many moles of base we need. 1.6L x 0.04M = 0.064 moles of acid. 0.63 = (mol of base) / 0.064 mol of base = 0.63 x 0.064 = 0.04032 moles Multiply this by the molar weight of sodium benzoate (144.105 g/mol) to get the amount of sodium benzoate needed (5.8 g))

C2D3 has a solubility product constant of 9.14×10−9. What is the molar solubility of C2D3?

molar solubility = 9.67×10^−3 M You are given the value of the solubility product constant Ksp. Substitute this value into the Ksp expression: Ksp = (2x)^2 * (3x)^3 = 108x^5 and solve for the molar solubility x.

The element gallium (Ga) freezes at 29.8 ∘C, and its molar enthalpy of fusion is ΔHfus=5.59kJ/mol. When molten gallium solidifies to Ga(s) at its normal melting point, is ΔS positive or negative?

negative In the liquid state, molecules move freely throughout the phase despite the intermolecular attractive forces. The process of solidification involves the loss of kinetic energy such that the molecules can no longer escape the intermolecular attraction, and they vibrate in place. Therefore, the system becomes more ordered when it freezes, and that corresponds to a decrease in entropy (ΔS<0). (Remember that ΔHfus is when a substance goes from solid to liquid (as stupid as that is). Because the question involves a liquid going to a solid that means our 5.59 value becomes -5.59. By plugging that into the ΔG = ΔH - TΔS equation we can see that ΔS must be a negative number (at equillibrium ΔG is 0). Or you can just think about how there is a decrease in entropy by going from a liquid to a solid.)

100. mL of 0.200 M HCl is titrated with 0.250 M NaOH. What is the pH of the solution after 50.0 mL of base has been added?

pH = 1.30 When conducting calculations involving a titration, the first step is to write the balanced chemical equation. Then, use the stoichiometric ratios developed from this equation to determine how many moles of each reagent are reacting. Consider the reaction HCl+NaOH→H2O+NaCl HCl and NaOH react in a 1:1 ratio. You should find the number of moles of each that you have, then determine whether any acid or base is left over after the two compounds react. (After determining what is left, remember that that is in moles so you need to convert to molarity by dividing by the total volume. Use that value to calculate the pH.)

A 30.0-mL volume of 0.50 M CH3COOH (Ka=1.8×10−5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH at 25 ∘C.

pH = 9.07 For a weak monoprotic acid titrated with strong base, the pH at the equivalence point is always greater than 7 because the anion of the weak acid is a base. When a weak acid such as CH3COOH reacts with an equal amount of the strong base, the reaction CH3COOH+NaOH→CH3COONa+H2O goes to completion. The product is a basic salt with the following equilibrium: CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq) Kb=[CH3COOH][OH−] / [CH3COO−] Calculate the salt concentration and use it as the initial concentration of CH3COO− when figuring out equilibrium concentrations. Then use the equilibrium OH− concentration to calculate [H3O+] and the pH.

A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. A solution is made by titrating 9.00 mmol (millimoles) of HA and 3.00 mmol of the strong base. More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 80.0 mL ?

pH = 9.15 The equivalence point occurs when the number of moles of acid equals the number of moles of base, in other words, when you have added 9.00 mmol of base to the 9.00 mmol of acid. Just as before, you should determine what is left over after the reaction by constructing the following table: Amount (mmol) ___ HA __ + __ OH− __ → __ A− __ + __ H2O initial: ___________ 9.00 _____ 9.00 _______ 0 change: ________ −9.00 ____ −9.00 _____ +9.00 final: _____________ 0 ________ 0 ________ ? The final pH depends on the final concentration of the conjugate base A− (not the number of moles or millimoles) and so you will have to use the given volume to find the molarity: molarity (M) = moles / liters = mmol / mL Then, treat the question as you would a typical weak base calculation by constructing an ICE table. Finally, recall that pOH = −log[OH] and that pH = 14.00 − pOH. (Don't forget to use Kb to find the value of x from the ICE table as you're now using the conjugate base, not the acid.)

Match each type of titration to its pH at the equivalence point for solutions at 25 ∘C. (What will the pH be in relation to 7) weak acid, strong base strong acid, strong base weak base, strong acid

pH less than 7: weak base, strong acid pH equal to 7: strong acid, strong base pH greater than 7: weak acid, strong base The equivalence point in an acid-base titration is the point at which stoichiometrically equivalent quantities of acid and base have been mixed together. At this point the reaction is complete because all analyte has been consumed by titrant. On a titration curve, the equivalence point is represented by the point of inflection (where the curve changes concavity). The figure (https://session.masteringchemistry.com/problemAsset/1244737/17/MC_1244737_graph.jpg) shows the titration of 40.0 mL of 0.100 M HCl with 0.100 M NaOH. When 40.0 mL of the NaOH solution is added, the acid-base neutralization reaction is complete. When analyzing titrations involving weak acids or bases, consider how the neutralization reaction will impact the pH of the system. For example, when titrating a weak acid with a strong base, at the equivalence point all the base has been used to neutralize the acid forming its conjugate weak base.

Here is the chemical reaction that represents solid lead chloride dissolving in water: PbCl2(s)⇌Pb2+(aq)+2Cl−(aq) Ksp=1.2×10−5 Which action would shift this reaction away from solid lead chloride and toward the dissolved ions? adding lead ions adding chloride ions removing chloride ions removing lead chloride

removing chloride ions The removal of lead ions or chloride ions would cause the reaction to shift to the right, thus dissolving more of the solid. The addition of a base would remove lead ions, whereas the addition of metal ions, such as silver ions, would remove chloride ions.

The chemical reaction that causes iron to corrode in air is given by 4Fe+3O2→2Fe2O3 in which at 298 K, ΔH∘rxn = −1684 kJ and ΔS∘rxn=−543.7J/K. What is the Gibbs free energy for this reaction at 3652 K ? Assume that ΔH and ΔS do not change with temperature.

ΔGrxn = 302 kJ The formula for Gibbs free energy is ΔGrxn = ΔHrxn − TΔSrxn. In this problem, ΔHrxn, ΔSrxn, and 3652 K are given to you. This positive value for ΔG means that, at extremely high temperatures, the corrosion reaction will actually go in reverse, converting iron(III) oxide into iron and oxygen.

Calculate the equilibrium constant, K, and ΔG∘ at 298 K for each of the following reactions. H2(g)+I2(g)⇌2HI(g) Calculate ΔG∘. Use the following data: ΔG∘f(H2(g)) = 0kJ/mol, ΔG∘f(I2(g)) = 19.37kJ/mol, ΔG∘f(HI(g)) = 1.79kJ/mol. Express your answer in kilojoules to two decimal places.

ΔG∘ = -15.79kJ (Use ΔG∘ = ∑nΔG∘f(products) - ∑nΔG∘f(reactants))

C2H5OH(g) ⇌ C2H4(g) + H2O(g) Calculate ΔG∘. Use the following data: ΔG∘f(C2H5OH(g)) = −168.5kJ/mol, ΔG∘f(C2H4(g)) = 68.3kJ/mol, ΔG∘f(H2O(g)) = −228.57kJ/mol. Express your answer in kilojoules to one decimal place.

ΔG∘ = 8.2kJ (Use ΔG∘ = ∑nΔG∘f(products) - ∑nΔG∘f(reactants))

The chemical reaction that causes iron to corrode in air is given by 4Fe+3O2→2Fe2O3 in which at 298 K, ΔH∘rxn = −1684 kJ and ΔS∘rxn = −543.7 J/K. What is the standard Gibbs free energy for this reaction? Assume the commonly used standard reference temperature of 298 K.

ΔG∘rxn = -1522 kJ The formula for Gibbs free energy is ΔG∘rxn=ΔH∘rxn−TΔS∘rxn. In this problem, ΔH∘rxn and ΔS∘rxn are given to you. Assume the standard reference temperature is 298 K. It makes sense that the rusting reaction has a negative Gibbs free energy value because rusting occurs spontaneously. However, notice that the entropy change for this reaction is not favored (7 moles going to 2 moles and a negative ΔS value). The reason that this reaction is nonetheless spontaneous at room temperature is the large negative value for enthalpy.

https://session.masteringchemistry.com/problemAsset/3238889/1/BLB10.19.3.jpg Predict the sign of ΔH accompanying this reaction on the figure. ΔH<0 ΔH>0 ΔH=0

ΔH>0 Energy must be put into the system in order to break a bond, and energy is released when a new bond forms.

Classify each process by its individual effect on the entropy of the universe, S. 1. a process run infinitesimally slowly at equilibrium and reversed to its original state 2. a constant composition mixture of solid and liquid water at STP (273.15 K and 1 atm) 3. motion of a frictionless pendulum 4. a bag of red marbles and a bag of green marbles dumped together on a table top 5. melting of ice cream from solid to liquid at 25∘C 6. sublimation of naphthalene (mothballs)

Increases S of the universe: 4. a bag of red marbles and a bag of green marbles dumped together on a table top 5. melting of ice cream from solid to liquid at 25∘C 6. sublimation of naphthalene (mothballs) Does not affect S of the universe: 1. a process run infinitesimally slowly at equilibrium and reversed to its original state 2. a constant composition mixture of solid and liquid water at STP (273.15 K and 1 atm) 3. motion of a frictionless pendulum Decreases S of the universe: (Blank)

Classify these descriptions by their effect on the entropy of the universe. 1. a system at equilibrium 2. a spontaneous process 3. a nonspontaneous process 4. a reversible process

Increases entropy: 2. a spontaneous process Has no effect on entropy: 1. a system at equilibrium 3. a nonspontaneous process 4. a reversible process

For the gaseous reaction shown here, the bins below represent three different possibilities for the equilibrium mixture. Classify each mixture by what it indicates about ΔG∘, K, and lnK for the reaction at standard conditions. https://session.masteringchemistry.com/problemAsset/1254059/23/1037714inst.jpg ΔG∘=0 ΔG∘>0 ΔG∘<0 K=1 K>1 K<1 lnK=0 lnK>0 lnK<0 Mixture 1: 1 square, 3 circles Mixture 2: 2 squares, 2 circles Mixture 3: 3 squares, 1 circle

Mixture 1: ΔG∘>0 K<1 lnK<0 Mixture 2: ΔG∘=0 K=1 lnK=0 Mixture 3: ΔG∘<0 K>1 lnK>0 (You can determine ΔG∘ based on how many reactants vs products there are. If there are more reactants than products then the reverse direction is spontaneous and so ΔG∘ > 0. If there are more products then the forward reaction is spontaneous and ΔG∘ < 0. If the number of reactants and products are the same the ΔG∘ = 0. You can find values for K and lnK by using the formula ΔG∘ = −RTlnK By plugging in a positive and negative value for ΔG∘ you can see what K and lnK look like.)

The chemical reaction that causes iron to corrode in air is given by 4Fe+3O2→2Fe2O3 in which at 298 K, ΔH∘rxn = −1684 kJ and ΔS∘rxn=−543.7J/K. At what temperature Teq do the forward and reverse corrosion reactions occur in equilibrium?

Teq = 3097 K Use the relation ΔG=ΔH−TΔS. A system is at equilibrium when the Gibbs free energy is equal to zero. Therefore, substitute zero for ΔG in the given formula and solve for Teq. This value means that 3097 K is the temperature at which this reaction changes from product-favored (spontaneous) to reactant-favored (nonspontaneous). As in this example, when ΔH and ΔS are both negative, the reaction is spontaneous at all temperatures below Teq. For a reaction in which ΔH and ΔS are both positive, the reaction is spontaneous at all temperatures above Teq.

True or False "Translational motion" of molecules refers to their change in spatial location as a function of time.

True

True or False The larger the number of atoms in a molecule, the more degrees of freedom of rotational and vibrational motion it likely has.

True

At constant temperature and pressure, which statement gives the relationship between the sign of ΔG and the spontaneity of the reaction? When ΔG = 0, the forward reaction is spontaneous. When ΔG > 0, the forward reaction is spontaneous. When ΔG < 0, the forward reaction is spontaneous. When ΔG < 0, the reverse reaction is spontaneous.

When ΔG < 0, the forward reaction is spontaneous.

https://session.masteringchemistry.com/problemAsset/3238452/1/Figure_P17.3.jpg If the acetic acid being titrated here were replaced by hydrochloric acid, would the pH at the equivalence point change?

Yes, the pH at the equivalence point will decrease.

What is the pH of 100 mL of a buffer that is 0.5 M HCNO and 0.5 M CNO− when 100 mL of a 0.2 M NaOH solution is added? Assume the volumes are additive. Ka of HCNO is 3.5 × 10−4.

3.82 In order to find the pH of this buffer after the addition of a strong base, you will need to find the concentrations of HCNO and CNO− after the addition. This is completed by assuming that all of the base is neutralized upon addition and calculating the moles of HCNO and CNO− after the addition. Once those are found the concentration is found using the new volume, and these are used in the Henderson-Hasselbalch equation. (The number of moles for HCNO and CNO- are both 5 prior to mixing in the NaOH. There are also 2 moles of NaOH prior to mixing. The 2 moles of NaOH will neutralize with 2 moles of HCNO and create an additional 2 moles of CNO-. Take the new mole amounts for HCNO and CNO- (3 and 7 respectively) and calculate their new concentration using the new volume (100 mL of buffer + 100 mL of NaOH solution = 200 mL total). Take those new concentrations and plug them into the HH equation.)

Calculate the percent ionization of 0.140 M lactic acid (Ka=1.4×10^−4).

3.2 % (3.1% was accepted. They got their answer by assuming that x is small enough that 0.14 - x = 0.14.) Start by writing out the balanced chemical equation for the ionization of the acid: HC3H5O3(aq)⇌C3H5O3−(aq)+H+(aq) Create an ICE table to find the value for x which is [H+]eq then plug this value into the percent ionization formula: Percent ionization = ([H+]equilibrium / [HC3H5O3]initial) × 100%

What is the entropy change of the system when 17.5 g of liquid benzene (C6H6) evaporates at the normal boiling point? The normal boiling point of benzene is 80.1°C and ΔH vap is 30.7 kJ/mol.

+19.5 J/K ΔS = qrev/T See Section 19.2 (qrev = (number of moles) * ΔH)

Using the standard molar entropies given in Table 19.1, and given that S∘ of CO2 (g) is 213.6 J/molK, calculate the change in the standard entropy of the system, ΔS∘, for the combustion of benzene according to the following equation: 2C6H6 (l) + 15O2 (g) ⇄ 6H2O (g) + 12CO2 (g) C6H6 (l) = 172.8 J/mol-K O2 (g) = 205.0 J/mol-K H2O (g) = 188.8 J/mol-K CO2 (g) = 213.6 J/mol-K)

+275.4 J/K (Use ΔS∘rxn = ∑nS∘(products) − ∑nS∘(reactants))

Here is the chemical reaction that represents solid lead chloride dissolving in water: PbCl2(s)⇌Pb2+(aq)+2Cl−(aq) Ksp=1.2×10−5 In which of the following would lead chloride be least soluble? pure water 1 M NaNO3 1 M KCl

1 M KCl In the previous part, you determined that the presence of extra chloride ions would shift the reaction toward the solid. Therefore, lead chloride will be least soluble in a solution that contains extra chloride ions. Lead chloride is least soluble in any solution that already contains chloride ions or lead ions. Furthermore, the higher the concentration of lead or chloride ions, the lower the solubility. That means that lead chloride would be even less soluble in 2 M KCl.

Here is the chemical reaction that represents solid lead chloride dissolving in water: PbCl2(s)⇌Pb2+(aq)+2Cl−(aq) Ksp=1.2×10−5 In which of the following would lead chloride be most soluble? 1 M KNO3 1 M HCl 1 M Pb(NO3)2 1 M CaCl2

1 M KNO3 You have already determined that the presence of extra chloride ions or lead ions would shift the reaction toward the solid. Therefore, lead chloride will be most soluble in a solution that does not contain any lead ions or chloride ions.

Calculate the percent ionization of 0.140 M lactic acid in a solution containing 0.0085 M sodium lactate. (Ka=1.4×10^−4)

1.3 % Start by writing out the balanced chemical equation for the ionization of the acid: HC3H5O3(aq) ⇌ C3H5O3−(aq) + H+(aq) The presence of lactate ions from the salt inhibits the amount of lactic acid that can ionize, which is known as the common ion effect. The following table allows you to determine the equation that relates the equilibrium constant to the concentrations of lactic acid ( HC3H5O3), H+, and lactate ion (C3H5O−3) at equilibrium: [HC3H5O3] _______ [H+] ___ [C3H5O−3] I: 0.140 ____________ 0 _______ 0.0085 C: −x _____________ +x ________ +x E: 0.140−x _________ x _______ 0.0085+x Substitute the expressions for the equilibrium concentrations into the expression for the equilibrium constant: Ka = [C3H5O3−][H+] / [HC3H5O3] = x(0.0085+x) / 0.140−x Substitute the value of the acid ionization constant, and rearrange the equation into the quadratic form. If you estimate all sums and differences as you had done in Part A, the value of x (or [H+]) would be 2.3×10^−3M . However, it would be more accurate to calculate [H+] using the quadratic formula based on the equilibrium expression derived from the tabulated data: Ka = x(0.0085+x) / 0.140−x When you solve for x using its quadratic formula x^2 + (0.0085+Ka)x − (0.140)(Ka)=0 the concentration is found to be 1.9×10^−3 M . Since x represents the concentration of lactic acid that ionizes in solution, you would determine the percent ionization in the same manner as you did in Part A using the more accurate value for x ([H+]). Percent ionization = 1.9×10^−3M / 0.140M × 100% = 1.3%

Glacial acetic acid is the concentrated form of acetic acid, the acid in vinegar. The term "glacial" refers to the appearance of the solid form, which resembles ice-like crystals. What is the melting point of this compound, in degrees Celsius, based on its thermodynamic data shown here? ΔHfus (kJ/mol) = 24.32 ΔSfus [J/(K⋅mol)] = 83.93

16.6 ∘C (My answer of 16.77 was accpeted. Mastering used 273.15 value to get C when I just used 273.) The melting point represents an equilibrium between solid and liquid, so the following equation can be applied: T = ΔHfus / ΔSfus In calculating the temperature, pay attention to units. The units of ΔHfus and ΔSfus must be compatible so that dividing them results in units of kelvins. The final step is to convert the temperature to the Celsius scale.

If ΔH = -80.0 kJ and ΔS = -0.500 kJ/K , the reaction is spontaneous below a certain temperature. Calculate that temperature.

160 K When ΔG is equal to zero, the reaction favors products and reactants equally. Thus, any temperature below this equilibrium temperature will cause the reaction to become spontaneous (favor products). Calculate when ΔG=0 using the following equation: ΔG=ΔH−TΔS

What is the pH of a buffer that is 0.6 M HF and 0.2 M NaF? The Ka of HF is 6.8 × 10−4.

2.69 (Use the Henderson-Hasselbalch equation: pH = pKa + log ([base] / [acid]). Note that pH = pKa - log ([acid] / [base]) can also be used, just be sure you put the acid/base in the numerator/denominator appropriately depending on whether you are adding the log or subtracting the log.)

You need to produce a buffer solution that has a pH of 5.09. You already have a solution that contains 10. mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74.

22 mmol acetate Buffer solutions can be produced by mixing a weak acid with its conjugate base or by mixing a weak base with its conjugate acid. The Henderson-Hasselbalch equation, pH = pKa + log([base] / [acid]) allows you to calculate the pH of a buffer. Note that molarity, moles, and millimoles are all proportional, so you can substitute the number moles, or millimoles, for the concentration terms in this formula. We want to find the quantity [base]/[acid]. Let's call that x, so pH=pKa+log(x) By rearranging the equation, we can solve for x, the concentration ratio: pH − pKa = log(x) =10^(pH−pKa) = x (Here x will equal 2.2 so that now [base] / [acid] = 2.2. Because [acid] is equal to 10 then solving for [base] gives 22.)

Which solution will have the highest pH? The Ka of HClO is 3.8 × 10−8. A 100 mL buffer solution of 0.1 M HClO, 0.1 M ClO− with 20 mL of 0.1 M HCl solution added. A buffer solution of 0.1 M HClO, 0.1 M ClO−. A 100 mL buffer solution of 0.1 M HClO, 0.1 M ClO− with 20 mL of 0.1 M NaOH solution added. A solution of 100 mL of water with 20 mL of 0.1 M NaOH added.

A solution of 100 mL of water with 20 mL of 0.1 M NaOH added. (The question is asking for which will have the highest pH. The pH will go up when a base is added. There are 2 options where a base is added (NaOH). In one option the base is being added to a buffer and in another it is being added to water. The base will have a much larger impact on pH when added to water than to a buffer.)

Here is the chemical reaction that represents solid lead chloride dissolving in water: PbCl2(s)⇌Pb2+(aq)+2Cl−(aq) Ksp=1.2×10−5 Which of the following actions would shift this reaction toward solid lead chloride? Add more lead chloride. Add more chloride ions. Remove chloride ions. Remove lead ions.

Add more chloride ions. Use Le Châtelier's principle.

Different salts have different solubilities in water. For some salts, the addition of H+ or OH− to the solution can actually increase its solubility. AgI, BaF2, and AgBr are all sparingly soluble salts. Which of these salts will be more soluble in an acidic solution than in water?

BaF2 To solve this problem, consider how each of the anions in the sparingly soluble salt will react with the H+ provided by the acidic solution. Keep in mind that in order for the solubility of the salt to increase, at least one of the ions of the sparingly soluble salt must be removed from the solution. This removal will drive the dissolution reaction of the salt to the right. The anion of the sparingly soluble salt is removed best if it is the conjugate base of a weak acid. The addition of H+ ions (from a strong acid such as HCl is useful to aid the solubility of BaF2 because the H+ ions react with the F− ions to form the weak acid HF. Because HF is a weak acid, it has a relatively small Ka. In other words, the equilibrium of the dissociation reaction HF ⇌ H+ + F− lies to the left. Thus the F− ions are effectively removed from solution, since they form HF with the H+ ions. This pulls the equilibrium of the dissociation reaction of the sparingly soluble salt, BaF2 ⇌ Ba2+ + 2F− to the right, increasing the solubility of BaF2.

Compare your result with that in the textbook (6.5×10−6). Suggest a reason for any differences you find between your value and the one in the textbook. (My result was Ksp = 5.30×10^−6) Because a change in pressure does change the value of an equilibrium constant, the solution may not have been kept at 1 atm. Because a change in volume does change the value of an equilibrium constant, the total volume of the solution may not have been 1 L. Because a change in temperature does change the value of an equilibrium constant, the solution may not have been kept at 25∘.

Because a change in temperature does change the value of an equilibrium constant, the solution may not have been kept at 25∘.

Which of the following mixtures may be act as a buffer solution? HF, NaF HI, NaI HCl, NaCl HBr, NaBr

HF, NaF A buffer resists dramatic changes in pH with the addition of acids and bases. This happens because the buffer contains an acid and a base, which can neutralize the addition. However, the acid and base in the buffer must not neutralize each other so a weak acid-conjugate base pair are used. (All of the options are strong acids except HF which is a weak acid. Therefore it is the only one that can act as a buffer.)

You just calculated that the heat of fusion for chloromethane is 6400 J/mol. The heat of fusion for hydrogen is 120 J/mol. Which of the following account for this difference? Check all that apply. Hydrogen molecules can pack more closely than chloromethane molecules. Hydrogen has stronger intermolecular forces than chloromethane. Chloromethane can absorb more energy at the same temperature. Chloromethane experiences dipole-dipole interactions. Chloromethane has a higher molar mass than hydrogen.

Chloromethane experiences dipole-dipole interactions. Chloromethane has a higher molar mass than hydrogen. Molar mass and intermolecular forces are two factors that impact melting points. Heavier molecules require more energy to break away from their neighbors. Molecules experiencing stronger intermolecular interactions also require more energy to break away. Dipole-dipole interactions, experienced by chloromethane, are significantly stronger than London dispersion forces.

You have to prepare a pH 3.60 buffer, and you have the following 0.10M solutions available: HCOOH, CH3COOH, H3PO4, HCOONa, CH3COONa, and NaH2PO4. Which solutions would you use? Check all that apply. HCOOH CH3COOH H3PO4 HCOONa CH3COONa NaH2PO4

HCOOH HCOONa (We are given 3 acids and their salts and we need to find which pair will have a buffer range that a pH of 3.60 falls into. Buffer Range = pH = pKa +/- 1 So find the pKa values of each of the acids then add and subtract 1 to them to find the Buffer Range. The range that includes the pH value that you want is the answer. Find pKa by looking up Ka values and doing pKa = -log (Ka))

Calculate the equilibrium constant, K, and ΔG∘ at 298 K for each of the following reactions. H2(g)+I2(g)⇌2HI(g) Calculate the equilibrium constant, K. (ΔG∘ = -15.79kJ)

K = 586 (Use K = e^-ΔG∘/RT)

Which equation is an expression of the second law of thermodynamics for a spontaneous process? A. ΔH univ = ΔH sys + ΔH surr < 0 B. ΔS univ = ΔS sys + ΔS surr = 0 C. ΔS univ = ΔS sys + ΔS surr < 0 D. ΔS univ = ΔS sys + ΔS surr > 0

D. ΔS univ = ΔS sys + ΔS surr > 0 The entropy of the universe increases in any spontaneous process. See Section 19.2

https://media.pearsoncmg.com/bc/bc_0media_chem/legacy_media/int_act_temp_dep/ In the activity, select H2. Click on the Run button, and observe the changes in entropy in the graph of molar entropy versus temperature. Observe the different phase changes the substance H2 undergoes, and classify the following phase changes according to whether entropy increases, entropy decreases, or entropy remains the same. H2(l)→H2(s) H2(s)→H2(l) H2(l)→H2(g) H2(g)→H2(l) H2(s)→H2(g) H2(g)→H2(s)

Entropy increases: H2(s)→H2(l) H2(l)→H2(g) H2(s)→H2(g) Entropy decreases: H2(l)→H2(s) H2(g)→H2(l) H2(g)→H2(s) Entropy remains the same: (None) (Phase changes from solid to liquid to gas increase in entropy, while phase changes from gas to liquid to solid decrease in entropy.)

The process of iron being oxidized to make iron(III) oxide (rust) is spontaneous. Which of these statements about this process is/are true? The reduction of iron(III) oxide to iron is also spontaneous. Equilibrium is achieved in a closed system when the rate of iron oxidation is equal to the rate of iron(III) oxide reduction. Because the process is spontaneous, the oxidation of iron must be fast. The oxidation of iron is endothermic. The energy of the universe is decreased when iron is oxidized to rust.

Equilibrium is achieved in a closed system when the rate of iron oxidation is equal to the rate of iron(III) oxide reduction. Equilibrium occurs when the rate of the forward reaction is equal to the rate of the reverse reaction. Thus, equilibrium is achieved in a closed system when the rate of iron oxidation is equal to the rate of iron(III) oxide reduction. Since the process of iron being oxidized is spontaneous, the reverse reaction would not be considered spontaneous. The energy of the universe will remain constant regardless, however, following the first law of thermodynamics, which states that energy is conserved. Two predictions we cannot make are regarding the speed and enthalpy of the reaction. Spontaneous reacts may occur at any rate, and may be endothermic or exothermic.

True or False "Rotational" and "vibrational" motions contribute to the entropy in atomic gases like He and Xe.

False (Rotational and vibrational motion can only be performed by atoms in molecules. It cannot be performed by lone atoms.)

True or False The third law of thermodynamics says that the entropy of a perfect, pure crystal at absolute zero increases with the mass of the crystal.

False. (The entropy of a pure crystalline substance at absolute zero is 0)

Rank these systems in order of decreasing entropy. Rank from highest to lowest entropy. To rank items as equivalent, overlap them. A. 1/2 mol of helium gas at 100 K and 20 L B. 1 mol of hydrogen peroxide gas at 273 K and 40 L C. 1 mol of helium gas at 273 K and 40 L D. 1/2 mol of liquid helium at 100 K E. 1 mol of helium gas at 273 K and 20 L F. 1 mol of fluorine gas at 273 K and 40 L G. 1/2 mol of helium gas at 273 K and 20 L

From greatest entropy to least entropy with none of them overlapping: B. 1 mol of hydrogen peroxide gas at 273 K and 40 L F. 1 mol of fluorine gas at 273 K and 40 L C. 1 mol of helium gas at 273 K and 40 L E. 1 mol of helium gas at 273 K and 20 L G. 1/2 mol of helium gas at 273 K and 20 L A. 1/2 mol of helium gas at 100 K and 20 L D. 1/2 mol of liquid helium at 100 K Things that increase entropy: increasing the number of independently moving particles increasing the volume increasing the temperature increasing the number of atoms per molecule (in general) At same conditions, 1 mol of hydrogen peroxide is more disordered than 1 mol of fluorine because hydrogen peroxide has more atoms than fluorine. Helium is less disordered than Fluorine since helium is monoatomic and fluorine isn't. A smaller volume of the same amount of substance is less disordered because molecules have less space to exist and be more disordered. Also, when the temperature goes down the kinetic energy of the molecules decreases thus the less energetic the molecules are. Lastly, the liquid state of a substance is in a more ordered state than when the substance is in a gaseous state thus less entropy.

https://media.pearsoncmg.com/bc/bc_0media_chem/legacy_media/int_act_temp_dep/ In the activity, select CH3Cl. Click on the Run button to observe how the molecular motions of CH3Cl vary with increasing temperature in the solid, liquid, and gaseous phases. Observe the following molecular motion diagrams and arrange them according to the average kinetic energy of the molecules. Rank the molecular motion diagrams from highest to lowest according to the kinetic energy of the molecules. To rank items as equivalent, overlap them. https://session.masteringchemistry.com/problemAsset/1247620/46/MC_1247620_Part-A_3.jpg (Gas Phase) https://session.masteringchemistry.com/problemAsset/1247620/46/MC_1247620_Part-A_1.jpg (Solid Phase) https://session.masteringchemistry.com/problemAsset/1247620/46/MC_1247620_Part-A_2.jpg (Liquid Phase)

High Kinetic Energy: (Gas Phase) Medium: (Liquid Phase) Low Kinetic Energy: (Solid Phase) Kinetic energy is the energy possessed by an atom or a molecule because of its motion. Solids have rigid structures with strong forces of attraction when compared to liquids, and thus, molecules move less randomly in the solid state. In gases, molecules have even more freedom than in liquids to move at random. The faster the molecules are moving, the more kinetic energy they possess.

Indicate whether ΔG increases, decreases, or does not change when the partial pressure of H2 is increased in each of the following reactions. 2HBr(g)→H2(g)+Br2(g) 2H2(g)+C2H2(g)→C2H6(g) N2(g)+2H2(g)→2NH3(g)

Increases: 2HBr(g)→H2(g)+Br2(g) Decreases 2H2(g)+C2H2(g)→C2H6(g) N2(g)+2H2(g)→2NH3(g) Does not change: (None) Hydrogen gas is either a reactant or a product in each reaction, so increasing its partial pressure will affect the value of the reaction quotient (Q) and ultimately affect the value of ΔG. The reaction quotient for a gas-phase reaction can be obtained as the product of the partial pressures of the products raised to their stoichiometric coefficients, divided by the partial pressures of the reactants raised to their stoichiometric coefficients. For example, the reaction quotient for N2(g)+2H2(g)→2NH3(g) is Q = (PNH3)^2 / (PN2)⋅(PH2)^2 Therefore, an increase in the partial pressure of the product will increase the value of Q, and an increase in the partial pressure of the reactant will decrease the value of Q. The relationship between free energy and the reaction quotient is ΔG=ΔG∘+RTlnQ, so increasing Q causes the value of ΔG to become more positive, and decreasing Q causes ΔG to become more negative. Thus, the reactions with hydrogen gas as a reactant become more spontaneous when the partial pressure of H2 is increased. Increasing the partial pressure of H2 in a reaction where it is a product will decrease the tendency for that reaction to occur. Reaction ________________ H2 classification _________ Direction ΔG N2(g)+2H2(g)→2NH3(g) ______ reactant _______ decreases, more spontaneous 2H2(g)+C2H2(g)→C2H6(g) ____ reactant _______ decreases, more spontaneous 2HBr(g)→H2(g)+Br2(g) _______ product _______ increases, less spontaneous

A chemist has some propene, H2C=CHCH3, in a compressed gas tank to convert to propane, H3CCH2CH3. The first thought is to combine the gas from the propene tank with that from a tank of a compressed hydrogen gas, H2. However, when this is tried nothing happens. The chemist finds some powdered palladium mixed in with inert charcoal in the chemical supply cabinet. Under conditions controlled for temperature and pressure, the chemist places the palladium into the flowpath from the hydrogen and propene tanks. As the mixture of propene and hydrogen flows past the palladium, a reaction occurs so that when the gas reaches the temperature controlled receiving tank, it is propane gas. After all the propene has been converted to propane, the curious chemist pours the palladium-charcoal mixture into a crucible. After analyzing the palladium, the chemist confirms that the chemical state has not changed. What word best describes the role that the palladium plays in the reaction between propene and hydrogen? It is a catalyst. It is an activation energy barrier. It is a solvent. It is a reagent.

It is a catalyst. Adding palladium to this reaction will increase the reaction rate because the platinum acts as a catalyst. A catalyst speeds up a chemical reaction by lowering the activation energy, but it is not used up in the reaction. In this case, the platinum helps break the hydrogen-hydrogen bonds in H2, thus lowering the energy barrier. It is important to note that catalysts do not change the thermodynamic properties of a reaction. A catalyst cannot change the equilibrium constant of a reaction. Some other common metal catalysts are copper, iron, nickel, platinum, rhodium, and ruthenium. To increase the accessible surface area, they can sometimes be mixed with an inert substance such as charcoal.

The reaction between carbon tetrachloride, CCl4, and water, H2O, to form carbon dioxide, CO2, and hydrogen chloride, HCl, has a ΔG∘ value of −232 kJ/mole, and so is thermodynamically favored. But when you mix carbon tetrachloride with water, no change is observed. What is a possible explanation for this? The reaction is not favored thermodynamically. It is a slow reaction kinetically. The activation energy of the reaction is too small. The reaction is not spontaneous.

It is a slow reaction kinetically. Start by defining each term in the problem. Thermodynamically favored reactions have a negative ΔG value, indicating that energy flows into the surroundings as the process continues. Spontaneous reactions readily occur. Kinetics refers to the rate, and has little to do with thermodynamic favorability or spontaneity. A reaction may occur quite slowly, but still continue readily over time and result in a release of energy. Activation energy is the energy required for the reaction to proceed. Generally, if the activation energy is lower, a reaction is considered to occur more rapidly. If ΔG∘ is negative for a given reaction, it means that there will be a net flow of free energy from the reacting system into the environment, and the reaction is favored thermodynamically. Such a reaction is said to be spontaneous. Be aware, however, that spontaneous does not mean instantaneous, and that thermodynamically favored reactions may still occur very slowly. For example, the hydrogenation of ethene, H2C=CH2, to form ethane, H3C−CH3, is a thermodynamically favored reaction for which ΔG∘=−125 kJ/mol. However, when hydrogen gas, H2, and ethene are mixed together at room temperature, they react spontaneously, but at a rate so slow as to be undetectable. From this example you can see that thermodynamics can predict whether a reaction is favored, but it says nothing about the rate at which the reaction will occur. To understand rates of reactions, we must turn to chemical kinetics. The reason that the above reaction occurs so slowly is that bonds must be broken before anything else can proceed. The bond-breaking event requires a certain amount of energy. This energy barrier is known as the activation energy of the reaction.

A chemist has some propene, H2C=CHCH3, in a compressed gas tank to convert to propane, H3CCH2CH3. The first thought is to combine the gas from the propene tank with that from a tank of a compressed hydrogen gas, H2. However, when this is tried nothing happens. The chemist finds some powdered palladium mixed in with inert charcoal in the chemical supply cabinet. Under conditions controlled for temperature and pressure, the chemist places the palladium into the flowpath from the hydrogen and propene tanks. As the mixture of propene and hydrogen flows past the palladium, a reaction occurs so that when the gas reaches the temperature controlled receiving tank, it is propane gas. After all the propene has been converted to propane, the curious chemist pours the palladium-charcoal mixture into a crucible. After analyzing the palladium, the chemist confirms that the chemical state has not changed. What did the palladium do to increase the rate of the reaction between the propene and hydrogen? It changed the thermodynamic equilibrium. It made the Gibbs free energy of the reaction more positive. It lowered the activation energy of the reaction. It increased the activation energy of the reaction.

It lowered the activation energy of the reaction. Gibbs free energy is related to the energy of the reactants and products, and will not change unless the energy of one of the two changes. The thermodynamic equilibrium would be related to the energy available to the reaction in relation to the overall energy of the reactants and products, thus it requires a change in energy for one of the two if it is to change. Activation energy is the energy required to reach the transition state and is the energy barrier for the reaction, only loosely related to the energy of the reactants and products.

Calculate the equilibrium constant, K. Express your answer to two significant figures. (ΔG∘ = 8.23kJ)

K = 3.6×10^−2 (Don't forget that ΔG∘ needs to be converted to J and also that there is a negative sign in the equation.) (Use K = e^-ΔG∘/RT)

Using the value of Ksp = 6×10^−51 for Ag2S, Ka1 = 9.5×10^−8 and Ka2 = 1×10^−19 for H2S, Kf = 1.0×10^11 for AgBr2-, calculate the equilibrium constant for the following reaction: Ag2S(s) + 4Br−(aq) + 2H+(aq) ⇌ 2AgBr2-(aq) + H2S(aq) Express the equilibrium constant to one significant figure.

Keq = 6×10^−3 (To solve write down the equations that add up to give the final equation in the question then solve by keeping in mind that k3 = K1K2. In short do (Ksp x Kf^2) / (Ka1 x Ka2) The overall process for this reaction can be viewed as the sum of four reactions: Ag2S(s) ⇌ 2Ag+(aq) + S2-(aq) S2-(aq + H+(aq) ⇌ HS-(aq) HS-(aq) + H+(aq) ⇌ H2S(aq) 2[Ag+(aq) + 2Br- ⇌ AgBr2^-] Ag2S(s) + 2H+(aq) + 4Br−(aq) ⇌ 2AgBr2^−(aq) + H2S(aq) The information given is applied to each of the four reactions to determine their individual equilibrium constants. The first reaction will use the Ksp value of Ag2S, the second will use 1/Ka2, the third will use 1/Ka1, and the fourth will use Kf^2. These four values are then combined to find K for the summation of the three reactions. K = Ksp×Kf^2 / Ka1×Ka2 = (6×10^−51)(1.0×10^11^)2 / (9.5×10^−8)(1×10^−19) = 6.31×10^−3 = 6×10−3

Chlorine gas, Cl2(g), reacts with nitric oxide, NO(g), to form nitrosyl chloride, NOCl(g), via the reaction Cl2(g)+2NO(g)⇌2NOCl(g) The thermodynamic data for the reactants and products in the reaction are given in the following table: Substance ____ ΔG∘f (kJ/mol) Cl2(g) __________ 0 NO(g) __________ 86.71 NOCl(g) ________ 66.30 Using standard free energy of formation values given in the introduction, calculate the equilibrium constant Kp of the reaction Cl2(g)+2NO(g)⇌2NOCl(g) The standard free energy of the reaction represents the drive the reaction has under standard conditions to move toward equilibrium from point A to point X in the diagram. Express the equilibrium constant numerically using three significant figures.

Kp = 1.43×10^7 Based on the calculations in Part A, our value for ΔG∘ should be rounded to one significant figure. However, this value is used in further calculations so we have kept more digits to prevent rounding errors. For a gaseous reaction, the K in this formula represents Kp. (Use ΔG∘rxn = ∑nΔG∘products − ∑mΔG∘reactants and ΔG∘ = −RT ln K. Don't forget to convert kJ to J.)

A saturated solution of barium fluoride, BaF2, was prepared by dissolving solid BaF2 in water. The concentration of Ba2+ ion in the solution was found to be 7.52×10−3 M . Calculate Ksp for BaF2.

Ksp = 1.70×10^−6 Under the assumption of complete dissociation, each mole of BaF2 produces one mole of Ba2+ and two moles of F−: BaF2 ⇌ Ba2+ + 2F− If solid BaF2 dissolves to produce 7.52×10^−3 M Ba2+, [ F− ] = 1.50×10^−2 M The balanced equation for the complete dissociation of BaF2 is BaF2(s)⇌Ba2+(aq)+2F−(aq) The expression for the solubility product is: Ksp = [Ba2+][F−]^2

Enter the the Ksp expression for C2D3 in terms of the molar solubility x.

Ksp = 108x^5 C2D3(s)⇌2C(aq)+3D(aq) Ksp = [C]^2[D^]3 Ksp = (2x)^2 * (3x)^3

AB2 has a molar solubility of 3.72×10−4 M. What is the value of the solubility product constant for AB2?

Ksp = 2.06×10^−10 You are given the value of the molar solubility x. Substitute this value into the Ksp expression: Ksp = x(2x)^2 = 4x^3 and solve for the solubility product constant Ksp.

Enter the Ksp expression for the solid AB2 in terms of the molar solubility x.

Ksp = 4x^3 The equation for the reaction when excess AB2 dissolves in water is: AB2(s)⇌A(aq)+2B(aq) The solubility product expression when excess AB2 dissolves in water is: Ksp = [A][B]^2 When excess solid YZ is put in water and allowed to dissolve the following equilibrium reaction occurs and the equilibrium table shown can be set up. _______________ YZ(s) ___________ Y+(aq) ______________ Z−(aq) [initial] _________ excess ___________ 0 _____________________ 0 [change] _________ -x _____________ +x ____________________ +x [equilibrium] _____ excess - x ________ x _____________________ x The amount of solid that dissolves is called the molar solubility, denoted here by x. Since the concentration of the solid is constant and is not included in the Ksp expression, the equilibrium expression for the reaction is Ksp=[Y][Z], or, in terms of molar solubility, Ksp=x2. (Because in this problem we have 2B, the change becomes +2x and when squared in the eq. expression it becomes 4x^2, which when multiplied by the other x becomes 4x^3.)

Excess Ca(OH)2 is shaken with water to produce a saturated solution. The solution is filtered, and a 50.00 mL sample titrated with HCl requires 11.23 mL of 0.0978 M HCl to reach the end point. Calculate Ksp for Ca(OH)2.

Ksp = 5.30×10^−6 (Convert the amount of HCl used into moles of HCl, convert that to moles of Ca(OH)2 keeping in mind that 2 mols of HCl gives 1 mole of Ca(OH)2, then convert moles of Ca(OH)2 to molarity using the 50.00 mL. This value is the molar solubility x. Plug that into the solubility equation of Ksp = [Ca2+][OH-]^2 = 4x^3)

https://session.masteringchemistry.com/problemAsset/1254059/23/MC_1254059_Set_4b.jpg The image represents a spontaneous, gaseous reaction at a constant temperature T K. Predict whether ΔH, ΔS, and ΔG for this reaction are positive, negative, or zero.

Positive: (None) Zero: (None) Negative: ΔS ΔG ΔH The sign of ΔG indicates whether or not the reaction is spontaneous. If the randomness (or disorder) of the system increases, the sign of ΔS is positive and vice versa. Once the signs of both ΔG and ΔS are known, the following equation can be used to predict the sign of ΔH: ΔG=ΔH−TΔS The reaction is spontaneous. What does this say about the value of ΔG? ΔG < 0 Examine the number of moles of gas in the reactants and products. What does this say about the change in disorder or randomness for the reaction? Randomness decreases.

Predict the sign of the entropy change, ΔS∘, for each of the reaction displayed. 2NaClO3(s) → 2NaCl(s) + 3O2(g) I2(s) → I2(g) C5H12(g) + 8O2(g) → 5CO2(g) + 6H2O(g) Ag+(aq) + Cl−(aq) → AgCl(s) 2Na(s) + Cl2(g) → 2NaCl(s) 2NO2(g) → N2(g) + 2O2(g)

Positive: I2(s)→I2(g) C5H12(g)+8O2(g)→5CO2(g)+6H2O(g) 2NO2(g)→N2(g)+2O2(g) 2NaClO3(s)→2NaCl(s)+3O2(g) Negative: Ag+(aq)+Cl−(aq)→AgCl(s) 2Na(s)+Cl2(g)→2NaCl(s) (Look at the coefficients and the states on each side of the reaction.)

Consider the following equilibrium: B(aq)+H2O(l)⇌HB+(aq)+OH−(aq) Suppose that B(aq) is added to a solution of B at equilibrium. Will the concentration of HB+(aq) increase, decrease, or stay the same?

The concentration of HB+(aq) will increase. The equilibrium-constant expression for this reaction is as follows: Kb=[HB+][OH−] / [B] Therefore, if the value of either [B] or [HB+] increases, the other must increase in order to maintain a constant value of Kb, since one is in the numerator and the other is in the denominator.

https://session.masteringchemistry.com/problemAsset/3238881/2/Figure_P19.7.jpg What major factor leads to a decrease in entropy as the reaction shown takes place? Increase in the temperature of system. The formation of new bonds decreases the number of degrees of freedom, or forms of motion, available to the atoms. The decrease in the number of molecules due to the formation of new bonds.

The decrease in the number of molecules due to the formation of new bonds. (The 2nd option would be in the case of a state change like condensation or freezing.)

Consider the following equilibrium: B(aq)+H2O(l)⇌HB+(aq)+OH−(aq) Suppose that B(aq) is added to a solution of B at equilibrium. Will the equilibrium constant for the reaction increase, decrease, or stay the same?

The equilibrium constant for the reaction will stay the same. The value of an equilibrium constant (K) can change depending on ambient conditions like temperature and pressure, but unlike the reaction quotient (Q), the value of K is independent of reactant and product concentrations. Therefore, adding either reactant or product to the equilibrium mixture will have no effect on the value of K.

Consider the following equilibrium: B(aq)+H2O(l)⇌HB+(aq)+OH−(aq) Suppose that B(aq) is added to a solution of B at equilibrium. Will the pH of the solution increase, decrease, or stay the same?

The pH of the solution will increase. The pH will change in response to adding either the base or its conjugate acid. Adding B to the solution would increase both HB+ and OH− as the reaction shifts toward the products, which would lower the pOH and raise the pH (both indicating the solution is becoming more basic). In contrast, adding HB+ to the solution would increase B but decrease OH− as it shifts toward the reactants, which would increase the pOH and decrease the pH (both indicating the solution is becoming more acidic). Under standard conditions, pH + pOH = 14.

https://session.masteringchemistry.com/problemAsset/3238889/1/BLB10.19.3.jpg Explain your choice. Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer. The reaction involves __________ five blue-blue and twenty blue-red bonds and then __________ twenty blue-red bonds. Enthalpies for bond breaking are always __________. In the depicted reaction, both reactants and products are assumed to be in the gas phase. There are _________ as many molecules of gas in the products, so ΔS is ___________ for this reaction. breaking forming positive negative twice half

The reaction involves *breaking* five blue-blue and twenty blue-red bonds and then *forming* twenty blue-red bonds. Enthalpies for bond breaking are always *positive*. In the depicted reaction, both reactants and products are assumed to be in the gas phase. There are *twice* as many molecules of gas in the products, so ΔS is *positive* for this reaction. The enthalpy change associated with breaking a bond is positive because energy must be put into the system in order to overcome the bond energy between atoms. An increased number of molecules increases the amount of possible randomness that exists in the system, resulting in an increase in entropy of the system. This increase is represented by a positive value.

Which statement corresponds to a reaction that has ΔH > 0 and ΔS > 0? The reaction is nonspontaneous at all temperatures. The reaction is spontaneous at all temperatures. The reaction is spontaneous at low temperatures. The reaction is spontaneous at high temperatures.

The reaction is spontaneous at high temperatures. When ΔH > 0, the reaction is endothermic. When ΔS > 0 the final state has more disorder than the initial state. See Section 19.6

Indicate whether each statement is true or false. The entropy change of the system is positive for any reversible process. The entropy of the system must increase in any spontaneous process. The entropy of the universe increases for any spontaneous process. The entropy change for an isothermal process depends on both the absolute temperature and the amount of heat reversibly transferred.

True" The entropy change for an isothermal process depends on both the absolute temperature and the amount of heat reversibly transferred. The entropy of the universe increases for any spontaneous process. False: The entropy of the system must increase in any spontaneous process. The entropy change of the system is positive for any reversible process. The second law of thermodynamics states that the entropy of the universe will increase with any spontaneous process. This is true if the net change in entropy for the combination of the system and its surroundings is positive. The change in entropy of the system can be negative, as long as it is paired with a large positive change in entropy of the surroundings. If the entropy change of the system is equal and opposite to that of the surroundings, the reaction is reversible. For an irreversible process, the entropy change is positive in one direction and negative in the other direction, so the process will only proceed in the direction that results in an increase in entropy. In the special case of an isothermal process, the change in entropy is equivalent to the amount of heat transferred, assuming a reversible process divided by the temperature at which the reaction occurs; however because entropy is a state function, the equation holds true for both reversible and irreversible processes. ΔS = qrev / T

Titration of a strong acid with a strong base. The pH curve for titration of 50.0 mL of a 0.100 M solution of hydrochloric acid with a 0.100 M solution of NaOH(aq). For clarity, water molecules have been omitted from the molecular art. https://session.masteringchemistry.com/problemAsset/3238451/1/replace.jpg What volume of NaOH(aq) would be needed to reach the equivalence point if the concentration of the added base were 0.250 M? Express your answer to three significant figures with the appropriate units.

V = 20.0 mL (1 mol of HCL reacts with 1 mol of NaOH. Because of this we can simply use M1V1 = M2V2. Therefore (0.1M)(50mL) = (0.25M)x x = 20.0 mL)

You have to prepare a pH 3.60 buffer, and you have the following 0.10M solutions available: HCOOH, CH3COOH, H3PO4, HCOONa, CH3COONa, and NaH2PO4. How many milliliters of HCOOH and HCOONa would you use to make approximately a liter of the buffer? Express your answers using two significant figures. Enter your answers separated by a comma. (Ka = 1.8x10^-4)

VHCOOH, VHCOONa = 580,420 mL (First use the HH equation to find what the ratio [base]/[acid] is equal to. In this case [base]/[acid] = 0.716593. Because both solutions are 0.10 moles / L and we need 1L of the buffer there will be a total of 0.10 moles of HCOOH and COOH- combined. So 0.10 = HCOOH + COOH-, therefore COOH- = 0.10 - HCOOH. Even though these are moles we can still plug them into the above ratio so: 0.716593 = (0.10 - HCOOH) / HCOOH Solving this equation results in: HCOOH = 0.058 mols To find the volume needed we take the moles and divide it by the concentration of 0.1 M to get 0.58 L = 580 mL of HCOOH To find the amount of HCOONa just take 1,000 - 580 = 420 mL.)

In which phase are water molecules least able to have rotational motion?

ice https://session.masteringchemistry.com/problemAsset/3238880/1/Figure_P19.6.jpg

Which correctly lists the following in order of increasing entropy? i. 1 mol of HCl (g) at 50 ° C ii. 1 mol of NaCl (s) at 25 ° C iii. 2 mol of HCl (g) at 50 ° C iv. 1 mol of HCl (g) at 25 ° C

ii. < iv. < i. < iii. A microstate is a single possible arrangement of the positions and kinetic energies of the molecules when the molecules are in a specific thermodynamic state. Entropy is the extent of randomness in a system, or the extent to which energy is distributed among various microstates of molecules in a system. See Section 19.3

As a chemist for an agricultural products company, you have just developed a new herbicide,"Herbigon," that you think has the potential to kill weeds effectively. A sparingly soluble salt, *Herbigon is dissolved in 1 M acetic acid* for technical reasons having to do with its production. *You have determined that the solubility product Ksp of Herbigon is 8.50×10−6*. Although the formula of this new chemical is a trade secret, it can be revealed that the formula for Herbigon is X-acetate (XCH3COO, where "X" represents the top-secret cation of the salt). It is this cation that kills weeds. Since it is critical to have Herbigon dissolved (it won't kill weeds as a suspension), you are working on adjusting the pH so Herbigon will be soluble at the concentration needed to kill weeds. *What pH must the solution have to yield a solution in which the concentration of X+ is 3.00×10−3 M ? The pKa of acetic acid is 4.76.*

pH = 2.21 To solve this problem, begin with the solubility product equation. The definition of Ksp for Herbigon (XAc) is Ksp = [X+][Ac−] Use the known Ksp value along with your desired X+ value to determine the maximum concentration of acetate ion allowed. Next, use the Ka equation for acetic acid to determine the quantity of H+ needed to get that desired concentration of acetate ion. Finally, convert the H+ value to pH. Using the equation pH = −log[H+] You have determined that the solubility product Ksp of Herbigon is 8.50×10^−6. Given the solubility product and desired X+ concentration of 3.00×10^−3 M, what is the maximum allowed [Ac−] (CH3COO−) to prevent precipitation of the salt? [CH3COO−] = 2.83×10^−3 M What is the needed [H+] to achieve the desired [Ac−]? Keep in mind that because acetic acid is a weak acid, adding extra H+ ions will drive the acid dissociation reaction CH3COOH⇌H++CH3COO− to the left. This will help bring the [CH3COO−] down to the concentration needed to keep Herbigon soluble. [H+] = 6.14×10^−3 M

You are asked to prepare a pH=4.00 buffer starting from 1.60 L of 0.0400 M solution of benzoic acid (C6H5COOH) and an excess of sodium benzoate (C6H5COONa). Given that the dissociation constant Ka for benzoic acid is equal to 6.3×10^−5. What is the pH of the benzoic acid solution prior to adding sodium benzoate?

pH = 2.81 (C6H5COOH ⇌ H+ + C6H5COO- Make an ICE table and use the equilibrium constant expression to find [H+] then use that to find pH.)

A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. A solution is made by titrating 9.00 mmol (millimoles) of HA and 3.00 mmol of the strong base. What is the resulting pH? Express the pH numerically to two decimal places.

pH = 4.95 First, create a table for the reaction between weak acid, HA, and strong base, OH−, to determine what is left over after they react completely: Amount (mmol) ___ HA __ + __ OH− __ → __ A− __ + __ H2O initial: __________ 9.00 ______ 3.00 _______ 0 change: _______ −3.00 _____ −3.00 _____ +3.00 final: ____________ ? ___________________ ? Since the resulting solution is a buffer, use the Henderson-Hasselbalch equation to determine the pH, pH = pKa + log[A−] / [HA] Note that the numbers of moles or millimoles may be used in place of concentrations in this formula since the total volume is the same for both species.

Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8×10−5) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant at 25 ∘C.

pH = 5.13 In a weak base-strong acid titration, the pH at the equivalence point is less than 7 because the salt produced is acidic. When a weak base such as NH3 reacts with an equal amount of the strong acid, the reaction NH3+HNO3→NH4NO3 goes to completion. The product is an acidic salt with the following equilibrium: NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) Ka = [H3O+][NH3] / [NH4+] Calculate the salt concentration at the equivalence point and use it as the initial concentration of NH4+ when figuring out the equilibrium concentrations. Use the equilibrium H3O+ concentration to find the pH. Given that Kb for NH3 is 1.8×10^−5, Ka = 5.56×10^−10 When 50.0 mL of 0.20 M NH3 is mixed with 50.0 mL of 0.20 M HNO3, the salt concentration = 0.10 M (0.01 moles of salt are produced and there is 100 mL of solution which results in 0.10 M) Calculate the concentration of H3O+ in a 0.10 M solution of NH4NO3. Ka for NH4+ is 5.56×10^−10. [H3O+] = 7.46×10^−6 M (In short, find how many moles of each you have, then convert back to molarity using the combined volume of 0.1 L. Create an ICE table for NH4+ = NH3 + H+ and use the Ka (not the Kb) value to find what [H+] is equal to. Then use that to find pH.)

What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid. Express the pH numerically to three decimal places. (The buffer consists of 0.809 mol of HA and 0.609 mol of NaA Ka of HA is 5.66×10−7.)

pH = 5.927 (Because the mols of HCl will react with the A- to make HA you need to subtract the HCl mols from the A- and add the mols of HCl to HA, then plug these new values into the HH equation.)

What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.609 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7. Express the pH numerically to three decimal places.

pH = 6.124 Since both the acid and base exist in the same volume, we can skip the concentration calculations and use the number of moles in the Henderson-Hasselbalch equation to calculate the pH. The answer will be the same. (pH = pKa + log([base] / [acid]))

What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base. Express the pH numerically to three decimal places. (The buffer consists of 0.809 mol of HA and 0.609 mol of NaA Ka of HA is 5.66×10−7.)

pH = 6.364 (Because the mols of NaOH will react with the HA to make A- you need to add the amount of NaOH mols to the amount of A- and subtract the mols of NaOH from HA, then plug these new values into the HH equation.)

A 64.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 32.0 mL of KOH at 25 ∘C.

pH = 7.00 In case of a strong acid-strong base titration, the salt produced is neutral. Therefore, the solution at the equivalence point has a pH of 7.00. The balanced equation for the titration reaction is HBr(aq)+KOH(aq)→KBr(aq)+H2O(l) Calculate the number of moles (or millimoles) of each reactant, then use stoichiometry to determine whether there is any reactant left over. If there is acid left over, the solution will be acidic. If there is base left over, the solution will be basic. If only products are left over (water and a neutral salt) the solution will be neutral. (In this case there is 0.016 moles of each so there is neither acid nor base left.)

100. mL of 0.200 M HCl is titrated with 0.250 M NaOH. What is the pH of the solution at the equivalence point?

pH = 7.00 The equivalence point is the point at which equal amounts of acid and base have reacted. In a strong acid with strong base titration, the products are completely neutral. Therefore, when all the acid has reacted with the base, the solution must be neutral. If more base were added, the pH of the solution would become basic. In this titration, each additional mL of base would add 2.5×10^−4 mol of base. If the equivalence point were overshot by even 1.0 mL, the pH of the solution would be 10.40.

The value of Ksp for silver chromate, Ag2CrO4, is 9.0×10−12. Calculate the solubility of Ag2CrO4 in grams per liter. Express your answer numerically in grams per liter.

solubility = 4.35×10^−2 g/L (My answer of 4.3x10^-2 was accepted) When we know the solubility product for a substance, we can calculate its solubility, which is actually the concentration of a saturated solution. At 25 ∘C, the concentration of a saturated solution of silver chromate is 1.3×10^−4 M or 4.3×10^−2 g/L. If we add 10 g of Ag2CrO4 to 1 L of water, only 4.3×10^−2 g of the salt will dissolve, and the rest will stay at the bottom of the beaker. The balanced equation for the complete dissociation of Ag2CrO4 is: Ag2CrO4(s)⇌2Ag+(aq)+CrO42−(aq) The expression for the solubility product is: Ksp=[Ag+]^2[CrO42−] If x represents the number of moles per liter of Ag2CrO4 that dissolves to form a saturated solution, then we can construct the following table: Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^2−(aq) excess __________ 0 ____________ 0 −x _____________ +2x ___________ +x Because Ksp = [Ag+]^2[CrO4^2−] Ksp = 9.0×10^−12 = 4x^3 x = 1.31×10^−4 M x = to the number of moles so multiply that by molar mass of Ag2CrO4 = 332 g/mol to find the answer.

What can be said about an exothermic reaction with a negative entropy change? spontaneous at all temperatures. spontaneous at high temperatures. spontaneous at low temperatures. spontaneous in the reverse direction at all temperatures. nonspontaneous in either direction at all temperatures.

spontaneous at low temperatures. The equation for Gibbs free energy is ΔG=ΔH−TΔS If ΔG<0, then the reaction is spontaneous. So consider the resulting sign of ΔG given negative values for both ΔH and ΔS. Calculate the change in Gibbs free energy using arbitrary values. If ΔH = −100kJ and ΔS = −1kJ/K, what is the value of ΔG at 0 K and at 900 K? Assume that ΔH and ΔS do not change with temperature. ΔG0K = -100 kJ ΔG900K = 800 kJ

What can be said about an endothermic reaction with a negative entropy change? spontaneous at all temperatures. spontaneous at high temperatures. spontaneous at low temperatures. spontaneous in the reverse direction at all temperatures. nonspontaneous in either direction at all temperatures.

spontaneous in the reverse direction at all temperatures. The equation for Gibbs free energy is ΔG=ΔH−TΔS If ΔG<0, then the reaction is spontaneous. So consider the resulting sign of ΔG given a positive value for ΔH and a negative value for ΔS. The spontaneity of a reaction depends both on the enthalpy change, ΔH, and entropy change, ΔS. Reactions that release energy produce more stable products, and the universe tends toward disorder. Thus, an exothermic reaction with a positive entropy change will always be spontaneous. Mathematically, this relationship can be represented as ΔG=ΔH−TΔS where ΔG is the change in Gibbs free energy and is the Kelvin temperature. If ΔG is negative, then the reaction is spontaneous. If ΔG is positive, then the reaction is nonspontaneous as written but spontaneous in the reverse direction.

A negative change in entropy indicates that the products have a *greater* number of available energy microstates than the reactants. the products have a *smaller* number of available energy microstates than the reactants.

the products have a *smaller* number of available energy microstates than the reactants. There are three moles of gas on the reactants side of this equation and only two moles of gas on the products side. This change in the number of gas molecules results in a decrease in the number of degrees of freedom (or forms of motion) available to the atoms in the system. Or stated another way, this results in fewer available energy microstates in the system and a corresponding decrease in entropy of the system.

https://session.masteringchemistry.com/problemAsset/3238452/1/Figure_P17.3.jpg If the acetic acid being titrated here were replaced by hydrochloric acid, would the amount of base needed to reach the equivalence point change?

will not change

Calculate the free energy ΔG at 25 ∘C for the nonstandard conditions at point B where the reaction quotient Q is 2.75×10−5. (ΔG∘ = -40.82 kJ)

ΔG = -66.8 kJ Under these conditions, the drive to form nitrosyl chloride is stronger than that at standard conditions. In solving Part A, we found that ΔG∘=−40.82 kJ. Use this value in the formula ΔG= ΔG∘ + RTlnQ to solve for ΔG, making sure all units are compatible. Since the overall goal is to have units of kilojoules, it will be useful to convert R to units of kilojoules per mole per kelvin.

Calculate the free energy ΔG at 25 ∘C for the nonstandard conditions at point C where the reaction quotient Q is 3.58×10^9. (ΔG∘ = -40.82 kJ)

ΔG = 13.7 kJ Under these conditions, the reverse reaction is favored with a free energy of −13.7 kJ. In solving Part A, we found that ΔG∘=−40.82 kJ. Use this value in the formula ΔG = ΔG∘ + RTlnQ to solve for ΔG, making sure all units are compatible. Since the overall goal is to have units of kilojoules, it will be useful to convert R to units of kilojoules per mole per kelvin.

https://media.pearsoncmg.com/bc/bc_0media_chem/legacy_media/int_act_temp_dep/ In the activity, select CH3Cl. Click on the Run button, and observe the graph of standard molar entropy versus temperature. Observe the equilibrium position, in which the solid and liquid phases exist together. Notice that the entropy increases from 79 J⋅mol−1 K−1 to 116 J⋅mol−1 K−1 as CH3Cl melts. Calculate the enthalpy of fusion (ΔHfus) at this equilibrium. Express the change in enthalpy (ΔHfus) to two significant figures.

ΔHfus = 6.4×10^3 J⋅mol−1 First, identify the temperature at which the solid and liquid phases of CH3Cl exist together, that is, the temperature at which CH3Cl melts. Next, find the change in entropy (ΔS) at this temperature. Notice that the entropy increases from 79 J⋅mol−1 K−1 to 116 J⋅mol−1 K−1 as CH3Cl melts. Finally, use the following formula to calculate the enthalpy of fusion, ΔHfus, ΔG=ΔHfus−TΔS where ΔG is the free energy, T is the temperature in Kelvin (K), and ΔS is the change in entropy. As CH3Cl melts, the entropy of the system increases. If the initial entropy of CH3Cl is 79 J⋅mol−1 K−1, and the final entropy is 116 J⋅mol−1 K−1, find the change in entropy (ΔS). ΔS=Sfinal−Sinitial

The element gallium (Ga) freezes at 29.8 ∘C, and its molar enthalpy of fusion is ΔHfus=5.59kJ/mol. Calculate the value of ΔS when 65.0 g of Ga(l) solidifies at 29.8 ∘C.

ΔS = -17.2 J/K (Don't forget the negative) At equilibrium, there is no change in the free energy, so ΔHeq=TΔSeq. When gallium freezes, it is losing heat, which signifies that ΔHfreezing = −5.59kJ/mol at the normal freezing point. 65.0 g of gallium is equivalent to 0.932 mol . Therefore, the change in entropy is determined by dividing the total amount of heat loss (in joules) by the absolute temperature (in kelvins).

https://session.masteringchemistry.com/problemAsset/3238889/1/BLB10.19.3.jpg Predict the sign of ΔS accompanying reaction on the figure. ΔS < 0 ΔS > 0 ΔS = 0

ΔS > 0 There are more product molecules than there are reactant molecules, which amounts to more disorder or randomness in the final state than in the initial state.

Calculate the standard entropy change for the reaction 2Na(s)+Cl2(g)→2NaCl(s) Express your answer to four significant figures and include the appropriate units. S∘ [J/(K⋅mol)] values: Na(s) = 51.30 Cl2(g) = 223.1 NaCl(s) = 72.10

ΔS∘ = -181.5 J/K (Use ΔS∘rxn = ∑nS∘(products) − ∑nS∘(reactants))

Calculate ΔS∘rxn for the reaction 2NO(g)+O2(g)→2NO2(g) Express your answer to one decimal place and include the appropriate units. Substance _____ S∘ (J/mol⋅K) NO2 _____________ 240.0 O2 ______________ 205.2 NO ______________ 210.8

ΔS∘rxn = -146.8 J/K The change in entropy for a reaction is equal to the sum of the entropy changes of the products minus the sum of the entropy changes for the reactants: ΔS∘rxn = ∑nS∘(productsnS∘) − ∑nS∘(reactants) A negative entropy value is not favored for spontaneity. However, the reaction is exothermic (negative enthalpy value), which is thermodynamically favored. This reaction is also enhanced by UV light from the sun.

A beaker with 1.00×10^2 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 4.40 mL of a 0.360 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

ΔpH = -0.28 (First determine the concentrations of CH3COOH and the CH3COO- by using the Henderson-Hasselbalch equation ([base] / [acid] can be represented as x / 0.1 -x which will give you the value for [base] then [acid] can be easily calculated). With the concentrations, find the moles of each as well as the moles of HCl. All of the HCl moles will neutralize with CH3COO- to create more CH3COOH so subtract HCL moles from CH3COO- moles to find the new amount of CH3COO- moles and add the number of HCl moles to the amount of CH3COOH moles to find the new amount of CH3COOH. At this point you can just plug the new mole values of CH3COO- and CH3COOH into the HH equation to find the new pH (leaving them as moles is okay, you don't need to convert them into Molarity). Then simply find the change in pH.

Calculate the standard free-energy change, ΔG∘, for a reaction for which ΔH∘ = 15.6 kJ and ΔS∘ = 125 J/K. Assume 25°C.

−21.7 kJ These state functions are related by the equation: .ΔG∘ = ΔH∘ - T ΔSo. See Section 19.5