Chemistry Test 3 (17, 19)

How does the entropy of the system change in the following processes? (a) the melting of ice cubes at 10 ∘C and 1 atm pressure; (b) separating a mixture of N2 and O2 into two separate samples, one that is pure N2 and one that is pure O2; (c) alignment of iron filings in a magnetic field; (d) the reaction of sodium metal with chlorine gas to form sodium chloride; (e) the dissolution of CH3OH(l) in water to form CH3OH(aq).

(a). Increases (b). Decreases (c). Decreases (d). Decreases (e). Increases

Classify each of the following reactions as one of the four possible types: 1. spontaneous at all temperatures; 2. nonspontaneous at all temperatures; 3. spontaneous at low T; nonspontaneous at high T; 4. spontaneous at high T; nonspontaneous at low T. (a) N2(g)+3F2(g)→2NF3(g);ΔH∘=−249kJ;ΔS∘=−278J/K (b) N2(g)+3Cl2(g)→2NCl3(g);ΔH∘=460kJ;ΔS∘=−275J/K (c) N2F4(g)→2NF2(g);ΔH∘=85kJ;ΔS∘=198J/K

(a). Spontaneous at low T; non-spontaneous at high T (b). Non-spontaneous at all temperatures (c). Spontaneous at high T; non-spontaneous at low T.

Equivalence Point

-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).

Spontaneous Processes

-Happen on their own, but sometimes require initial push to get going -PRODUCT FAVORED (Occur in definite direction: Towards the formation of product) AT 25 Deg C: -H2O (s) to H2O (l) -CO2 (s) to CO2 (g) -4Fe (s) + 3O2 (g) to 2Fe2O3 (s) -Burning paper at 233 DegC -At any given temperature and pressure, processes are spontaneous only in one direction!!!! -Direction may depend on temperature -Many (but not all) are exothermic (ΔH < 0 combustion that releases heat)

Titrations

-If the acid is unknown, a known amount of base is added and the pH is monitored. -If the base is unknown, known amount of acid is added and pH is monitored -A plot of pH as a function of the volume of titrant added is called a pH titration curve. Prior to the titration, the pH is determined by the concentration of the analyte. When the titrant is added, it begins to neutralize the analyte. The decrease in the analyte concentration changes the pH. At the equivalence point, equivalent quantities of acid and base have been mixed together such that the acid-base reaction is complete, and the pH is determined by the product. -In the case of a strong acid-strong base titration, the salt formed is neutral, and pH=7. -In a weak acid-strong base titration, a basic salt is produced, and pH>7. -In a weak base-strong acid titration, the salt is acidic, and pH<7. After the equivalence point, the pH is determined by the concentration of excess titrant. *Weak base graph goes down, strong base graph goes up

Standard Entropy of Reaction Notes

-Molecular systems tend to move spontaneously to a state of maximum randomness or disorder. -As a state function, entropy change, ΔS, depends only on initial and final states. ΔS has a positive value when disorder increases and a negative value when disorder decreases. -The following conditions usually result in an increase in entropy: -A change of phase: solid→liquid→gas, -An increase in the number of gas molecules -A solid dissolving to form a solution. Although the sign of the entropy change can be predicted as described above, the actual value of ΔS∘ must be calculated from the absolute entropy values, S∘, of the reactants and products: ΔS∘=S∘(products)−S∘(reactants) 1. Unlike enthalpies of formation, standard molar entropies of elements at the reference temperature of 298 K are not zero. 3. Standard molar entropies generally increase with increasing molar mass. 4. Standard molar entropies generally increase with an increasing number of atoms in the formula of a substance.

Third Law of Thermodynamics

-No system can reach absolute zero (K=0) -At absolute zero, there would be no molecule movement (Only one micro states)

Buffer Capacity

... is a measure of the amount of H3O+ / OH- ions that can be absorbed without a change in pH ... is related to the concentration of [acid] and [base] in the buffer solution

Determine the concentration of an acid if it required 22.7 mL of 0.115M NaOH to neutralize 25.0 mL of the acid.

0.104 M Macid*Volume(acid)=Mbase*Volume(base) Macid= 0.115M*22.7mL/25.0mL

Predict the sign of the entropy change, ΔS∘, for each of the following reactions. 1. Ag+(aq)+Cl−(aq)→AgCl(s) 2. CaCO3(s)→CaO(s)+CO2(g) 3. 2NO2(g)→N2(g)+2O2(g) 4. 2Mg(s)+O2(g)→2MgO(s) 5. C4H8(g)+6O2(g)→4CO2(g)+4H2O(g) 6. H2O(l)→H2O(g)

1. Negative (Reactants are more random/disordered, thus, entropy decreases in the change from reactants to products.) 2. Positive (Products are more random/disordered, thus, entropy increases in the change from reactants to products.) 3. Positive (Products are more random/disordered, thus, entropy increases in the change from reactants to products.) 4. Negative (Reactants are more random/disordered) 5. Positive (Products are more random/disordered, thus, entropy increases in the change from reactants to products.) 6. Positive (Products are more random/disordered, thus, entropy increases in the change from reactants to products.)

The Ag+ concentration in a saturated solution of AgCl is 1.3 x 10-5 M. What is the solubility product of AgCl ?

1.7*10^-10 AgCl (s) = Ag+ (aq) + Cl- (aq) Ksp = [Ag+][Cl-] = (1.3 x 10-5)2 = 1.7 x 10-10

What is the equilibrium constant K for the following reaction at 298 K? H2 (g) + Br2 (l) = 2 HBr (g) ΔGo = -106.4 kJ/mol

4.48 x 10^18 ΔGo = -RT ln Keq Keq = e^(-Delta G^o/RT)

Formation of solutions involves:

A change in enthalpy ( ΔH ) a change in entropy ( ΔS ) -Enthalpy: Intermolecular interactions: Heat evolved or absorbed in reaction -Entropy: Reflects the degree of randomness/disorder -the large increase in entropy upon mixing drives the formation of most solutions

For each of the following pairs, choose the substance with the higher entropy per mole at a given temperature: A) Ar(l) or Ar(g) B) He(g) at 3 atm pressure or He(g) at 1.5 atm pressure C) 1 mol of Ne(g) in 15.0 L or 1 mol of Ne(g) in 1.50 L D) CO2(g) or CO2(s)

A) Ar (g) B) He(g) at 1.5 atm pressure C) 1 mol of Ne(g) in 15.0 L D) CO2(g)

The value of Ka for nitrous acid (HNO2) at 25 ∘C is 4.5×10−4. A) Write the chemical equation for the equilibrium that corresponds to Ka. B) By using the value of Ka, calculate ΔG∘ for the dissociation of nitrous acid in aqueous solution. C) What is the value of ΔG when [H+] = 6.0×10−2 M , [NO−2] = 6.2×10−4 M , and [HNO2] = 0.20 M?

A) HNO2(aq)⇌H+(aq)+NO−2(aq) Reverse initial reaction (Dissociation) B) ΔG∘ = 19.1 kJ =-RT*ln(Keq) =(-.008314*298)*ln(4.5*10^-4) C) ΔG = -2 kJ ?

What are the criteria for spontaneity: A) in terms of entropy? B) in terms of free energy?

A) In any spontaneous process the entropy of the universe increases. B) In any spontaneous process operating at constant temperature, the free energy of the system decreases.

Consider the accompanying diagram, which represents how ΔG for a hypothetical reaction responds to temperature (Linear graph with downward slope with Delta G on Y-axis and temp on X-axis, Delta G= 0 at 250K) A) Is ΔH positive or negative? B) Is ΔS positive or negative?

A) Positive B) Positive

For a particular reaction, ΔH = -35 kJ and ΔS = -90 J/K . Assume that ΔH and ΔS do not vary with temperature. A) At what temperature will the reaction have ΔG=0? B) If T is increased from that in part A, will the reaction be spontaneous or nonspontaneous?

A) T = 390 K deltaG = deltaH - TdeltaS B) Non-Spontaneous

How many milliliters of 0.105 MHCl are needed to titrate each of the following solutions to the equivalence point? A) 50.0 mL of 9.50×10−2 MNaOH B) 22.0 mL of 0.119 MNH3 C) 125 mL of a solution that contains 1.40 g of NaOH per liter

A) V=45.2 mL B) V=24.9 mL C) V=41.67 mL

Calculate the pH of the following solutions: A) A solution that is 6.0×10−2 M in potassium propionate (C2H5COOK or KC3H5O2 ) and 8.5×10−2 M in propionic acid (C2H5COOH or HC3H5O2 ). The Ka of propionic acid is 1.3×10-5. B) A solution that is 7.0×10−2 M in trimethylamine, (CH3)3N, and 0.11 M in trimethylammonium chloride, (CH3)3NHCl. The Kb of trimethylamine is 6.4×10-5. C) A solution that is made by mixing 50.0 mL of 0.16 M acetic acid and 50.0 mL of 0.23 M sodium acetate. The Ka of acetic acid is 1.8×10-5

A) pH=4.7 pH=pKa+log[A^-/HA] *HA is weak acid, A^- is conjugate base pH=-log[1.3*10^-5]+log[(6.0*10^-2)/(8.5*10^-2)] B) pH=9.6 1.0*10^-14/6.4*10^-5=Ka= 1.5625*10^-10 -log[1.5625*10^-10]+log[7.0*10^-2/0.11] C) pH=4.90 0.050*0.16M = .008 0.050*0.23M = .0115 pH=-log[1.8*10^-5]+log[.0115/.008]

A) What is the pH of a buffer prepared by adding 0.607 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. B) 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. C) 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.

A) pH=6.248 pH=pKa+log[A−]/[HA] pH= 6.247+ log[.305/.304] B) pH=6.030 pH= -log[5.66*10^-7]+ log[.2295/.3785] C) pH= 6.537 pH= -log[5.66*10^-7]+log[.402/.206]

From the values given for ΔH∘ and ΔS∘ , calculate ΔG∘ for each of the following reactions at 298 K. A) 2PbS(s)+3O2(g)→2PbO(s)+2SO2(g); ΔH∘=−844kJ; ΔS∘=−165J/K B) 2POCl3(g)→2PCl3(g)+O2(g); ΔH∘=572kJ; ΔS∘=179J/K C) If the reaction is not spontaneous under standard conditions at 298 K, at what temperature (if any) would the reaction become spontaneous?

A) ΔG∘ = -795 kJ (Spontaneous) B) ΔG∘ = 519 kJ (Non-Spontaneous) C) T = 3200 K

Determine pH of buffer solution which is initially .15M acetic acid and .25M sodium acetate, before and after adding .05 mol of HCl. Ka for acetic acid = 1.8*10^-5. B) What would the pH have been if the HCl had been added directly to water without any buffer?

Before: 4.97 pH = pKa + log[base]/[acid] pH=4.74 + log(0.25/0.15) = 4.97 After: 4.74 pH=4.74 + log0.25-0.05/ 0.15 + 0.05 B) HCl is a strong acid pH = -log [strong acid] pH = -log [0.05 M] pH = 1.3

Henderson-Hasselbalch Equation

Calculates pH of a buffer pH=pKa+log[A−]/[HA]

Questions

Chapter 19, page 1: How do we know that DeltaS is greater than 0 (There is an equal number of molecules)- Mixing = increase in energy

Calculate the solubility from Ksp:

Consider the following equilibrium between a solid salt and its dissolved form (ions) in a saturated solution: CaF2(s)⇌Ca2+(aq)+2F−(aq) At equilibrium, the ion concentrations remain constant because the rate of dissolution of solid CaF2 equals the rate of the ion crystallization. The equilibrium constant for the dissolution reaction is Ksp=[Ca2+][F−]2 Ksp is called the solubility product and can be determined experimentally by measuring the solubility, which is the amount of compound that dissolves per unit volume of saturated solution.

Dissolving a gas in a liquid ___________ the entropy

Decreases (Gas molecules are confined/more organized)

First law of thermodynamics

DeltaE^tot=Q*W Q=heat W=Work

Second Law of Thermodynamics

For any spontaneous process, the entropy of the universe increases -In a larger container, gases have a larger entropy (more space to spread out = more disorganization- The number of ways to arrange molecules increases with container size) -Larger molecules/atoms generally have a larger entropy than smaller ones -Dissolving a solid or liquid will increase the entropy -An increase in gaseous molecules leads to an increase in entropy -Increase in entropy when gases form from liquids or solids, or when liquids or aqueous solutions form from solids -Lowest temp=higher order= lowest entropy -Higher temp= lower order= higher entropy

Adding which of these substances to the following equilibrium will cause the pH to increase? HF (aq) + H2O (l) → F- (aq) + H3O+ (aq) HF H2O KF NaCl

KF When a strong electrolyte containing a common ion is added to a solution of a weak electrolyte, the weak electrolyte ionizes less. If you added HF, it would shift the equilibrium to the right, decreasing the pH.

A saturated solution of magnesium fluoride , MgF2, was prepared by dissolving solid MgF2 in water. The concentration of Mg2+ ion in the solution was found to be 1.18×10−3 M . Calculate Ksp for MgF2.

Ksp = 6.60×10−9 MgF2⇌Mg2++2F^− [F−] = 2.36×10−3 M Ksp=[Mg2+][F−]^2

If a patient ingests 260 mL of a saturated barium sulfate solution, how much toxic Ba2+ ion has the patient consumed? The solubility product Ksp of BaSO4 is 1.1×10−10.

Mass of Ba2+ = 3.74×10^−4 g BaSO4(s)⇌Ba2+(aq)+SO42−(aq) 1.10*10^-10=x^2 x= Ba2+ = 1.05×10^−5 M 1.05*10^-5 M * 137.3 g/mol = .001437 g/L

In which of the following aqueous solutions would AgCl have the greatest solubility? 0.015 M NaCl pure water 0.020 M AgNO3 0.020 M HCl

Pure Water Factors that affect the solubility of ionic compounds in aqueous solutions include pH and the common ion effect.

What is the solubility of 0.3 M HF (how much HF dissociates to F-) in the presence of 0.1 M KF ? The Ka of HF is 7.2 x 10-4 What is the pH

Solubility = 2.16x10-3 M pH= 2.67 7.2 x 10^-4 = 0.1 x/ 0.3 x= 0.3×7.2×10^−4/0.1 x is: - amount of HF that dissociates - extra F- formed - amount of H3O+ formed pH = -log(2.16 x 10-3) = 2.67

Buffer Solutions

Solutions that contain a weak acid, HA, and its conjugate base, A−, are called buffer solutions because they resist drastic changes in pH. When a solution contains a weak acid and its conjugate base or a weak base and its conjugate acid, it will be a buffer solution. For buffers containing a weak acid, the principal reaction is: HA(aq)+H2O(l)⇌H3O+(aq)+A−(aq) or HA(aq)⇌H+(aq)+A−(aq) Equilibrium concentrations of HA and A− are approximately equal to their initial concentrations. You can visualize a buffer solution containing approximately equal concentrations of HA and A− in action using the following equation, where Ka is the equilibrium constant: Ka=[H3O+][A−]/[HA] [H3O^+]=Ka[HA]/[A−] When the concentrations of HA and A− are relatively large, the addition of a small amount of acid or base changes their concentration negligibly. The result is that the ratio [HA]/[A−] maintains a value approximately equal to 1, and the [H3O+] is approximately equal to the Ka value under buffer conditions. Buffer solutions consist of approximately equal amounts of a weak acid (or base) and its conjugate base (or acid) added as a salt Buffer solutions are most effective (buffer best) at pH values close to their pKa

DeltaH=-36 kJ and DeltaS= -95 J/K. At what temperature will the reaction have deltaG = 0?

T= 380 DeltaG = DeltaH-TDeltaS 0 = DeltaH - TDeltaS T=(-36kJ)/(-0.095 kJ)

Le Chatelier's Principle for HA(aq)⇌H+(aq)+A−(aq)

The buffer will follow Le Châtelier's principle. If acid is added, the reaction shifts to consume the added H+, forming more HA. When base is added, the base will react with H+, reducing its concentration. The reaction then shifts to replace H+ through the dissociation of HA into H+ and A−. In both instances, [H+] tends to remain constant.

Based on the molecular structures shown, which of these isomers would you expect to have the higher standard molar entropy at 25 ∘C?

The less symmetrical/ less organized molecular structure has the higher entropy

If ΔH∘is positive and ΔS∘is negative......

The reaction is non-spontaneous at all temperatures!

Consider the following reaction: 2H2 (g) + O2 (g) → 2H2O(l) ΔS = -326.3 J/K, ΔH = -571.7 kJ , and ΔG = -475.3 kJ at 25oC. Does the the decomposition of water ever become spontaneous?

Yes: T=1755 K Reverse Reaction

Indicate whether ΔG increases, decreases, or does not change when the partial pressure of H2 is increased in each of the following reactions. (a) N2(g)+3H2(g)→2NH3(g) (b) 2HBr(g)→H2(g)+Br2(g) (c) 2H2(g)+C2H2(g)→C2H6(g)

a) Decreases b) Increases c) Decreases

What is the pH of a buffer mixture made of 0.12 M acetic acid (CH3CO2H or HOAc) and 0.1 M sodium acetate (NaCH3CO2 or NaOAc)? Kb for acetate = 5.6 x 10-10

pH= 4.7 HOAc (aq) + H2O (l) = OAc- (aq) + H3O+ (aq) Ka = 1.8*10^-5 = [OAc−][H3O+]/(HOAc) Henderson-Hasselbalch Equation pH = pKa + log[base]/[acid] pH=-log(1.8×10−5) + log (0.1/0.12) = 4.74 + (-0.079) = 4.7

Calculate the pH of the solution made by adding 0.50 mol of HOBr and 0.30 mol of KOBr to 1.00 L of water. The value of Ka for HOBr is 2.0×10^−9.

pH=8.48 HOBr(aq)+H2O(l)⇌H3O+(aq)+OBr−(aq) 2.0*10^-9= 0.30x/0.50 -log[3.3*10^-9] M *When the concentrations of acid and base in a buffer solution are equal, [H3O+]=Ka and pH= pKa. When there is more acid than base, [H3O+]>Ka and pH< pKa. The pKa for HOBr is 8.70. As expected, the calculated pH is less than pKa.

Determination of Ka from Titration Curves

pKa = pH Ka = 10-pKa

The value of Ksp for silver sulfate, Ag2SO4, is 1.20×10−5. Calculate the solubility of Ag2SO4 in grams per liter.

solubility = 4.50 g/L Ag2SO4(s)⇌2Ag+(aq)+SO42−(aq) Ksp=(2x)^2*x=4x^3 x = 1.44×10−2 M (1.44*10^-2M)*312 g/mol 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 sulfate is 1.4×10−2 M or 4.5 g/L. If we add 10 g of Ag2SO4 to 1 L of water, only 4.5 g of the salt will dissolve, and the rest will stay at the bottom of the beaker.

Gibbs Free Energy of a Reaction

ΔG = ΔGo + RT ln Q R= 8.314 J/mol-K ΔG^o = Gibbs free energy under "standard conditions" (298 K, 1 atm, 1M [soln]) Q = reaction quotient: experimental product & reactant concentration - can be anything ΔG^o is the Gibbs free energy change under "standard conditions" ΔG = ΔG^o + RT ln ([products]/[reactants]) ΔG = ΔG^o + RT ln 1/1 ΔG =ΔG^o "Standard conditions": • 1atm • 1 M concentrations Large values of K (at room temperature): ΔGo < 0 Small values of K (at room temperature): ΔGo > 0

Gibbs Free Energy Change

ΔG = ΔH-TΔS A reaction is spontaneous if ΔG is negative -A measure of the amount of "useful work" a system can perform -Endergonic: If ΔG>0, the reaction in the forward direction is non-spontaneous but the reverse reaction is spontaneous.

In a chemical reaction two gases combine to form a solid. What do you expect for the sign of ΔS?

ΔS<0