Chem 1308 - Dr. M Jiang (Spring 2020) Ch 11 - 13
The Henry's law constant at 25 °C for CO₂ in water is 0.035 M/atm. What is the solubility of CO₂, in molarity units, in 1.0 L of water when the partial pressure of CO₂ is 2.3 atm?
0.0805 M Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .Henry's Law states that the concentration of a gas (its solubility) is dependent on the Henry's Law constant (kHkH) and the partial pressure of the gas (PgasPgas):Cgas=kH×PgasCgas=0.035 Matm×2.3 atm=0.081 M
A nonvolatile solute is dissolved in benzene and the resulting solution has a vapor pressure of 18.5 torr. What is the mole fraction of the solute? (P° of benzene is 26.5 torr).
0.302 Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. . Raoult's Law states that the vapor pressure of a solution consisting of a volatile solvent (benzene) and nonvolatile solute is equal to the product of the vapor pressure of the pure solvent and the mole fraction of the solvent.Psolution=χbenzenePobenzenePsolution=χbenzenePbenzeneoThe mole fraction of the benzene can be calculated from the vapor pressure of the solution (18.5 torr) and the vapor pressure of benzene (26.5 torr).χbenzene=PsolutionPobenzene=18.526.5=0.698χbenzene=PsolutionPbenzeneo=18.526.5=0.698The mole fraction of the solute can be calculated by subtracting the mole fraction of the benzene from one.χsolute=1−0.698=0.302
You have a solution of two volatile liquids, X and Y. Pure liquid X has a vapor pressure of 410.0 torr and pure liquid Y has a vapor pressure of 150.0 torr at the temperature of the solution. The mole fraction of X in the vapor at equilibrium above the solution is 1.5 times the mole fraction of liquid X in the solution. What is the mole fraction of liquid X in the solution?
0.474 Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .The pressure of liquid X and liquid Y in the vapor can be expressed in terms of their mole fraction in solution and the vapor pressure of the pure solvents.PX=XX×P0X=XX×410.0torrPXX=XXX×PXX0=XXX×410.0torrPY=XY×P0Y=XY×150.0torr=(1-XX)×150.0torrPXY=XXY×PXY0=XXY×150.0torr=(1-XXX)×150.0torrThe total vapor pressure is the sum of PXPX and PYPYPtot=XX×410.0torr+(1-XX)×150.0torrPXtot=XXX×410.0torr+(1-XXX)×150.0torrThe mole fraction of liquid X in the vapor is given by its pressure in the vapor divided by the total vapor pressure, and is equal to 1.5 times the mole fraction in the liquid, or 1.5×XX1.5×XX . Since the units are all in torr, they will cancel out. PXPtot=XX×410.0XX×410.0+(1-XX)×150.0=1.5×XXPXXPXtot=XXX×410.0XXX×410.0+(1-XXX)×150.0=1.5×XXX Solve for the value of XXXX.410.0XX(410.0−150.0)XX+150.0=1.5×XX410.0XXX(410.0−150.0)XXX+150.0=1.5×XXX410.0XX(260)XX+150.0=1.5×XX410.0XXX(260)XXX+150.0=1.5×XXX410.0XX=390X2X+225XX410.0XXX=390XXX2+225XXX390X2X=410.0XX−225XX=185XX390XXX2=410.0XXX−225XXX=185XXXX2X=185XX390XXX2=185XXX390XX=185390=0.474
NaCl has a van't Hoff factor of i = 1.9. What is the concentration of particles in a 0.4501 M solution of NaCl?
0.855 M Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .When an ionic compound dissociates, the ions are separated from each other and surrounded by water molecules. In this compound, the sodium ion separates from the chloride ion:NaCl(aq)→Na+(aq)+Cl−(aq)NaCl(aq)→NaX+(aq)+ClX−(aq)Instead of assuming complete dissociation, this problem presents the van't Hoff factor, which tells us that the ions only mostlymostly separate from each other, since the factor is 1.9 rather than 2:1.9×0.4501 M=0.86 M
Determine K for a reaction at 200 K if ∆G° =22.7 kJ/mol. (R = 8.314 J/mol ･ K)
1.18 × 10⁻⁶ Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .Solve for K in the following equation: ∆G°=−RTlnK 22.7kJ/mol = −8.314(J/mol･K) × (1 kJ/10^3 J) ×(200 K×ln(K)) 22.7kJ/mol = (−1.663kJ/mol) ×ln(K) ln(K) = −22.7/1.633 = −13.65 K=e^−13.65 = 1.18×10^−6
Consider the following reactions: A ⇌ B, K₁=3.76 A ⇌ C, K₂=2.00 What is K for the reaction C ⇌ B?
1.88 Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .When there are multiple equations, K=K1×K2K=K1×K2. Add the equations together to achieve the target equation. When you reverse the equation, you have to "flip" the K. A−⇀↽−B3.76A↽−−⇀B3.76C−⇀↽−A12.00----------------------C↽−−⇀A12.00_C−⇀↽−B1.88
A 251 mL aqueous solution containing 0.400 g of an unknown polymer has an osmotic pressure of 26.9 torr at 30.0 °C. What is the molar mass of the polymer?
1119.5 g/mol Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. . Osmotic pressure, ΠΠ, of a solution is calculated from the molar concentration, M, by the equationΠ=MRTΠ=MRTwhere R is the gas constant (0.08206 L⋅atmmol⋅KL⋅atmmol⋅K) and T is the absolute temperature. First rearrange the equation to calculate the molarity of the solution which has an osmotic pressure of 26.9 torr at 30.0 °C, changing the temperature from °C to K:M=ΠRT=26.9torr×1 atm760torr0.08206L⋅atmmol⋅K⋅(273+30.0)K=0.001423molLM=ΠRT=26.9torr×1 atm760torr0.08206L⋅atmmol⋅K⋅(273+30.0)K=0.001423molLThen calculate the number of moles in 251 mL of solution:0.001423molL×251 mL×1 L1000 mL=0.0003573 mol0.001423molL×251 mL×1 L1000 mL=0.0003573 molThere were 0.400 g of the polymer in the 251 mL of solution, so the molar mass can be calculated:0.400 g0.0003573 mol=1120gmol
A solution is made using 200.0 mL of methanol (density 0.792 g/mL) and 300.0 mL of water (density 1.000 g/mL). What is the molality methanol in water?
16.47 m Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .Molality (m) is mole of solute divided by kg of solvent (mole solutekg solvent)(mole solutekg solvent). First, calculate grams of the solute (methanol):200.0 mL methanol×0.792 g ml=158.4 g methanol200.0 mL methanol×0.792 g ml=158.4 g methanolNow, calculate the moles of methanol:158.4 g methanol ×1 mole32.04 g=4.944 moles methanol158.4 g methanol ×1 mole32.04 g=4.944 moles methanolFinally, calculate the molality of methanol:4.944 mole methanol300 g water×1000 g1 kg=16.5 m4.944 mole methanol300 g water×1000 g1 kg=16.5 m(NOTE: since water is 1.000 g/mL, then 300.0 m: = 300.0 g)
At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.100 M?
2.125 M Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .Henry's Law states that the concentration of a gas (its solubility) is dependent on the Henry's Law constant (kHkH) and the partial pressure of the gas (PgasPgas):Cgas=kH×PgasCgas=kH×PgasUse the given information to solve for kHkH:kH=CgasPgas=0.080 M1.7 atm=0.04706 M/atmkH=CgasPgas=0.080 M1.7 atm=0.04706 M/atmNow use the value of kHkH and the new pressure to solve for the solubility of the gas:Pgas=CgaskHPgas=0.100 M0.04706 M/atm=2.1 atm
A nonvolatile solute is dissolved in benzene so that it has a mole fraction of 0.141. What is the vapor pressure of the solution? (P° of benzene is 26.5 torr)
22.8 torr Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .Raoult's Law states that the vapor pressure of a solution consisting of a volatile solvent (benzene) and nonvolatile solute is equal to the product of the vapor pressure of the pure solvent and the mole fraction of the solvent.Psolution=χsolventPosolventPsolution=χsolventPsolvento Since the mole fraction of the solute is given, the mole fraction of benzene can be calculated by subtracting from one.χsolvent=1−0.141=0.859χsolvent=1−0.141=0.859 Now vapor pressure can be calculated.Psolution=(0.859)(26.5 torr)=22.8 torr
What is ∆G° for the reaction CH₃OH(g) → CO(g) + 2 H₂(g) at 25°C?
24.84 kJ/mol Solution: The free energy can be calculated using the equation below. ΔGo=ΔHo−TΔS0ΔGo=ΔHo−TΔS0 Plug the given values into the problem and solve for ΔGoΔGo. Remember to convert J to kJ and convert T into Kelvin. ΔGo=(90.7)−(25+273K)(0.221)=24.8
At what temperature, in °C, is a certain reaction at equilibrium if ∆H = +87.9 kJ/mol and ∆S = +170.2 J/mol ･ K?
243.45 °C Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .A reaction is at equilibrium when ∆G = 0. For a reaction where ΔH=+87.9kJmol∆H=+87.9kJmol and ΔS°=+170.2Jmol･K∆S°=+170.2Jmol･K one can calculate the temperature at which ∆G becomes 0.ΔG=ΔH−TΔS∆G=∆H−T∆SΔG=+87.9kJmol−T×170.2Jmol･K∆G=+87.9kJmol−T×170.2Jmol･KWhen ∆G=0T×170.2Jmol･K=+87.9kJmol×1000 J1 kJT×170.2Jmol･K=+87.9kJmol×1000 J1 kJT=8.79×104 K170.2=516 K=516−273=243 °C
A solution is made using 16.7 percent by mass CH₂Cl₂ in CHCl₃. At 30 °C, the vapor pressure of pure CH₂Cl₂ is 490 mm Hg, and the vapor pressure of pure CHCl₃ is 260 mm Hg. The normal boiling point of CHCl₃ is 61.7 °C. a) What intermolecular forces are shared between CH₂Cl₂ and CHCl₃? b) Would you expect this solution to form an ideal solution? c) What is the mole fraction of CH₂Cl₂ in the solution? d) What is the vapor pressure of the solution? e) What is the molality of CH₂Cl₂ in the solution? f) what is the boiling point of the solution? (Kb for CHCl₃ is 3.67 °C/m)
3.63 m 1.35 mol 111.11 g/mol Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .The molality of the solution can be found using the change in the freezing point (18.6 °C and the Kf of benzene:ΔT=Kf⋅mΔT=Kf⋅m18.6°C=(5.12°C/m)⋅m18.6°C=(5.12°C/m)⋅mm=3.63mol/kg Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .The solution has a molality of 3.63 mol/kg. We know we have 425.0 mL of benzene, so we can use the density of benzene to find the kg of the solvent, which will then allow us to solve for the moles.(425.0mL)(0.877g1mL)(1kg1000g)=0.373kg(425.0mL)(0.877g1mL)(1kg1000g)=0.373kg(3.63mol/kg)(0.373kg)=1.35mol Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .A molar mass is the grams of a compound divided by the moles of the compound. We have 150.0 grams of the unknown and 1.35 mol.150.0g1.35mol=111g/mol
Determine ∆G° for the reaction N₂O₄(g) ⇌ 2 NO₂(g) (K= 0.144 at 25°C).
4.8 kJ/mol Solution: Solve for ∆G° in the following equation:ΔG°=−RTlnK ΔG°=−8.314(J/mol･K)× (1 k/J10^3 J)×298 K×ln(0.144) ∆G°=−2.478kJ/mol×ln(0.144) ΔG°=−2.478kJ/mol×−1.938=4.80kJ/mol
When 2.63 g of a polypeptide is dissolved in 101 mL of water, the resulting solution is found to have an osmotic pressure of 0.125 atm at 37.0 °C. What is the molar mass of the polypeptide? (Assume the volume doesn't change when the polypeptide is added.)
5299.3 g/mol Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .Solve for molarity (M) using the temperature (converted to Kelvin) and pressure given in the problem.Π=MRTΠ=MRT(0.125atm)=(M)(0.08206L⋅atmmol⋅K)(37+273.15K)(0.125atm)=(M)(0.08206L⋅atmmol⋅K)(37+273.15K)Then, using the relationship between moles and liters in molarity, solve for the number of moles using the given volume (converted to L).M=0.125atm(1)(0.08206L⋅atmmol⋅K)(37+273.15K)=4.911416×10−3M=0.125atm(1)(0.08206L⋅atmmol⋅K)(37+273.15K)=4.911416×10−3M=molLM=molL4.911416×10−3=x.1014.911416×10−3=x.101x=4.960530×10−4x=4.960530×10−4Then, solve for the molar mass using the relationship between grams and moles using the given grams and the calculated moles.molarmass=gmolmolarmass=gmolmolarmass=2.63g4.960530×10−4=5300gmol
What is the osmotic pressure of a solution made from 22.3 g of methanol (MM = 32.04 g/mol) that was added to water to make 251 mL of solution at 25.0 °C?
67.8 atm Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .Osmotic pressure, ΠΠ, of a solution is calculated from the molar concentration, M, by the equationΠ=MRTΠ=MRTwhere R is the gas constant (0.08206 L⋅atmmol⋅KL⋅atmmol⋅K) and T is the absolute temperature. For the solution of 22.3 g of methanol in 251 mL of water at 25°C we can calculate ΠΠ as follows:Π=22.3 g ×1 mol32.04 g251 mL×1 L1000 mL×0.08206L⋅atmmol⋅K×298 K=67.8 atm
What is the entropy when 1.20 moles of CCl₂F₂ vaporize at 25°C? [∆H(vap) = 17.2 kJ/mol at 25°C]
69.262 J/K Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .First, calculate the heat for 1.20 moles. 17.2kJmol×1.20mol=20.64kJ17.2kJmol×1.20mol=20.64kJThen convert this value from kJ to J.20.64kJ×1000J1kJ=20,640J20.64kJ×1000J1kJ=20,640JThe entropy can be found using the equation below, with temperature in Kelvin. Plug in the known values into the equation and solve for S.ΔS=heatT=20,640J(273+25K)=69.3JK
A 83.5 g sample of a nonelectrolyte is dissolved in 250.1 g of water. The solution is determined to have a boiling point of 102.3 °C. What is the molar mass of the compound? (Kb for water is 0.510 °C/m).
74.09 g/mol Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. . The molality of the solution can be determined from the boiling point elevation (102.3°C -100°C = 2.3 °C) and the Kb of water (0.510 °C/m).2.3 °C0.510 °Cm=4.5098 m=4.5098mol1000 gwater2.3 °C0.510 °Cm=4.5098 m=4.5098mol1000 gwaterThere were 83.5 g of the compound in 250.1 g water. Next determine the concentration of the unknown compound per 1000 g of water.83.5 g250.1 g water×1000 g water=333.87 g83.5 g250.1 g water×1000 g water=333.87 g Then calculate the molar mass:333.87 g4.5098 mol=74gmol
The percent by mass of methanol (MM = 32.04 g/mol) in an aqueous solution is 22.3%. What is the molality of the methanol solution?
8.96 m Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .Percent by mass is x gsolute100 gsolutionx gsolute100 gsolution and the mass of the solution is the mass of the solute plus the mass of the solvent. Therefore:22.3 % methanol=22.3 gmethanol100 gsolution100 gsolution=22.3 gmethanol+x gsolventx gsolvent=100 gsolution−22.3 gmethanol=77.7 gsolvent22.3 % methanol=22.3 gmethanol100 gsolution100 gsolution=22.3 gmethanol+x gsolventx gsolvent=100 gsolution−22.3 gmethanol=77.7 gsolventNow, use the ratio of solute to solvent to convert to molality (m), which is molsolutekgsolventmolsolutekgsolvent:22.3 gmethanol77.7 gsolvent×1 molmethanol32.04 gmethanol×1000 g1 kg=8.96 m
Solve the following expression for x. What is(are) the value(s) of the positive root(s)?
A) 0.197 Solution: Using the quadratic formula, you can solve for the positive value of x. The quadratic equation is,x=−b±√b2−4ac2ax=−b±b2−4ac2awhen ax2+bx+c=0ax2+bx+c=0. Using the values from this problem, solve for x.x2=(12.0)(0.200−x)x2=(12.0)(0.200−x)x2=2.4−12.0xx2=2.4−12.0xx2+12.0x−2.4=0x2+12.0x−2.4=0x=−(12)±√(12)2−4(1)(−2.4)2(1)=0.197
Consider the following equilibrium reaction: A (g) + B (s) ⇌ C (g) If Kp = 6 × 10⁻³, which species will have the highest partial pressure at equilibrium?
A) A Solution: The equilibrium constant is calculated by dividing the partial pressures of the products by the partial pressure of the reactants. Coefficients become exponents. Compounds which are solids and liquids have constant composition and therefore do not appear in the equilibrium expression. Kp=PCPAKp=PCPA If the KpKp is less than 1, this indicates that the partial pressure of A is larger than the partial pressure of C.
Consider the following energy diagram and determine which of the following statements is true.
A) At equilibrium, we expect [Reactants] < [Products]. Solution: At equilibrium, we would expect the concentration of the products to get greater than the concentration of the reactants because the activation energy is lower for the forward reaction. In this case, kfkr>1kfkr>1 meaning that KeqKeq>1. This indicates that products are favored at equilibrium.
The following picture represents the dissociation of HF in aqueous solution (water molecules not shown). Based on this image,
A) HF is a weak electrolyte Solution: A strong electrolyte would be fully dissociated into its ions in water and a nonelectrolyte would not dissociate at all in water. This shows some dissociation of ions and is therefore modeling a weak electrolyte.
Which of the following statements are always true for a reaction at equilibrium? I. The rate of the forward and reverse reactions are equal. II. The concentrations of the reactants and the products remain constant. III. The amount of reactants is equal to the amount of products.
A) I and II Solution: When a system is at dynamic equilibrium, the rates of the forward and reverse reactions are equal. This also means that the concentrations of reactants and products will appear to remain constant.
How does the third law of thermodynamics allow absolute entropies of substances to be determined?
A) It defines a reference point by which entropy changes can be measured and assigned as an absolute entropy for a substance. Solution: The third law of thermodynamics defines a reference point by which entropy changes can be measured and assigned as an absolute entropy for a substance.
For a solution to be ideal, which of the following statements is true?
A) The solute and solvent must have similar strengths of intermolecular forces. Solution: An ideal solution is one which follows Raoult's Law, which says that the vapor pressure of the solvent in the solution is equal to the mole fraction of the solvent times the vapor pressure of the pure solvent. For this condition to be true the solute and solvent must have similar strengths of intermolecular forces.
If Kp is the equilibrium constant for the reaction 2 A (g) + B (g) ⇌ 4 C (g). What is equilibrium constant (in terms of Kp) for the reaction 2 C (g) ⇌ A (g) + ½ B (g)?
A) see problem image Solution: Equilibrium constants are the products divided by the reactants. Coefficients become exponents. Because of this, multiplying the coefficients by 1/2 results in an exponent of 1/2. Because the equation is reversed, the Kp needs to be inverted.
Which of the following is the correct expression for the pressure based equilibrium constant for the reaction: 4 NH₃ (g) + 7 O₂ (g) ⇌ 4 NO₂ (g) + 6 H₂O (g)
A) see problem image Solution: Equilibrium constants are the products divided by the reactants. Coefficients become exponents. In this case, the equilibrium constant would be represented as: Kp=(PNO2)4(PH2O)6(PNH3)4(PO2)7
When NaCl dissolves in water
A) the Na⁺ ion will be attracted to the oxygen atom in water. Solution: NaCl dissolves in water to form an Na+ and a Cl-. The Na+ is attracted to the negatively charged oxygen atom in water.
Determine ∆G° and the spontaneity of the reaction A₂(g) + 3 X₂(g) → 2 AX₃(g) at 591 K.
B) +196 kJ/mol, nonspontaneous Solution: The free energy is calculated using the following equation: ΔGo=ΔHo−TΔSoΔGo=ΔHo−TΔSo Plug in the given values to find the free energy. Convert J to kJ. ΔGo=(−86)−(591×(−0.478))=196ΔGo=(−86)−(591×(−0.478))=196 This value is positive and indicates that the reaction is nonspontaneous.
Which of the following has the smallest standard molar entropy (∆S°)?
B) C(diamond) Solution: Perfect crystalline structures have lower entropy than all other states of matter because they are highly ordered. As molecules get larger, they have more entropy. Liquids by comparison have less order and have higher entropies than solids.
Based only on intermolecular forces, which of the following pairs of molecules would be expected to form a nearly ideal solution?
B) CH₃CH₂CH₂CH₂CH₃ and CH₃CH₂CH₂CH₃ Solution: An ideal solution is one which follows Raoult's Law, which says that the vapor pressure of the solvent in the solution is equal to the mole fraction of the solvent times the vapor pressure of the pure solvent. For this condition to be true the solute and solvent must have similar strengths of intermolecular forces. Only CH₃CH₂CH₂CH₂CH₃ and CH₃CH₂CH₂CH₃ would have nearly identical intermolecular forces with each other.
Which of the following statements is true about the solubility of CO₂ (g) in H₂O (l)?
B) CO₂ will be more soluble at lower temperatures. Solution: The solubility of gases in water generally increase as the water temperature decreases. Therefore, CO₂ will be more soluble at lower temperatures.
Which of the following reactions would have the most positive ∆S° value?
B) CO₂(s) ⟶ CO₂(g) Solution: The most positive change in entropy would indicate that there is more disorder at the end of the reaction. Option B goes from one mole of a highly ordered solid to one mole of a very disordered gas.
Which of the following best describes the third law of thermodynamics?
B) S° = 0 for perfect Li(s) at 0 K Solution: The third law of thermodynamics states that the entropy for a perfect crystal is 0 at absolute zero temperature in Kelvin.
For a reaction for which ∆H = -142 kJ/mol and ∆S = -178 J/mol･K, which of the following statements is true?
B) The reaction is spontaneous below 798 K. Solution: Spontaneity is determined by the free energy. When ΔGΔG is negative, it is spontaneous. ΔG=ΔH−TΔSΔG=ΔH−TΔS Plug in the given values. ΔG=(−142000)−(798)(−178)ΔG=(−142000)−(798)(−178)ΔG=44ΔG=44At values greater than 798, this reaction is not spontaneous. Now try at a lower value.ΔG=(−142000)−(797)(285)ΔG=(−142000)−(797)(285)ΔG=−134ΔG=−134 This reaction is spontaneous at temperatures less than 798K.
For the reaction below, the partial pressure of NO₂ is 0.50 atm and the partial pressure of N₂O₄ is 0.25 atm. What must happen for the reaction to reach equilibrium? 2 NO₂ (g) ⇌ N₂O₄ (g) Kp = 0.25
B) The reaction needs to shift in the reverse direction. Solution: The reaction quotient is calculated by dividing the partial pressures of the products by the partial pressures of the reactants. Coefficients become exponents. Use the given pressures to calculate the reaction quotient and then compare to the given equilibrium constant. Qp=PN2O4(PNO2)2=(0.25)(0.50)2=1Qp=PN2O4(PNO2)2=(0.25)(0.50)2=1 This is the larger than the given KPKP indicating that the reaction has more products and needs to shift to the reverse direction to reach equilibrium.
Consider the following acidic equilibrium: H₂CO₃(aq) + H₂O(l) ⇌ HCO₃⁻(aq) + H₃O⁺(aq). If you add NaHCO₃ to this solution, which of the following will occur?
B) The reaction will shift in the reverse direction. Solution: If you add NaHCO₃ to this solution, the reaction will shift in the reverse direction to reestablish equilibrium.
Which of the following is the correct expression for the pressure based equilibrium constant for the reaction: CaCO₃ (s) ⇌ CO₂ (g) + CaO (s)
B) see problem image Solution: Compounds which are solids and liquids have constant composition and therefore do not appear in the equilibrium expression. Equilibrium constants are the products divided by the reactants. Coefficients become exponents. In this case, the equilibrium constant would be Kp=PCO2Kp=PCO2.
Which of the following is the correct expression for the concentration based equilibrium constant for the reaction: 2 C (aq) + D (aq) ⇌ 2 A (aq) + B (aq)
B) see problem image Solution: Equilibrium constants are the products divided by the reactants. Coefficients become exponents. In this case, the equilibrium constant would be represented as: Kc=[A]2[B][C]2[D]Kc=[A]2[B][C]2[D].
Which of the following aqueous solutions would have the highest vapor pressure?
C) 0.10 M CH₃OH Solution: According to Raoult's Law, the vapor pressure of a solution is proportional to the mole fraction of solvent, and therefore, inversely proportional to the total solute concentration. So, the lowest molarity of ions will have the highest vapor pressure. Since ionic compounds dissociate in aqueous solution and molecular compounds do not, the total concentration of ions is lowest in CH₃OH (0.10 M). The other solutions (all ionic) have total solute concentrations greater than 0.10M.
A supersaturated solution can be made to precipitate out by:
C) Both (A) and (B) Solution: A supersaturated solution is unstable. Adding more solute will provide the seed crystals for the extra solute to crystallize from the solution. Simply agitating the solution an also start the crystallization process, producing the seed crystals for more solute to crystallize from the solution.
Which of the following solutions would be most likely to show a positive deviation from Raoult's law?
C) CH₃CH₂CH₂OCH₃ and CH₃CH₂CH₂CH₂OH Solution: CH₃CH₂CH₂OCH₃ has only dipolar and dispersion forces, while CH₃CH₂CH₂CH₂OH also has hydrogen bonding forces in addition. This pair has the biggest difference in forces than the other three pair, in which the forces are similar in both compounds. Therefore CH₃CH₂CH₂OCH₃ and CH₃CH₂CH₂CH₂OH is most likely to show a deviation from Raoult's law.
Based only on intermolecular forces, which of the following would be the least soluble in CH₃CH₂OH?
C) CH₃CH₃ Solution: Molecules with a C-O or O-H bond can experience both dipole-dipole and hydrogen bonding interaction with CH₃CH₂OH, while CH₃CH₃ can only experience dispersion forces, so CH₃CH₃ will be the least soluble in CH₃CH₂OH.
Which of the following reactions results in an increase in the entropy of the system?
C) NH₄Cl(s) → HCl(g) + NH₃(g) Solution: Option C shows one mole of a highly ordered solid goin got two moles of very disordered gases. This results in an increase in entropy.
Which of the following statements best describes what occurs at equilibrium?
C) The rate of the reaction in the forward direction is equal to the rate of the reaction in the reverse direction. Solution: When a system is at dynamic equilibrium, the rates of the forward and reverse reactions are equal.
Which of the following will decrease the concentration of A if the reaction below is at equilibrium? A (aq) + B (s) ⇌ 3 C (aq) ΔH < 0
C) decrease temperature. Solution: A negative ΔH indicates that this is an exothermic reaction with energy being released in the forward reaction. Decreasing the temperature would shift the reaction to the right to create more products to reach equilibrium.
Consider the reaction below. Which of the following would increase the partial pressure of B at equilibrium? A (s) + B (g) ⇌ C (g) ΔH < 0
C) increasing the temperature. Solution: A negative ΔH indicates that this is an exothermic reaction with energy being released in the forward reaction. Increasing the temperature would shift the reaction to the left to create more reactants to reach equilibrium.
If a chemical reaction has a positive ∆H and a positive ∆S, then
C) it will be spontaneous at high temperatures and non-spontaneous at low temperatures. Solution: A negative ∆G means a process is spontaneous, and a positive ∆G is always non-spontaneous. ∆G = ∆H - T∆S and T is always positive since it is in kelvin. A reaction where ∆H and ∆S are both positive will be non-spontaneous at low temperatures when ∆H dominates and will be spontaneous at high temperatures when T∆S dominates.
Which of the following is the correct expression for the pressure based equilibrium constant for the reaction: 2 NO (g) + Br₂ (l) ⇌ 2 NOBr (g)
C) see problem image Solution: Compounds which are solids and liquids have constant composition and therefore do not appear in the equilibrium expression. Equilibrium constants are the products divided by the reactants. Coefficients become exponents. In this case, the equilibrium constant would be Kp=(PNOBr)2(PNO)2Kp=(PNOBr)2(PNO)2.
Consider the following diagrams which show the progress for the reaction A(blue) ⇌ B (red). The equilibrium constant (K) for this reaction is 0.8. At which point does the reaction reach equilibrium?
C) see problem image Solution: The equilibrium constant is the products divided by the reactants. A K value of 0.8 is consistent with image choice C where the ratio of product to react is 4:5
Which of the following is true for a reaction when K is much less than one?
C) ∆G° = large positive number Solution: When K is much less than one, the equilibrium lies far to the left and the reaction is very unfavorable. Since ∆G° = - RTlnK, ∆G° will be large and positive.
Which set of results applies to a reaction that is spontaneous at 250 K but is nonspontaneous at 350 K?
C) ∆H < 0 and ∆S < 0 Solution: The free energy can be calculated using this formula. ΔG=ΔH−TΔSΔG=ΔH−TΔS If the reaction is spontaneous at low temperatures and nonspontaneous at high temperatures, this indicates that the both the ∆H and ∆S are less than 0.
Which of the following aqueous solutions would have the highest boiling point?
D) 1.0 m CaCl₂ Solution: The dissolved solute will raise the boiling point. The more "things" that are dissolved (ions or molecules), the higher the boiling point will be. The CaCl₂ will break up into one Ca²⁺ ion and two Cl⁻ ions and there will be a concentration of 3.0 m of "things" in solution. The other choices will have fewer "things" in solution and will raise the boiling point less.
Which of the following describes a colloid?
D) A dispersion of large particles evenly distributed through a mixture. Solution: A colloid is a dispersion of large particles that are evenly distributed through a mixture.
Which of the following phase changes is not spontaneous at room temperature? Assume only the first phase is initially present.
D) H₂O(g) → H₂O(s) Solution: The requirement for a spontaneous process or reaction is that ΔΔG is negative, because that means that ΔΔSuniverseuniverse is positive. When the signs of ΔΔH and ΔΔS for the process are opposite, then it is easy to determine the sign of ΔΔG, but when the signs of ΔΔH and ΔΔS for the process are the same, we must either calculate the value of ΔΔG at that temperature, or we can make a qualitative judgement based on the phase change temperatures for the material. For all these processes, the signs of ΔΔH and ΔΔS are all the same. For example, changing from gas to solid has a negative ΔΔH and a negative ΔΔS. We know that the temperature at which water changes from solid to liquid is 0°C. Using this information, we can determine that changing from gas to solid must also occur at 0°C or below, and that this process will not be spontaneous at room temperature (25°C). Applying this same analysis to the other processes, we can see that they must all be spontaneous at room temperature.
Which one of the following will dissociate into the largest number of ions in solution?
D) Na₃PO₄ Solution: Na₃PO₄ will dissociate into the largest number of ions in solution. Sodium forms ions with a +1 charge. There are three sodium ions for every one phosphate ion. This dissociates into four ions.
What will the sign on ∆S be for the following reaction and why?
D) Negative, because there are more moles of gas on the reactant side than the product side. Solution: The sign of ∆S can be determined by comparing the number of moles of gas on the reactant and product side. Since the number of moles of gas is decreasing, entropy is decreasing, and thus is negative.
Consider the energy diagram for a solution formed between molecules A and B. What can be said about this solution?
D) The solution formed is ideal. Solution: There is no change in enthalpy when the solution is formed, which indicates that the solute-solvent attractions are the same as the solute-solute and solvent-solvent attractions, so the solution is ideal.
Which law states that the entropy of a perfect crystalline substance is zero at the absolute zero of temperature?
D) Third law of thermodynamics Solution: The third law of thermodynamics states that the entropy of a perfect crystalline substance is zero at the absolute zero of temperature.
Which of the following steps in solution formation is exothermic?
D) establishing the solute-solvent intermolecular forces. Solution: It takes energy to overcome any type of intermolecular forces, so overcoming intermolecular forces would be endothermic. Establishing forces such as those between solute and solvent, on the other hand, release energy and are exothermic.
If a chemical reaction has a negative ∆H and a negative ∆S, then
D) it will be spontaneous at low temperatures and non-spontaneous at high temperatures. Solution: A negative ∆G means a process is spontaneous, and a positive ∆G is always non-spontaneous. ∆G = ∆H - T∆S and T is always positive since it is in kelvin. A reaction where ∆H and ∆S are both negative will be spontaneous at low temperatures when ∆H dominates and will be non-spontaneous at high temperatures when T∆S dominates.
If the reaction quotient (Q) is equal to the equilibrium constant (K) for a reaction then
D) the reaction is at equilibrium and the reaction will proceed at equal rates in the reverse and forward direction. Solution: Q is a ratio of products and reactants at a given time and K is the ratio of products and reactants at equilibrium. If Q is equal to K, then the reaction is at equilibrium; however, chemical equilibrium is a dynamic process. Both forward and reverse reactions are occuring at the same rate.
A reaction will be spontaneous at all temperatures if _____
D) ∆H is negative and ∆S is positive Solution: The free energy can be calculated using this formula. ΔG=ΔH−TΔSΔG=ΔH−TΔS A reaction will be spontaneous at all temperatures if ∆H is negative and ∆S is positive.
Which of the following are spontaneous processes? 1. Ice melting at 273 K (assume only ice is initially present). 2. Heat flowing from a hot object to a cold object. 3. An iron bar rusting.
E) 1, 2, and 3 Solution: All three of these processes are spontaneous and occur naturally.
Which of the following ions will contribute most to elevating the boiling point of H₂O?
E) All will contribute equally Solution: Anything that is dissolved will elevate the boiling point (and lower the freezing point) of the solvent. The amount of solute, not the characteristics, increases the degree of change.
The following reaction is at equilibrium in a closed 4L vessel: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) What will happen if the vessel volume is reduced to 3L? (Assume constant temperature.)
E) The equilibrium will shift to the right, producing more NH₃(g). Solution: At constant temperature, reducing the volume will initially result in increased pressure. The system will re-establish equilibrium by reducing the pressure, and this will require shifting the equilibrium to the side containing the smallest number of moles of gas. There are four moles of gas on the reactant side and only two moles on the product side. Reducing the volume (initially increasing pressure) will shift the equilibrium to the right.
Which of the following sets of conditions is true for an exothermic reaction that is spontaneous at all temperatures?
E) ∆H < 0, ∆S > 0, ∆G < 0 Solution: The free energy can be calculated using this formula. ΔG=ΔH−TΔSΔG=ΔH−TΔS A reaction with the following conditions would be spontaneous at all temperatures: ∆H < 0, ∆S > 0, ∆G < 0. The ∆G < 0 indicates that the reaction is spontaneous.
A reaction vessel initially contains only 3.5 atm of N₂O₄. When the system reaches equilibrium according to the reaction below, there is 1.7 atm of NO₂ in the vessel. What is Kp for this reaction? 2 NO₂ (g) ⇌ N₂O₄ (g)
Solution:
A solution is made initially with 0.200 M HIO₃ (Kc = 0.170). Once the equilibrium below is established, what is the equilibrium concentration of H⁺ ions? HIO₃ (aq) ⇌ H+ (aq) + IO₃⁻ (aq)
Solution:
A solution made by dissolving 150.0 g of an unknown compound in 425.0 mL of benzene to make a solution which has a freezing point 18.6 °C lower than that of pure benzene. Kf for benzene is 5.12 °C/m and the density of benzene is 0.877 g/mL. What was the molality of the solution?
Solution:
A student initially adds 1.00 mol of A to an empty 1.0 L flask. After 10 minutes, there are 0.83 mol of A left. What is the reaction quotient? Equation: A(aq) → 2 B(aq)
Solution:
An empty steel container is filled with 2.0 atm of H₂ and 1.0 atm of F₂. The system is allowed to reach equilibrium according to the reaction below. If Kp = 0.45 for this reaction, what is the equilibrium partial pressure of HF? H₂ (g) + F₂ (g) ⇌ 2 HF (g)
Solution:
At 355 K, ∆G = -5.7 kJ/mol for the reaction A (g) → 2 B (g). If the partial pressures of A and B are 2.41 atm and 0.110 atm respectively, what is the standard free energy for this reaction at this temperature?
Solution:
At equilibrium, a reaction vessel contains 4.50 atm of Br₂ and 1.10 atm of NBr₃. According to the reaction: 2 NBr₃ (g) ⇌ N₂ (g) + 3 Br₂ (g) Kp = 4.8 Determine the equilibrium partial pressure of N₂.
Solution:
Consider the following reaction at 25 °C: 5 SO₃(g) + 2 NH₃(g) → 2 NO(g) + 5 SO₂(g) + 3 H₂O(g)
Solution:
Consider the reaction: N₂ (g) + 3 Br₂ (g) ⇌ 2 NBr₃ (g) At equilibrium, the concentrations of N₂ and Br₂ are 0.34 M and 0.70 M respectively, and the concentration of NBr₃ is 0.090 M. What is Kc for this equilibrium?
Solution:
Ethylene glycol (C₂H₆O₂) is used as an additive to the water in your automobile to lower its freezing point. A solution of ethylene glycol in water has a freezing point of -5.80°C. What is the mole fraction of ethylene glycol in this solution? (Kf for water is 1.86°C･kg/mol).
Solution:
For each of the following processes, determine the sign on the entropy and the best explanation for that sign.
Solution:
For the equilibrium A + B ⇌ C + D K was measured to be 1.51 × 10⁻² at 25°C and 55.0 at 600°C. What is ∆S° for the reaction? Assume ∆H° and ∆S° are temperature independent.
Solution:
For the following reaction 3O₂(g) ⇌ 2O₃(g) Kc = 2.10 × 10⁻⁷ at a certain temperature. If [O₂] = 0.0495 M when at equilibrium, what is the equilibrium O₃ concentration?
Solution:
For the following reaction NO₂(g) + SO₂(g) ⇌ NO(g) + SO₃(g) K = 4.15 at a particular temperature. If NO₂ and SO₂ had initial concentrations of 0.650 M, and NO and SO₃ had initial concentrations of 0.450, calculate the equilibrium concentration of NO.
Solution:
For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?
Solution:
For the reaction A (g) → 3 B (g), Kp = 80100 at 298 K. When ∆G = -14.2 kJ/mol, what is the partial pressure of A when the partial pressure of B is 2.00 atm for this reaction at 298 K.
Solution:
For the reaction CCl₄(g) ⇌ C(s) + 2Cl₂(g) Kp = 0.76 at 700°C. What initial pressure of pure carbon tetrachloride will produce a TOTAL equilibrium pressure of 1.50 at 700°C?
Solution:
For the reaction below, Kc = 4.60 × 10⁻⁶. What is the equilibrium concentration of B if the reaction begins with 0.700 M A? A (g) ⇌ 3 B (g)
Solution:
For the reaction below, Kc = 9.2 × 10⁻⁵. What is the equilibrium concentration of D if the reaction begins with 0.48 M A? A (aq) + 2 B (s) ⇌ C (s) + 2 D (aq)
Solution:
For the reaction: 2 A (g) + B (s) ⇌ 2 C (s) + D (g) Kp = 8210 At 298 K in a 10.0 L vessel, the known equilibrium values are as follows: 0.025 atm of A, 0.22 mol of B, and 10.5 mol of C. What is the equilibrium partial pressure of D?
Solution:
Given the following reaction CO(g) + 2H₂(g) ⇌ CH₃OH(g) You have a reaction flask that contains a mixture of 0.1 M CH₃OH, 0.152M CO, and 0.250M H₂. Find the reaction quotient, Q, for this system.
Solution:
When 0.200 mol of CaCO₃(s) and 0.300 mol of CaO(s) are placed in an evacuated, sealed 10.0-L container and heated to 1039 K, PCO₂ = 0.220 atm after equilibrium is established. CaCO₃(s) ⇌ CaO(s) + CO₂(g) Additional CO₂(g) is pumped into the container to raise the pressure to 0.550 atm. After equilibrium is re-established, what is the total mass (in g) of CaCO₃ in the container?
Solution:
You dissolve a 30.0 g mixture of glucose (C₆H₁₂O₆) and NaCl in 1.00 kg water. The freezing point of this solution is found to be -0.851°C. Assuming ideal behavior, what is the mass of glucose in the mixture? (Kf for water is 1.86°C･kg/mol).
Solution:
What is the molality of lithium ions in a 0.302 m solution of Li₃PO₄ assuming the compound dissociates completely?
Solution: 0.906 mol Li/kg solvent
What is the molarity of an aqueous solution that contains 0.072 g C₂H₆O₂ per gram of solution . The density of the solution is 1.04 g/mL
Solution: 1.21 mol/L
What is ∆G for a reaction where ∆G° = -4.5 kJ/mol and Q = 4.0 at 295 K? (R = 8.314 J/mol ･ K)
-1.099 kJ/mol Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .Solve for ∆G in the following equation:ΔG=ΔG°+RTlnQ∆G=∆G°+RTlnQΔG°=−4.5kJmol+8.314Jmol･K×1 kJ103 J×295 K×ln(4.0)∆G°=−4.5kJmol+8.314Jmol･K×1 kJ103 J×295 K×ln(4.0)ΔG°=−4.5kJmol+2.453kJmol×ln(4.0)ΔG°=−4.5kJmol+2.453kJmol×ln(4.0)ΔG°=−4.5kJmol+2.453kJmol×1.386ΔG°=−4.5kJmol+2.453kJmol×1.386ΔG°=−4.5kJmol+3.400kJmol=−1.1kJmol
FeCl₃ has a van't Hoff factor of 3.4. What is the freezing point of an aqueous solution made with 0.81 m FeCl₃? (Kf for water is 1.86 °C/m)
-5.12 °C Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .The change in freezing point is related to the molality of the solute, the freezing point constant of the solvent (KfKf), and (since the solute is ionic), the van't Hoff factor (i). ΔTf=Tf(solvent)−Tf(solution)=Kf×m×iΔTf=Tf(solvent)−Tf(solution)=Kf×m×i ΔTf=1.86 ˚C m ×0.81 m ×3.4 =5.122 ˚C ΔTf=1.86 ˚C m ×0.81 m ×3.4 =5.122 ˚C Now, solve for the freezing point of the solution:Tf(solvent)−ΔTf=Tf(solution)0.00 ˚C −5.122 ˚C =−5.1 ˚C
What is the molarity of a solution of a polypeptide which has an osmotic pressure of 0.125 atm at 37.0 °C?
0.004914 M Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. . Osmotic pressure, ΠΠ, of a solution is calculated from the molar concentration, M, by the equationΠ=MRTΠ=MRTwhere R is the gas constant (0.08206 L⋅atmmol⋅KL⋅atmmol⋅K) and T is the absolute temperature. To calculate the molarity of a polypeptide solution which has an osmotic pressure of 0.125 atm at a temperature of 37.0 °C, rearrange the equation and substitute the values, changing the temperature from °C to K:M=ΠRT=0.125 atm0.08206L⋅atmmol⋅K⋅(273+37.0)K=0.00491molL=0.00491 M
If the equilibrium constant of a given reaction is 2.54, what is the equilibrium constant of its reverse reaction?
0.394 Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .If a reaction is reversed, the equilibrium constant expression is inverted.12.54=0.394
What is the entropy change when 315 J of energy is reversibly transferred to a sample of water at 25°C?
1.06 J/K Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .The entropy change can be calculated using the formula below. ΔS=heatTΔS=heatTConvert the given T into Kelvin and plug the numbers into the formula. ΔS=31525+273=1.06JK
A solution is made using 200.0 mL of methanol (density 0.792 g/mL) and 301.1 mL of water (density 1.000 g/mL). What is the mass percent of the methanol?
34.47 % Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .Mass percent is defined as: mass of componenttotal mass×100mass of componenttotal mass×100. First, calculate the mass of the components of the solution:200.0 mL methanol×0.792 gmL=158.4 g methanol301.1 mL water×1.000 gmL=301.1 g water200.0 mL methanol×0.792 gmL=158.4 g methanol301.1 mL water×1.000 gmL=301.1 g waterNow, calculate the mass % of methanol:mass % methanol=mass of methanoltotal mass×100=158.4 g of methanol(158.4 g methanol+301.1 g water)×100=35.47 % methanol
Two volatile substances are mixed with one another so that substance A has a mole fraction of 0.73. Given the information in the table below, what is the vapor pressure of the solution?
88 torr Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .You can calculate the vapor pressure of the solution by using Raoult's Law. If the mole fraction of A is 0.73, then the mole fraction of B is (1-0.73). P=XAPoA+XAPoBP=(42.5)(0.73)+(1−0.73)(210)P=XAPoA+XAPoBP=(42.5)(0.73)+(1−0.73)(210)P=88torr
A solution is an example of
A) a homogenous mixture Solution: A solution is defined as a homogeneous mixture.
If Kp = 7.3 × 10⁻⁶ for the reaction below at 400 K, then what is the value of Kc? 2 NBr₃ (g) ⇌ N₂ (g) + 3 Br₂ (g)
Solution:
The Kp for the reaction A (g) ⇌ 2 B (g) is 0.0450. What is Kp for the reaction 4 B (g) ⇌ 2 A (g)?
Solution:
The neutralization of a strong acid and strong base has an enthalpy change, ∆H°, = -55.9 kJ/mol. The net ionic equation for this reaction is H⁺(aq) + OH⁻(aq) ⇌ H₂O(l) which is the reverse of the autoionization of water, and therefore ∆H° for the autoionization of water = +55.9 kJ/mol. Kw for the autoionization of water is 1.0 × 10⁻¹⁴ at 25°C. Distilled water has a [H⁺] = 1.0 × 10⁻⁷ M and a neutral pH of 7.0. Calculate the neutral pH of distilled water at 65°C.
Solution:
What is the osmotic pressure of a 0.201 M solution of MgCl₂ at 37.0 °C? (assume complete dissociation).
What is the osmotic pressure of a 0.201 M solution of MgCl₂ at 37.0 °C? (assume complete dissociation). Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. . Osmotic pressure, ΠΠ, of a solution is calculated from the molar concentration, M, by the equationΠ=MRTΠ=MRTwhere R is the gas constant (0.08206 L⋅atmmol⋅KL⋅atmmol⋅K) and T is the absolute temperature. Assuming complete dissociation, a 0.201 M solution of MgCl₂ would have a particle molarity of 0.201 ××3 = 0.603 M. A temperature of 37.0 °C would be 273 + 37.0 = 310 K. Substituting these values in the equation, we can calculate ΠΠ as follows: Π=0.603molL×0.08206L⋅atmmol⋅K×310 K=15.3 atm