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This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 6.0 points Consider the reaction below:
2NH 3 (g) ←→ 3H 2 (g) + N 2 (g)
If K is 10−^80 which of the following is a good estimate of equilibrium concentrations of NH 3 , H 2 and N 2 , respectively?
- 5 M, 7.5 M and 2.5 M, respectively
- 0 M, 15 M and 5 M, respectively
- 5 M, 0 M and 0 M, respectively correct
- 1 M, 9 M and 4.5 M, respectively
Explanation: The magnitude of K is sufficiently small that we can infer that none of the reactants will convert into products.
002 6.0 points
Rank the following liquids in increasing order of solubility in water (H 2 O): CH 3 F, (CH 3 ) 2 NH, CH 3 Cl, CBr 4.
- CH 3 Cl < CH 3 F < (CH 3 ) 2 NH < CBr 4
- CH 3 F < (CH 3 ) 2 NH < CBr 4 < CH 3 Cl
- CBr 4 < CH 3 Cl < CH 3 F < (CH 3 ) 2 NH correct
- (CH 3 ) 2 NH < CBr 4 < CH 3 Cl < CH 3 F
Explanation: According the axiom that ”like dissolves like,” the least miscible will be the non-polar molecule CBr 4. The most miscible will be (CH 3 ) 2 NH, which exhibits hydrogen bonding just as water does. Between the two polar molecules, CH 3 F is the more polar and thus the more miscible because of the greater elec- tronegativity difference.
003 6.0 points
The combustion of octane has a standard change in enthalpy, ∆H = − 5 , 500 kJ · mol−^1. If the reaction were per- formed at T 1 = 200 K and T 2 = 300 K, which pair of equilibrium constants might be cor- rect?
- K 1 = 10^300 , K 2 = 1 correct
- K 1 = 100, K 2 = 10
- K 1 = 1, K 2 = 10^300
- K 1 = 10, K 2 = 100
Explanation: Since the reaction is exothermic, the value of K should be inversely proportional to T. So the higher temperature must have the smaller value of K. This eliminates two of the an- swer choices. Since ∆H = − 5 , 500 kJ · mol−^1 is a very large number, it also means that dramatic changes in K are expected in re- sponse to changes in T , eliminating two an- swer choices, and leaving only one that satis- fies both criteria.
004 6.0 points Calculate the vapor pressure at 25◦C of a mixture of benzene and toluene in which the mole fraction of benzene is 0.650. The vapor pressure at 25◦C of benzene is 94.6 torr and that of toluene is 29.1 torr.
- 61.5 torr
- 51.3 torr
- 71.7 torr correct
- 124 torr
- 84.4 torr
Explanation:
005 6.0 points What would be the pOH of a solution of
Sr(OH) 2 (strontium hydroxide) prepared by dissolving 122 g of the base into 10 L of pure water (H 2 O)?
- 13
- 1
- 0.7 correct
Explanation:
122 g Sr(OH) 2 ×
1 mol 122 g
= 1 mol Sr(OH) 2
1 mol Sr(OH) 2 10 L H 2 O
= 0.1 M Sr(OH) 2
Sr(OH) 2 (aq) ↔ Sr2+(aq) + 2 OH−(aq)
[OH−] = Cb · 2 = 0.1M · 2 = 0.2M
pOH = − log[OH−] = − log(0.2) = 0. 7
006 6.0 points
Write the equilibrium expression for the fol- lowing reaction:
2 N O(aq) + 4 H 2 O(ℓ) ←→ 3 H 2 (aq) + 2 H+(aq) + 2 N O 3 − (aq)
1. K =
[N O]^2
[H+]^2 · [N O− 3 ]^2
2. K =
[N O]^2
[H 2 ]^3 · [H+]^2 · [N O− 3 ]^2
3. K =
[H 2 ]^3 · [H+]^2 · [N O− 3 ]^2
[N O]^2 · [H 2 O]^4
4. K =
[H 2 ]^3 · [H+]^2 · [N O− 3 ]^2
[N O]^2
correct
5. K =
[[H+]^2 · [N O− 3 ]^2
[N O]^2
Explanation: Set up K, products in the numerator, reac- tants in the denominator, all raised to respec- tive stoichiometric coefficients.
007 6.0 points
Which of the following statements is/are true concerning Kw and temperature (T)? I) Kw gets larger as T increases
II) Kw = 10−^12 for water at T = 273 K
III)
Kw [OH−]
= [H+] for water at any T
- I only
- I, II
- II, III
- III only
- II only
- I, III correct
- I, II, III
Explanation: All equilibrium processes are temperature dependent, and because auto-protolysis is en- dothermic, Kw increases as temperature in- crease. Thus statement I is true, but state- ment II is the opposite of the actual case. Statement III is the definition of Kw.
008 6.0 points What would be the osmotic pressure exerted on a semi-permeable membrane separating two chambers, one with a NO− 3 concentration of 1.4 M and the other with a Ca2+^ concen- tration of 0.4 M. (Note: at room temperature the product of the gas constant and tempera- ture is roughly equal to 25 L · atm · mol−^1 .)
- 10 atm
- 0 .01 atm
- 3 .5 atm
- 0 .035 atm
011 6.0 points An animal cell assumes its normal volume when it is placed in a solution with a total solute molarity of 0.3 M. If the cell is placed in a solution with a total solute molarity of 0.1 M,
- the escaping tendency of water in the cell increases.
- no movement of water takes place.
- water leaves the cell, causing contrac- tion.
- water enters the cell, causing expansion. correct
Explanation:
012 6.0 points What would be pH of a 0.001 M solution of malonic acid if malonic acid has a Ka = 1 × 10 −^5?
- 10
- 7
- 8
- 4 correct
- 14
Explanation: The equation
[H+] = (Ka · Ca)^0.^5
[H+] = (1 × 10 −^5 · 0 .001 M)^0.^5 = 0.0001 M
pH = 4
013 6.0 points Suppose we put 1.0 mol of HI(g), 1.0 mol of H 2 (g), and 1 mol of I 2 (g) in a 2.0 liter reaction vessel and the following equilibrium is established:
2 HI(g) ⇀↽ H 2 (g) + I 2 (g)
If Kc = 10 for this reaction at the tempera- ture of the equilibrium mixture, compute the
equilibrium concentration of HI.
- 0.429 M
- 0.240 M
- 0.071 M
- 0.260 M
- 0.102 M
- 0.295 M
- 0.145 M
- 0.205 M correct Explanation: [HI]ini =
1 mol 2 L
= 0.5 M
[H 2 ]ini =
1 mol 2 L
= 0.5 M
[I 2 ]ini =
1 mol 2 L
= 0.5 M
Q =
[H 2 ] [I 2 ]
[HI]^2
(0.5)^2
= 1 < Kc = 10
Therefore equilibrium moves to the right. 2 HI ⇀↽ H 2 (g) + I 2 (g) ini, M 0. 5 0. 5 0. 5 ∆, M − 2 x x x eq, M 0. 5 − 2 x 0 .5 + x 0 .5 + x
(0.50 + x) (0.50 + x) (0. 50 − 2 x)^2
0 .5 + x
- 50 − 2 x
0 .5 + x = 0. 50
10 x x = 0. 148
[HI] = 0. 5 − 2 x = 0.205 M
014 6.0 points
How many grams of water at 80.0◦C must be added to 10.0 grams of ice at − 20 ◦C to result in ONLY liquid water at 0◦C?
- 25.0 g
- 74.0 g
- 109 g
- 11.2 g correct
- 1.25 g
- 21.2 g
- 68.8 g
- 56.6 g
- 22.4 g
Explanation:
015 6.0 points Which combination of ∆G◦^ and K is possible at standard conditions?
- ∆G◦^ = − 76 .8 kJ, K = 0. 961
- ∆G◦^ = 53 kJ, K = 1. 04
- ∆G◦^ = 99.9 J, K = 2. 2 × 1013
- ∆G◦^ = − 66 .2 J, K = 1. 03 correct
- ∆G◦^ = −105 J, K = 6. 21 × 10 −^10
Explanation: If ∆G◦^ is positive, K must be less than 1; if ∆G◦^ is negative, K must be greater than 1. Also, if ∆G◦^ is small (compared to RT), then K will be close to 1, and if ∆G◦^ is large, K will be much smaller or greater than 1.
016 6.0 points In the derivation of the Clausius-Clapeyron equation, the ∆S vap◦ term is present in early steps but isn’t present in the final equation. Why is this?
- We assume that ∆Svap = ∆S vap◦ , and we are able to eliminate the term altogether.
- We assume ideal behavior for the gas, and the ∆S vap◦ term reduces to a constant and is eliminated. correct
- The ∆S vap◦ term actually is present in the final equation.
- We assume that ∆S vap◦ is independent of temperature and are able to ignore it.
Explanation: Because we assume that the gas behaves ideally, ∆S◦ vap is the same regardless of the gas in question, and taking its base e expo- nent results in constant which can be readily eliminated.
017 6.0 points
Which of the following statements is/are true concerning auto-protolysis of water? I) For pure water at equilibrium, [OH−] al- ways equals [H+]. II) The equilibrium value of [H+] can be zero at room temperature. III) Auto-protolysis is endothermic.
- I, II
- III only
- II, III
- II only
- I, III correct
- I, II, III
- I only
Explanation: [OH−] equals [H+] only when water un- dergoes auto-protolysis in the absence of any other acidic or basic species. [H+] can ap- proach zero, but it can never be zero - state-
I) decreasing the pressure of N 2 (g) II) increasing the temperature of the sample III) changing to a more polar solvent
- III only
- I and III
- I and II
- II only
- I only
- II and III
- none of these correct
Explanation: The solubility of a gas is directly propor- tional to the gas’ pressure, inversely propor- tional to the temperature of the solvent. Con- sidering the principle of like dissolve like, a non-polar species such as N 2 (g) will be more soluble in non-polar solvents. The volume of the solvent does not impact the solubility, which is an intensive property; this is evident when one considers the unity of solubility such as grams per liter, etc.
022 6.0 points
Consider the combustion of glucose:
C 6 H 12 O 6 (s)+6 O 2 (g) ←→ 6 CO 2 (g)+6 H 2 O(g)
∆H for this reaction is −3 kJ/mol. Which of the following conditions would result in the production of additional water vapor? I) Adding more carbon dioxide. II) Performing the reaction in a hotter envi- ronment. III) Decreasing the pressure of the system. IV) Adding a reagent that irreversibly binds glucose and prevent it from further reac- tions.
- III only correct
- I, III
- II only
- III, IV
- I only
- I, II, III
Explanation: Adding more carbon dioxide would shift the reaction to the left, reducing the amount of water vapor produced. Likewise, since the reaction is already exothermic (negative ∆H), doing the reaction is a hotter envi- ronment would also shirt the reaction to the left. Adding a reagent that sequesters glu- cose would reduce glucose concentration, thus shifting the reaction to the left. The only cor- rect choice is decreasing the pressure of the system. Since there are six moles of gas on the left and 12 on the right, decreasing pres- sure would shirt the reaction to the right.
023 6.0 points What is the pH of a 4 M solution of potassium generate? Ka for the generic acid (RCOOH) is 5 × 10 −^7.
- 10.45 correct
Explanation: MRCOO− = 4 M Ka = 5 × 10 −^7 It’s a salt of a weak generic acid. Get it? Generic acid makes generate ions. Ha! This means you need a Kb for the weak base A−. Use Kb =
Kw Ka
and you’ll get the Kb =
2 × 10 −^8. You CAN use the approximation for the equilibrium which means that
[OH−] =
Kb · CA− = 0.000282843 M
pH = 14 − pOH = 14 + log(0.000282843) = 10. 4515
024 6.0 points
Consider the reaction,
A(aq) + B(aq) ←→ C(aq)
The equilibrium constant, K, is 2. If the concentrations of A, B and C are 2 M, 2 M and 10 M, respectively, which of the following would occur?
- the reaction would move right
- nothing would occur
- not enough information
- the reaction would move left correct
Explanation:
Q =
[C]
[A] · [B]
Since Q > K, the reaction would move left.
025 6.0 points Consider two closed containers. Container X is a 2 L container that contains 0.5 L of acetone. Container Y is a 3 L container that contains 1.8 L of acetone. Both containers and contents are at 28◦C. Which of the following is true?
- The vapor pressure in container Y is greater.
- The vapor pressure in container X is greater.
- You would need information about the shape of the containers to be able to answer this question.
- The vapor pressures in both containers are equal. correct Explanation:
026 6.0 points Rank the following salts from least to greatest molar solubility: A) KIO 4 Ksp = 0. 000371 B) CdS Ksp = 3. 6 × 10 −^29 C) CuS Ksp = 8. 5 × 10 −^45 D) CaCr 2 O 4 Ksp = 0. 023
- B < C < A < D
- C < B < A < D correct
- B < C < D < A
- C < D < A < B Explanation: All of the salts are composed of a single cation and anion, so the solubility of each is the square root of its Ksp.
027 6.0 points For a solution with an [OH−] of 10−^5 M, what would be the value of [H+] , pH and pOH, respectively?
- 10 −^9 M, 9, 14
- 109 M, 9, 5
- 10 −^14 M, 9, 5
- 10 −^9 M, 5, 9
- 10 −^9 M, 9, 5 correct
- 10 −^14 M, 5, 9 Explanation:
[H+] =
10 −^14
[OH−]
10 −^14
10 −^5
= 10−^9
pH = −log 10 ([H+]) pOH = −log 10 ([OH−]) = 14 − pH
