test 1
Version 017 – Exam 1 – Laude – (52455)
This print-out should have 30 questions.
Multiple-choice questions may continue on the next column or page – find all choices before answering.
001
1
003 6.0 points
What would be pH of a 1 M solution of a very expensive weak base, unobtainamine, whose pKb is 3.4?
6.0 points
1. 12.3 correct
Consider a reaction with ∆Hrxn =
997 kJ · mol−1 . Which of the following pairs of K values and temperatures is possible for this reaction?
1. K1 = 15.7, T1 = 138 K;
K2 = 1.57 × 1038 , T2 = 153 K correct
2. K1 = 11.5, T1 = 128 K;
K2 = 3.83, T2 = 145 K
3. K1 = 3.1, T1 = 103 K;
K2 = 3.1, T2 = 340 K
2. 14
3. 1.7
4. 7
5. 0
Explanation:
The equation
[OH− ] = (Kb · Cb )0.5 can be algebraically rearranged to pOH = 0.5(pKb + pCb )
4. K1 = 9.76, T1 = 102 K;
K2 = 3.74 × 10−61 , T2 = 116 K
Explanation:
For and endothermic reaction, K is directly related to T. For an exothermic reaction, K is inversely related to T. K is constant with temperature only for ∆Hrxn = 0 kJ.
002 6.0 points
Which combination of ∆G and K is possible?
1. ∆G = −29.4 J, K = 1.01 correct
2. ∆G = −77.6 kJ, K = 0.981
3. ∆G = 47.3 J, K = 3.02 × 1013
4. ∆G = 54.7 kJ, K = 1.03
5. ∆G = −82.5 J, K = 3.15 × 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.
= 0.5(3.4 + 0) = 1.7 pH = 14 − pOH = 12.3
004 6.0 points
Consider the chemical equation below:
N2 O4 (g) + 4H2 (g) ←→ 4H2 O(g) + N2 (g)
What would Keq be for this reaction?
1. Keq =
2. Keq
[H2 O]4 [N2 ] correct [N2 O4 ][H2 ]4
[N2 O4 ][H2 ]4
=
[H2 O]4
3. Keq =
[H2 O]4
[N2 O4 ][H2 ]4
4. Keq =
[N2 O4 ][H2 ]4
[H2 O]4 [N2 ]
Explanation:
For a reaction of the form: aA(s) + bB(aq) ↔ cC(ℓ) + dD(g)
The equilibrium constant will take the form: [D]d
Keq =
[B]b
This print-out should have 30 questions.
Multiple-choice questions may continue on the next column or page – find all choices before answering.
001
1
003 6.0 points
What would be pH of a 1 M solution of a very expensive weak base, unobtainamine, whose pKb is 3.4?
6.0 points
1. 12.3 correct
Consider a reaction with ∆Hrxn =
997 kJ · mol−1 . Which of the following pairs of K values and temperatures is possible for this reaction?
1. K1 = 15.7, T1 = 138 K;
K2 = 1.57 × 1038 , T2 = 153 K correct
2. K1 = 11.5, T1 = 128 K;
K2 = 3.83, T2 = 145 K
3. K1 = 3.1, T1 = 103 K;
K2 = 3.1, T2 = 340 K
2. 14
3. 1.7
4. 7
5. 0
Explanation:
The equation
[OH− ] = (Kb · Cb )0.5 can be algebraically rearranged to pOH = 0.5(pKb + pCb )
4. K1 = 9.76, T1 = 102 K;
K2 = 3.74 × 10−61 , T2 = 116 K
Explanation:
For and endothermic reaction, K is directly related to T. For an exothermic reaction, K is inversely related to T. K is constant with temperature only for ∆Hrxn = 0 kJ.
002 6.0 points
Which combination of ∆G and K is possible?
1. ∆G = −29.4 J, K = 1.01 correct
2. ∆G = −77.6 kJ, K = 0.981
3. ∆G = 47.3 J, K = 3.02 × 1013
4. ∆G = 54.7 kJ, K = 1.03
5. ∆G = −82.5 J, K = 3.15 × 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.
= 0.5(3.4 + 0) = 1.7 pH = 14 − pOH = 12.3
004 6.0 points
Consider the chemical equation below:
N2 O4 (g) + 4H2 (g) ←→ 4H2 O(g) + N2 (g)
What would Keq be for this reaction?
1. Keq =
2. Keq
[H2 O]4 [N2 ] correct [N2 O4 ][H2 ]4
[N2 O4 ][H2 ]4
=
[H2 O]4
3. Keq =
[H2 O]4
[N2 O4 ][H2 ]4
4. Keq =
[N2 O4 ][H2 ]4
[H2 O]4 [N2 ]
Explanation:
For a reaction of the form: aA(s) + bB(aq) ↔ cC(ℓ) + dD(g)
The equilibrium constant will take the form: [D]d
Keq =
[B]b