CHEM 36

General Chemistry
EXAM #2
March 13, 2002

Name: Key

INSTRUCTIONS: Read through the entire Page Possible Points

exam before you begin. Answer all of the Points Earned
questions. For questions involving calculations, 2 25 25
show all of your work -- HOW you arrived at
a particular answer is MORE important than 3 30 30
the answer itself! Circle your final answer to 4 25 25
numerical questions. 5 45 45
6 25 25
The entire exam is worth a total of 150 points.
Attached are a periodic table and a formula TOTAL: 150 150
sheet jam-packed with useful stuff. Good
Luck!
1. For the dissociation of acetic acid in water:

CH 3COOH (aq) D H+ (aq) + CH 3COO- (aq)

at 25.0oC, Ka = 1.76 x 10-5.

a. [10 pts] Calculate ∆Go (kJ/mol) for this reaction.

∆ Go = -RTLnK
= -(8.3145 J/mol-K)(298.15 K)Ln(1.76 x 10-5)
= 2.7139 x 104 J/mol x 10-3 kJ/J = 27.139 kJ/mol

b. [5 pts] At equilibrium, what is the value of ∆G?

For any system at equilibrium: ∆ G = 0

2. [10 pts] Calculate the molarity of an aqueous acetic acid solution that is 36.0%
acetic acid (by mass). The density of the solution is 1.045 g/mL and the molecular
weight of acetic acid is 60.05 g/mol.

36.0 g HAc x 1 mol HAc x 1.045 g soln x 1000 mL = 6.2648 mol HAc/L
100.0 g soln 60.05 g HAc mL soln L

= 6.26 M HAc

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3. [15 pts] Given equilibrium constants for reactions 1 and 2, compute K3:

(1) H 2O (l) + H 2O(l) D H3O+ (aq) + OH - (aq) K1 = 1.0 x 10-14

(2) NH 4+ (aq) + H 2O (l) D NH3 (aq) + H 3O+ (aq) K2 = 5.7 x 10-10

(3) NH 3 (aq) + H 2O (l) D NH4+ (aq) + OH - (aq) K3 = ???

Reaction (1) – Reaction (2) = Reaction (3)

H2O (l) + H2O(l) D H3O+ (aq) + OH - (aq) K1

NH3 (aq) + H3O+ (aq) D NH4+ (aq) + H2O (l) 1/K2
NH3 (aq) + H2O (l) D NH4+ (aq) + OH - (aq) K3

So: K3 = K1 (1/K2) = 1.0 x 10-14/5.7 x 10-10 = 1.754 x 10-5 = 1.8 x 10-5

4. Chloroacetic acid dissociates in aqueous solution to produce the chloroacetate and

hydronium ions:

CH 2ClCOOH (aq) + H 2O (l) D CH 2ClCOO- (aq) + H 3O+ (aq)

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a. [10 pts] Compute K at 25.0 C. given the following set of equilibrium
concentrations:

[CH 2ClCOOH] = 0.0888 M

[CH 2ClCOO-] = 0.0112 M
[H3O+] = 0.0112 M

K = [CH2ClCOO-][H3O+] = (0.0112)(0.0112) = 1.4126 x 10-3 = 1.41 x 10-3

[CH2ClCOOH] (0.0888)

b. [5 pts] Calculate the pH of this solution.

pH = -Log[H3O+] = -Log(0.112) = 1.95078 = 1.95

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5. Consider the following endothermic (∆Ho = 41.18 kJ/mol) reaction:

CO2 (g) + H 2(g) D CO (g) + H 2O (l)

At 25.0 oC, K = 3.2 x 10-4.

If we start with 1.0 atm of each of the gas phase species in a closed vessel containing
some water:

a. [10 pts] Is the system at equilibrium? If not, will the reaction proceed to
the left or to the right in order to reach equilibrium?

Q = PCO = (1.0) = 1.0 > > K

PCO2 PH2 (1.0)(1.0)

If Q > K, then there is an excess of CO and reaction will shift to the:

Left Right No Change (circle one)

b. [15 pts] Calculate the partial pressure of H2 (g) for the system at
equilibrium.

CO2 (g) + H2(g) D CO (g) + H2O (l)

Initial 1.00 1.00 1.00
Change +x +x -x
Equilibrium 1.00+x 1.00+x 1.00-x

3.2 x 10-4 = (1.00 –x)

(1.00+x)(1.00+x)

A quadratic: 3.2 x 10-4 x2 + 1.00064 x – 9.9968 x 10-1 = 0

x = 0.998714

PH2 = 1.00 + x = 1.998714 = 2.0 atm

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 c. [5 pts each] How will the equilibrium partial pressure of H 2 (g) be affected
by the following:
i. CO2 (g) is added to the system (at constant volume).

Increase Decrease No Change (circle one)

ii. The volume of the reaction vessel is halved.

Increase Decrease No Change (circle one)

iii. Argon gas is added to the system (at constant volume).

Increase Decrease No Change (circle one)

iv. The temperature of the system is increased.

Increase Decrease No Change (circle one)

6a. [10 pts] Arrange the following acids in order of increasing strength (briefly
explain your reasoning):
CH 3CH 2OH, HClO4, CH 3COOH, H 2O

CH3CH2OH < H2O < CH3COOH < HClO4

strong acid (most # O)

only one EN O OH- can stablize two EN O

to accommodate charge with one O to accommodate
negative charge and a hydrogen negative charge

b. [10 pts] Identify the corresponding conjugate bases for these acids:

Acid Conjugate Base

CH 3CH 2OH CH3CH2O-

HClO4 ClO4-

CH 3COOH CH3COO-

H3O+ H2O

c. [5 pts] Arrange the conjugate bases in order of increasing strength.

ClO4- < H2O < CH3COO- < CH3CH2O-

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7. [5 pts each] Given the Ka values for the following acids:

CH 2ClCOOH (Chloroacetic Acid): Ka = 1.4 x 10-3

C6H5COOH (Benzoic Acid) Ka = 6.5 x 10-5
HCN (Hydrocyanic Acid) Ka = 6.2 x 10-10

a. Which acid is the weakest?

HCN has the smallest Ka, so it is the weakest acid

b. Write the acid dissociation equilibrium reaction for the acid you identified in
part a.

HCN (aq) + H2O (l) D CN - (aq) + H3O+ (aq)

c. Which of the conjugate bases for these acids is the weakest?

The conjugate base of the strongest acid will be the weakest:

CH2ClCOOH is the strongest acid,

so CH2ClCOO- is the weakest conjugate base

d. Calculate the value of Kb for the base you identified in part c.

For a conjugate acid/base pair: KaKb = Kw

So: Kb = Kw/Ka = 1.0 x 10-14/1.4 x 10-3

= 7.14 x 10-12
= 7.1 x 10-12

e. Write the base dissociation (Kb) equilibrium reaction for the base you
identified in part c.

CH2ClCOO- (aq) + H2O (l) D CH2ClCOOH (aq) + OH - (aq)

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Extra Credit! -- 10 pts

The triple bond in the N2 molecule is very strong, but at high enough temperatures
even it breaks down. At 5000 K, when the total pressure exerted by a sample of
nitrogen is 1.00 atm, N2 (g) is 0.65% dissociated at equilibrium:

N2 (g) D 2 N (g)

At 6000 K, with the same total pressure, the proportion of N2 (g) dissociated at
equilibrium rises to 11.6%. Calculate the ∆H of this reaction.

We need K at 5000 K (T1) and at 6000 K (T2):

At 5000 K: N2 D 2N
I 1.00 -
C -0.0065 +2(0.0065)
E 0.9935 0.0130

So: K1 = (PN)2 = (0.0130)2 = 1.70 x 10-4

PN2 0.9935

At 6000 K: N2 D 2N
I 1.00 -
C -0.116 +2(0.116)
E 0.884 0.232

So: K2 = (PN)2 = (0.232)2 = 6.09 x 10-2

PN2 0.884

Now, plug into Van’t Hoff equation and solve for ∆ H:

∆ Ho/R)[(1/T 2) – (1/T 1)]

Ln(K2/K1) = -(∆

∆ Ho (-4.0091 x 10-6)
5.8812 = -∆

∆ Ho = 1.467 x 106 J → 1.5 x 103 kJ