Free Response
N2(g) + 3 H2 (g)  2 NH3 (g)
1) For the reaction represented abov, the value of the equilibrium constant, Kp, is
3.1 * 10^-4 at 700K.
(a) Write the expression for the equilibrium constant, Kp, for the rxn.
(b) Assume that the initial partial pressures of the gases are as follows:
P(N2) = 0.411 atm, P(H2) = 0.903 atm, and P(NH3) = 0.224 atm.
i. Calculate the value of the rxn quotient, Q, at there initial
conditions.***Ch13***
ii. Predict the direction in which the rxn will proceed at 700 K if the
initial partial pressures are those given above. Justify your ans.
(c) Calculate the value of the eq. Constant, Kc, given that the value of the Kp
for the rxn at 700 K is 3.1 * 10^-4.
(d) The value of Kp for the rxn represented below is 8.3 * 10 ^ -3 at 700 K.
NH3 (g) + H2S (g)  NH4HS (g)
Calculate the value of Kp at 700 K for each of the rxns represented below.
(i) NH4HS (g)  NH3 (g) + H2S (g)
(ii) 2 H2S (g) + N2 (g) + 3 H2 (g)  2 NH4HS (g)
1. According to the phase diagram shown, where does a mixture of solid and liquid exist
at equilibrium?
(A) along line MN (B) along line KN
(C) along line LN (D) in the region KNL
2. A homogeneous liquid reaction mixture is often heated to
increase the rate of reaction. This is best explained by the
fact that raising the temperature
(A) increases the heat of reaction.
(B) decreases the energy of activation.
(C) increases the vapor pressure of the liquid
(D) increases the average kinetic energy of the reactants.
3. What is the [H+] in a 0.10 M
solution of ascorbic acid,
C6H8O6?
-5
-3 M
(C) 4.0 x 10-3 M
4. A 0.10 M solution of which salt is the most acidic?
(A) NH4C2H3O2 (B) NaCN
(C) KNO3 (D) AlCl3
5. A student is asked to prepare a buffer solution with a pH of 4.00. This can be
accomplished by using a solution containing which of the following?
Ka = HNO2 4.5 × 10-4
Ka = HCN 4.9 × 10-10
(A) HNO2 only (B) HCN only
(C) HNO2 and NaNO2 (D) HCN and NaCN
1. According to the phase
diagram shown, where
does a mixture of solid
and liquid exist at
equilibrium?
(A) along line MN (B) along line KN
(C) along line LN (D) in the region KNL
2. A homogeneous liquid reaction mixture is often heated to
increase the rate of reaction. This is best explained by the
fact that raising the temperature
(A) increases the heat of reaction.
(B) decreases the energy of activation.
(C) increases the vapor pressure of the liquid
(D) increases the average kinetic energy of the reactants.
3. What is the [H+] in a 0.10 M
solution of ascorbic acid,
C6H8O6?
-5
-6 M
-3 M
(C) 4.0 x 10-3 M
4. A 0.10 M solution of which salt is the most acidic?
(A) NH4C2H3O2 (B) NaCN
(C) KNO3 (D) AlCl3
5. A student is asked to prepare a buffer solution with a pH of 4.00. This can be
accomplished by using a solution containing which of the following?
Ka = HNO2 4.5 × 10-4
Ka = HCN 4.9 × 10-10
(A) HNO2 only (B) HCN only
(C) HNO2 and NaNO2 (D) HCN and NaCN
Free Response- ANSWERS
N2(g) + 3 H2 (g)  2 NH3 (g)
1. For the reaction represented abov, the value of the equilibrium constant, Kp, is
3.1 * 10^-4 at 700K.
a. Write the expression for the equilibrium constant, Kp, for the rxn.
Kp = (NH3)p2 / (N2)p x (H2)p3
2pts.
b. Assume that the initial partial pressures of the gases are as follows:
P(N2) = 0.411 atm, P(H2) = 0.903 atm, and P(NH3) = 0.224 atm.
i. Calculate the value of the rxn quotient, Q, at there initial
conditions.***Ch13***
Q = (NH3)p2 / (N2)p x (H2)p3
= (0.224)2 / (0.411)(0.903)3
Q = 0.166
1 pt.
ii. Predict the direction in which the rxn will proceed at 700 K if the
initial partial pressures are those given above. Justify your ans.
Since Q>Kp, the numerator must decrease and the denominator must increase, so the rxn
must proceed from right to left to establish equilibrium.
2pt.
c. Calculate the value of the eq. Constant, Kc, given that the value of the Kp
for the rxn at 700 K is 3.1 * 10^-4.
Kp = Kc(RT). n
.n = 2 – 4 = -2
Kp = Kc(RT)-2
3.1 × 10-4 = Kc(0.0821 L atm mol K × 700 K)-2
3.1 104 = Kc(57.5)2
3.1 104 = Kc(3.0 104)
1.0 = Kc
2pt.
d. The value of Kp for the rxn represented below is 8.3 * 10 ^ -3 at 700 K.
NH3 (g) + H2S (g)  NH4HS (g)
Calculate the value of Kp at 700 K for each of the rxns represented below.
(j) NH4HS (g)  NH3 (g) + H2S (g)
Kp = 1/ 8.3 x 10 ^ -3
= 1.2 10 ^ 2
1 pt.
(ii) 2 H2S (g) + N2 (g) + 3 H2 (g)  2 NH4HS (g)
2 [NH3(g) + H2S(g) NH4HS(g)] Kp = (8.3 103)2
N2(g) + 3 H2(g) 2 NH3(g) Kp = 3.1 104
2 H2S(g) + N2(g) + 3 H2(g) 2 NH4HS(g)
Kp = (8.3 103)2 (3.1 104) = 2.1 108
2pt.