2004 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)
CHEMISTRY
Section II
(Total time—90 minutes)
Part A
Time—40 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A.
CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS.
It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if
you do not. Attention should be paid to significant figures.
Be sure to write all your answers to the questions on the lined pages following each question in the booklet with the
goldenrod cover. Do NOT write your answers on the lavender insert.
Answer Question 1 below. The Section II score weighting for this question is 20 percent.
→ 2 NH (g)
N2(g) + 3 H2(g) ←
3
1. For the reaction represented above, the value of the equilibrium constant, Kp , is 3.1 × 10 – 4 at 700. K.
(a) Write the expression for the equilibrium constant, Kp , for the reaction.
(b) Assume that the initial partial pressures of the gases are as follows:
pN = 0.411 atm, pH = 0.903 atm, and pNH = 0.224 atm.
2
2
3
(i) Calculate the value of the reaction quotient, Q, at these initial conditions.
(ii) Predict the direction in which the reaction will proceed at 700. K if the initial partial pressures are
those given above. Justify your answer.
(c) Calculate the value of the equilibrium constant, Kc , given that the value of Kp for the reaction at 700. K is
3.1 × 10 – 4.
(d) The value of Kp for the reaction represented below is 8.3 × 10 –3 at 700. K.
→ NH HS(g)
NH3(g) + H2S(g) ←
4
Calculate the value of Kp at 700. K for each of the reactions represented below.
→ NH (g) + H S(g)
(i) NH4HS(g) ←
3
2
→ 2 NH HS(g)
(ii) 2 H2S(g) + N2(g) + 3 H2(g) ←
4
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AP® CHEMISTRY
2004 SCORING GUIDELINES (Form B)
Question 1
→ 2 NH (g)
N2(g) + 3 H2(g) ←
3
1. For the reaction represented above, the value of the equilibrium constant, Kp , is 3.1 × 10 – 4 at 700. K.
(a) Write the expression for the equilibrium constant, Kp , for the reaction.
Kp =
p 2 NH3
1 point for pressure expression
1 point for correct substitution
pN 2 × p 3H 2
(b) Assume that the initial partial pressures of the gases are as follows:
pN = 0.411 atm, pH = 0.903 atm, and pNH = 0.224 atm.
2
2
3
(i) Calculate the value of the reaction quotient, Q, at these initial conditions.
Q =
p 2 NH3
pN 2 × p 3H 2
=
(0.224) 2
(0.411)(0.903)3
1 point for calculation of Q with correct mass action expression
Note: must be consistent with part (a)
Q = 0.166
(ii) Predict the direction in which the reaction will proceed at 700. K if the initial partial pressures are
those given above. Justify your answer.
Since Q > Kp , the numerator must decrease and the
denominator must increase, so the reaction must
proceed from right to left to establish equilibrium.
1 point for direction or for stating that Q > Kp
1 point for explanation
(c) Calculate the value of the equilibrium constant, Kc , given that the value of Kp for the reaction 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
1 point for calculating ∆ n
L atm
× 700 K)−2
mol K
3.1 × 10−4 = Kc(57.5)−2
1 point for correct substitution and value of Kc
3.1 × 10−4 = Kc(3.0 × 10−4)
1.0 = Kc
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AP® CHEMISTRY
2004 SCORING GUIDELINES (Form B)
Question 1 (cont’d.)
(d) The value of Kp for the reaction represented below is 8.3 × 10 –3 at 700. K.
→ NH HS(g)
NH3(g) + H2S(g) ←
4
Calculate the value of Kp at 700. K for each of the reactions represented below.
→ NH (g) + H S(g)
(i) NH4HS(g) ←
3
2
Kp =
1
8.3 × 10
−3
= 1.2 × 102
1 point for the calculation of Kp
→ 2 NH HS(g)
(ii) 2 H2S(g) + N2(g) + 3 H2(g) ←
4
→ NH HS(g)]
2 × [NH3(g) + H2S(g) ←
4
→ 2 NH (g)
N (g) + 3 H (g) ←
2
2
3
Kp = (8.3 × 10−3)2
Kp = 3.1 × 10− 4
→ 2 NH HS(g)
2 H2S(g) + N2(g) + 3 H2(g) ←
4
Kp = (8.3 × 10−3)2 (3.1 × 10−4) = 2.1 × 10− 8
1 point for squaring Kp for NH4HS or for
multiplying Kp’s
1 point for correct Kp
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AP® CHEMISTRY
2004 SCORING COMMENTARY (Form B)
Question 1
Sample: 1A
Score: 10
This response earns a perfect score of 10 points: 2 points for part (a), 1 point for part (b)(i), 2 points for part
(b)(ii), 2 points for part (c), 1 point for part (d)(i), and 2 points for part (d)(ii).
Sample: 1B
Score: 8
Only 1 out of 2 points is earned in part (c): 1 point is earned for correctly calculating n, but the wrong R is
used. Only 1 out of 2 points is earned in part (d)(ii) because one of the Kp values is multiplied by two
when it should be squared.
Sample: 1C
Score: 6
Only 1 out of 2 points is earned in part (a) because molar concentrations are used in the Kp expression, as
indicated by the brackets. No points are earned in part (b)(ii) because the prediction and explanation are
incorrect. Only 1 out of 2 points is earned in part (c): 1 point is earned for correctly calculating n, but there
is a math error in the calculation of Kc .
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