Question at position 6 Consider the following reaction: 2NOBr(g) ⇌ 2NO(g) + Br2(g) A 1.0 L vessel was filled with pure NOBr, at a pressure of 4.0 atm, at 300 K. After equilibrium was established, the partial pressure of NOBr was 2.5 atm. Calculate Kp for the reaction. 0.270.121.250.450.62单项选择题
A
0.27
B
0.12
C
1.25
D
0.45
E
0.62

Question

Question at position 6 Consider the following reaction: 2NOBr(g) ⇌ 2NO(g) + Br2(g) A 1.0 L vessel was filled with pure NOBr, at a pressure of 4.0 atm, at 300 K. After equilibrium was established, the partial pressure of NOBr was 2.5 atm. Calculate Kp for the reaction. 0.270.121.250.450.62单项选择题

A

0.27

B

0.12

C

1.25

D

0.45

E

0.62

BlackTom AI · Accepted Answer

For the reaction 2NOBr(g) ⇌ 2NO(g) + Br2(g), the equilibrium constant in terms of pressure is Kp = (P_NO)^2 (P_Br2) / (P_NOBr)^2. Relevant concepts: use partial pressures and stoichiometry to build an ICE (Initial-Change-Equilibrium) table, relate the change in NOBr pressure to the amounts of NO and Br2 formed, then substitute equilibrium pressures into the Kp expression.

Step-by-step reasoning:
1. Initial pressures: P_NOBr = 4.0 atm; P_NO = 0, P_Br2 = 0 (pure NOBr initially).
2. At equilibrium P_NOBr = 2.5 atm, so the change in NOBr is ΔP_NOBr = 2.5 − 4.0 = −1.5 atm (consumed 1.5 atm).
3. Relate this change to stoichiometry. Let x be the extent where 2x is the amount of NOBr consumed. Since 2x = 1.5 atm, x = 0.75 atm.
4. Using stoichiometry: P_NO produced = 2x = 2(0.75) = 1.5 atm. P_Br2 produced = x = 0.75 atm.
5. Equilibrium pressures: P_NOBr = 2.5 atm, P_NO = 1.5 atm, P_Br2 = 0.75 atm.

Calculation (showing all intermediate steps):
Kp = (P_NO)^2 (P_Br2) / (P_NOBr)^2
= (1.5^2 × 0.75) / (2.5^2)
1.5^2 = 2.25
2.25 × 0.75 = 1.6875
2.5^2 = 6.25
Kp = 1.6875 / 6.25 = 0.27

Why this is correct: pressures were computed consistently with the 2:2:1 stoichiometry and substituted into the standard Kp expression for this heterogeneous gas-phase equilibrium, yielding 0.27.

Why the other choices are incorrect:
- 0.12 and 0.45 likely come from arithmetic or stoichiometry errors (e.g., not squaring correctly or misassigning changes).
- 1.25 would correspond to inverting numerator/denominator or using wrong pressure values.
- 0.62 is roughly double the correct value and implies an incorrect product or denominator calculation.

Conclusion: Kp = 0.27, which matches the provided correct answer.

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