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Multiple Choice
The decomposition of ammonia is: 2NH3(g) -> N2(g) + 3H2(g). If the pressure of ammonia is 1.0 × 10^-3 atm, and the pressures of N2 and H2 are each 0.20 atm, what is the value of Kp' at 400°C for the reverse reaction?
A
4.0 × 10^3
B
2.5 × 10^3
C
1.0 × 10^3
D
5.0 × 10^3
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1
Identify the reverse reaction: The given reaction is 2NH3(g) -> N2(g) + 3H2(g). The reverse reaction is N2(g) + 3H2(g) -> 2NH3(g).
Write the expression for Kp for the forward reaction: Kp = (P_N2 * (P_H2)^3) / (P_NH3)^2, where P_N2, P_H2, and P_NH3 are the partial pressures of N2, H2, and NH3, respectively.
Calculate Kp for the forward reaction using the given pressures: Substitute P_N2 = 0.20 atm, P_H2 = 0.20 atm, and P_NH3 = 1.0 × 10^-3 atm into the Kp expression.
Determine Kp' for the reverse reaction: Kp' is the reciprocal of Kp for the forward reaction. Therefore, Kp' = 1 / Kp.
Calculate Kp' using the value obtained for Kp from the forward reaction to find the equilibrium constant for the reverse reaction at 400°C.