Question

The decomposition of ammonia is: 2 NH3(g) ⇌ N2(g) + 3 H2(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 for Kp' at 400°C for the reverse reaction?

a) 1.0 × 10-3
b) 0.40
c) 2.5 × 10-3
d) 0.008

Answer 1

To calculate Kp' for the reverse reaction of the decomposition of ammonia, use the formula Kp = (P_products)^coefficients / (P_reactants)^coefficients. By substituting the given pressures and simplifying, Kp' is found to be 2.5 × 10^-3, which matches option (c).

Step-by-step explanation:

The question involves the calculation of the equilibrium constant Kp for the reverse reaction of the decomposition of ammonia into nitrogen and hydrogen. Using the formula for Kp, which is Kp = (Pproducts)coefficients / (Preactants)coefficients, with the coefficients coming from the balanced equation, we calculate Kp' for the reverse reaction as follows:

Kp' = (PNH3)2 / (PN2 * PH2)3

Substituting the given partial pressures into the equation, we have:

Kp' = (1.0 × 10-3 atm)2 / (0.20 atm * (0.20 atm)3)

This simplifies to Kp' = (1.0 × 10-6 atm2) / (0.20 atm * 0.008 atm3)

After multiplying out the denominators and simplifying, we find that Kp' = 2.5 × 10-3, which corresponds to option (c).