Question

At 500°C the equilibrium constant, Kp, is 4.00 × 10–4 for the equilibrium: 2HCN(g) H2(g) + C2N2(g) What is Kp for the following reaction? H2(g) + C2N2(g) 2HCN(g)

Answer 1

Answer: The value of $K_p$ for the equation is $2.5\times 10^3$

Explanation:

The given chemical reaction is:

$2HCN(g)\rightleftharpoons H_2(g)+C_2N_2(g);K_1$

The chemical equation for which the equilibrium constant is to be calculated follows:

$H_2(g)+C_2N_2(g)\rightleftharpoons 2HCN(g);K_p$

As, the equation is the result of the reverse of given reaction. So, the equilibrium constant for the equation will be the inverse of equilibrium constant for the given reaction.

The value of equilibrium constant for the equation is:

$K_p=\frac{1}{K_1}$

We are given:

$K_1=4.00\times 10^{-4}$

Putting values in above equation, we get:

$K_p=\frac{1}{4.00\times 10^{-4}}=2.5\times 10^3$

Hence, the value of $K_p$ for the equation is $2.5\times 10^3$