2003 AIPMT EquilibriumCombining Equilibrium Constants Medium Question 92 of 104 in Equilibrium

The following equilibria are given :
$N_2 + 3H_2 \rightleftharpoons 2NH_3$, $K_1$
$N_2 + O_2 \rightleftharpoons 2NO$, $K_2$
$H_2 + \dfrac{1}{2}O_2 \rightleftharpoons H_2O$, $K_3$
The equilibrium constant of the reaction $2NH_3 + \dfrac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $K_1$, $K_2$ and $K_3$ is :

A $K_1K_2K_3$
B $\dfrac{K_1K_2}{K_3}$
C $\dfrac{K_1K_3^2}{K_2}$
D $\dfrac{K_2K_3^3}{K_1}$

The answer is D.

Solution

Build the required reaction from the given ones.
Reverse the first reaction: $2NH_3 \rightleftharpoons N_2 + 3H_2$, equilibrium constant $\dfrac{1}{K_1}$
Take the second reaction as it is: $N_2 + O_2 \rightleftharpoons 2NO$, $K_2$
Multiply the third reaction by 3: $3H_2 + \dfrac{3}{2}O_2 \rightleftharpoons 3H_2O$, $K_3^3$
Adding: $2NH_3 + O_2 + \dfrac{3}{2}O_2 \rightleftharpoons 2NO + 3H_2O$, i.e. $2NH_3 + \dfrac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$ ($N_2$ and $3H_2$ cancel)
When reactions are added, their equilibrium constants are multiplied:
$K = \dfrac{1}{K_1} \times K_2 \times K_3^3 = \dfrac{K_2K_3^3}{K_1}$

Concept Manipulating equilibrium constants

Equilibrium in past papers

104 questions from this chapter have appeared across 27 exam years.