Question

The following equilibria are given: $N_2$ +3$H_2$ $\leftrightharpoons$ $2N{H}_3$ $K_1$ ; $N_2$ + $O_2$ $\leftrightharpoons$ 2NO $K_2$ ;
$H_2$ + $\frac{1}{2}$ $O_2$ $\leftrightharpoons$ $H_{2}O{K}_3$ ;

The Equilibrium constant for the reaction $2N{H}_3$ + $\frac{5}{2}$ $O_2$ $\leftrightharpoons$ 2NO + 3$H_2$ O in terms of $K_1{K}_2$ and $K_3$ is:

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

There are three reactions given, each with its own unique equilibrium constant.

$N_2$ +3$H_2$ $\leftrightharpoons$ $2N{H}_3$ $K_1$ ; $N_2$ + $O_2$ $\leftrightharpoons$ 2NO $K_2$ ;$H_2$ + $\frac{1}{2}$ $O_2$ $\leftrightharpoons$ $H_{2}O{K}_3$;

Now we need to figure out the equilibrium constant of $2N{H}_3$ + $\frac{5}{2}$ $O_2$ $\leftrightharpoons$ 2NO + 3$H_2$ O in terms of $K_1{K}_2$ and $K_3$ is:

The required equilibrium constant is $K_c$ = $\frac{\left[NO{]}^2\right[H_{2}O{]}^2}{\left[N{H}_3{]}^2\right[O_2{]}^{\frac{5}{2}}}$

We can achieve this by manipulating the the given three reactions:

First let us reverse the first reaction

2$N{H}_3$ $\leftrightharpoons$ $N_2$ +3$H_2$ ($\frac{1}{K_1}$) { since reversing the equilibrium makes the new equilibrium constant for the now new reaction the reciprocal of the old one; try deriving it}

Now add the equation $N_2$+ $O_2$ $\leftrightharpoons$ 2NO $K_2$ to above and further, add

3 $\times$ [ $H_2$ + $\frac{1}{2}$ $O_2$ $\leftrightharpoons$ $H_{2}O$ ] { now try and derive mathematically the new equilibrium constant for this reaction}

The equilibrium constant for 3$H_2$ + ($\frac{3}{2}$) $O_2$ $\leftrightharpoons$ 3$H_{2}O$ this reaction is $K_4$ = $\frac{[H_{2}O{]}^3}{\left[H_2{]}^3\right[O_2{]}^{\frac{3}{2}}}$ = ${\left(K_3\right)}^3$

Adding the three equations while maintaining the lengthy equilibrium expression gives us the required equilibrium constant:

$K_c$ =$\frac{\left[NO{]}^2\right[H_{2}O{]}^3}{\left[N{H}_3{]}^2\right[O_2{]}^{\frac{5}{2}}}\frac{1}{K_1}$) ($K_2$) ${K_3}^3$

Hence (c) $K_2$ $\frac{\left({K_3}^3\right)}{K_1}$