Question

Given:
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right);K_1$
$N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right);K_2$
$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons H_{2}O\left(g\right);K_3$
where $K_1,K_2,K_3$ are equilibrium constants.
The equilibrium constant for the reaction,
$2N{H}_3\left(g\right)+\frac{5}{2}{O}_2\left(g\right)\rightleftharpoons 2NO\left(g\right)+3H_{2}O\left(g\right)$ will be:

• A. $K_1{K}_2{K}_3$
• B. $\frac{K_1{K}_2}{K_3}$
• C. $\frac{K_1{K}_3^2}{K_2}$
• D. $\frac{K_2{K}_3^3}{K_1}$

Answer 1

The correct option is D $\frac{K_2{K}_3^3}{K_1}$

For the given Equations:
eq.1,
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right);K_1$

eq.2,
$N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right);K_2$

eq.3,
$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons H_{2}O\left(g\right);K_3$

We have to fine the equilibrium constant for
$2N{H}_3\left(g\right)+\frac{5}{2}{O}_2\left(g\right)\rightleftharpoons 2NO\left(g\right)+3H_{2}O\left(g\right)$ will be:

Reversing the eq. 1,
we get , eq. 4
$2N{H}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right);K_4=\frac{1}{K_1}$

Multiplying eq. 3 by 3,
we get , eq. 5
$3H_2\left(g\right)+\frac{3}{2}{O}_2\left(g\right)\rightleftharpoons 3H_{2}O\left(g\right);K_5=(K_3{)}^3$

Adding equation 2, 4, and 5, we get the required equation,
$2N{H}_3\left(g\right)+\frac{5}{2}{O}_2\left(g\right)\rightleftharpoons 2NO\left(g\right)+3H_{2}O\left(g\right)$

$K=K_2\times K_4\times K_5$

putting values,

$K=K_2\times \frac{1}{K_1}\times (K_3{)}^3=\frac{K_2\times K_3^3}{K_1}$



Theory:

Characteristics of $K$ (Equilibrium constant) :
Equilibrium constant $\left(K\right)$ for the backward reaction is calculated by taking the reciprocal of equilibrium constant $\left(K\right)$ for the forward reaction and vice versa .

Suppose for the given reaction :

$A+B\rightleftharpoons C$ then $K_c=\frac{\left[C\right]}{\left[A\right]\left[B\right]}$
$C\rightleftharpoons A+B$ then $K_c^{{}^{\prime }}=\frac{\left[A\right]\left[B\right]}{\left[C\right]}=\frac{1}{K}$
If the equilibrium reaction stoichiometry is changed, the power of the equilibrium constant also gets changed by the same quantity.

Suppose for the given reaction :

$A+B\rightleftharpoons C$ ; $K_{eq}=K$
$2A+2B\rightleftharpoons 2C$; $K_{eq}^{{}^{\prime }}=K^2$

if,
$A+B\rightleftharpoons C$ ; $K_{eq}=K$
$\frac{1}{2}A+\frac{1}{2}B\rightleftharpoons \frac{1}{2}C$ then $K_{eq}^{{}^{\prime \prime }}=K^{\frac{1}{2}}$

In case stepwise multiple equilibria leading to the final products, the equilibrium constant of the net equilibrium = product of each stepwise equilibrium constants.
Note:- whenever we add the reactions we multiply the K values and when we subtract the reaction we need to divide the K values.

Example 1 :
$A+B\rightleftharpoons C$; $K_1$
$C\rightleftharpoons D$; $K_2$
for $A+B\rightleftharpoons D;K=K_1\times K_2$

Example 2 :
$H_2\left(g\right)+2C\left(s\right)+\frac{5}{2}{O}_2\left(g\right)\rightleftharpoons 2C{O}_2\left(g\right)+H_{2}O\left(l\right)\Rightarrow K_1$

$C_2{H}_2\left(g\right)+\frac{5}{2}{O}_2\left(g\right)\rightleftharpoons 2C{O}_2\left(g\right)+H_{2}O\left(l\right)\Rightarrow K_2$

$H_2\left(g\right)+2C\left(s\right)\rightleftharpoons C_2{H}_2\left(g\right)\Rightarrow K=\frac{K_1}{K_2}$