Question

$K_C$ for $N_2+O_2\rightleftharpoons 2NO$ at certain temperature is $1.6\times {10}^{-3}$, then $K_P$ for $NO\rightleftharpoons 1/2N_2+1/2O_2$ at same temperature will be:

• A. $2\times {10}^{-2}$
• B. $25$
• C. $5$
• D. $1.6\times {10}^{-3}$

Answer 1

Answer: (B)

$25$

The given reaction is :-

$N_2+O_2\rightleftharpoons 2NO;K_C=1.6\times {10}^{-3}$

$\Rightarrow K_C=\frac{{\left[NO\right]}^2}{\left[N_2\right]\left[O_2\right]}=1.6\times {10}^{-3}\to \left(I\right)$

Now, $NO\rightleftharpoons \frac{1}{2}{N}_2+\frac{1}{2}{O}_2$

$\Rightarrow K_C=\frac{{\left[N_2\right]}^{1/2}{\left[O_2\right]}^{1/2}}{\left[NO\right]}$

$={\left(\frac{1}{K_C}\right)}^{1/2}={\left(\frac{1}{1.6\times {10}^{-3}}\right)}^{1/2}$

$\Rightarrow {K_C}^{{}^{\prime }}=\frac{{10}^2}{4}=25$

Now, ${K_P}^{{}^{\prime }}={K_C}^{{}^{\prime }}{\left(RT\right)}^{△n};△n=\frac{1}{2}+\frac{1}{2}-1=0$

$\Rightarrow {K_P}^{{}^{\prime }}={K_C}^{{}^{\prime }}=25$