Question

For the reversible reaction, $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$ at ${500}^{o}C$, the value of $K_p$ is $1.44\times {10}^{-5}$ when partial pressure is measured in the atmosphere. The corresponding value of $K_c$ with concentration in mol litre${}^{-1}$, is :

• A. $1.44\times {10}^{-5}/(0.082\times 500{)}^{-2}$
• B. $1.44\times {10}^{-5}/(8.314\times 773{)}^{-2}$
• C. $1.44\times {10}^{-5}/(0.082\times 773{)}^2$
• D. $1.44\times {10}^{-5}/(0.082\times 773{)}^{-2}$

Answer 1

The correct option is D $1.44\times {10}^{-5}/(0.082\times 773{)}^{-2}$

The relationship between $K_p$ and $K_c$ is $K_p=K_c(RT{)}^{\Delta n}$

For the reaction $N_2\left(g\right)+3H_2\left(g\right)\iff 2N{H}_3\left(g\right)$, the value of $\Delta n$ is $2-(1+3)=-2$

Substitute this value in the above expression.

$K_p=K_c(RT{)}^{\Delta n}=K_c(RT{)}^{-2}$

But $K_p=1.44\times {10}^{-5},R=0.082\text{L atm /mol. K}$

$T=500+273=773K$

Substitute values in the above expression.

$1.44\times {10}^{-5}=K_c(0.082\times 773{)}^{-2}$

$K_c=1.44\times {10}^{-5}./(0.082\times 773{)}^{-2}$