Question

For the reversible reaction
$N_2\left(g\right)+3H_2\left(g\right)\iff 2N{H}_3\left(g\right)$ at ${500}^{\circ }C$ the value of $K_p$ is $1.44\times {10}^{-5}$ when partial pressure is measured in atmospheres. The corresponding value of $K_c$ with concentration of $mole.li{t}^{-1}$ is:

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

Answer 1

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

The relationship between $K_p$ and $K_c$ is $K_p=K_c(RT{)}^{\Delta n}$.
For the reaction $N_2\left(g\right)+3H_2\left(g\right)\iff 2N{H}_3\left(g\right)+heat$, 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.
$1.44\times {10}^{-5}=K_c(0.082\times 773{)}^{-2}$
$K_c=\frac{1.44\times {10}^{-5}}{(0.082\times 773{)}^{-2}}$.