Question

For the reversible reaction, $N_2\left(g\right)+{3H}_2\left(g\right)\rightleftharpoons {2NH}_3\left(g\right)$ at $500K$, the value of $K_p=1.44\times {10}^{-5}$ when partial pressure is measured in atmospheres. The corresponding value of $K_c$, with concentration in $mole{L}^{-1}$, 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. $\frac{1.44\times {10}^{-5}}{{(0.082\times 773)}^2}$
• D. $\frac{1.44\times {10}^{-5}}{{(0.082\times 773)}^{-2}}$

Answer 1

The correct option is A $\frac{1.44\times {10}^{-5}}{{(0.082\times 500)}^{-2}}$

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

and $\Delta n=-2$

$\therefore K_c={\frac{K_p}{RT}}^{\Delta n}$

$K_c=\frac{1.44\times {10}^{-5}}{(0.082\times 500{)}^{-2}}$

Hence, the correct option is A.