Question

For the reversible reaction, $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$ 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 in mole $litr{e}^{-1}$, 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. $\frac{1.44\times {10}^{-5}}{(0.082\times 773{)}^{-3}}$

Answer 1

Answer: (C)

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

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$

$△n=2-3-1=-2$

$K_p=1.44\times {10}^{-5}T={500}^{0}C$

we know that, $=773K$

$K_p=K_C{\left(RT\right)}^{△n}$

$K_C=\frac{K_P}{{\left(RT\right)}^{△n}}=\frac{1.44\times {10}^{-5}}{{(0.082\times 773)}^{-2}}$

$=\frac{1.44\times {10}^{-5}}{{(0.0821\times 773)}^{-2}}$