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 $1.44\times {10}^{-5}at{m}^{-2}$ when partial pressure is measured in atmosphere. The corresponding value of $K_c$, with concentration in $mol/L$ 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 D $\frac{1.44\times {10}^{-5}}{(0.082\times 773{)}^{-2}}$

For any given equilibrium, $K_P=K_C(RT{)}^{\Delta ng}$

where, $\Delta ng$= (gaseous moles of products)-(gaseous moles of reactants)

In given reaction, $\Delta ng=2-(3+1)=-2$

$\therefore K_P=\frac{K_C}{(RT{)}^2}$

$\therefore K_C=K_P(RT{)}^2=\frac{K_P}{(RT{)}^{-2}}=\frac{1.44\times {10}^{-5}}{(0.082\times 773{)}^{-2}}mol/L$