Q. For the reversible reaction, $N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2{NH}_{3}(g)$at $500℃$, 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 $mol\ L^{- 1}$, is
A) $\frac{1.44 \times 10^{- 5}}{(0.082 \times 773)^{- 2}}$
B) $\frac{1.44 \times 10^{- 5}}{(0.082 \times 773)^{2}}$
C) $\frac{1.44 \times 10^{- 5}}{(8.314 \times 773)^{- 2}}$
D) $\frac{1.44 \times 10^{- 5}}{(0.082 \times 500)^{- 2}}$
Explanation
(a)
$N_{2} + 3H_{2} \rightleftharpoons 2{NH}_{3}$
$\underset{4}{\overset{1\ \ \ \ \ \ \ \ \ 3}{︸}}$ 2
$\mathrm{\Delta}n_{g} = 2 - 4 = - 2$
$K_{p} = K_{c}(RT)^{\mathrm{\Delta}n_{g}}$
$\therefore K_{p} = K_{c}(RT)^{- 2}$
$\therefore K_{c} = \frac{K_{p}}{(RT)^{- 2}} = \frac{1.44 \times 10^{- 5}}{(0.082 \times 773)^{- 2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.