Q. For the reversible reaction, $N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)$ 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) $\frac{1.44 \times 10^{- 5}}{(0.082 \times 773)^{- 3}}$
B) $\frac{1.44 \times 10^{- 5}}{(0.082 \times 500)^{- 2}}$
C) $\frac{1.44 \times 10^{- 5}}{(8.314 \times 773)^{2}}$
D) $\frac{1.44 \times 10^{- 5}}{(0.082 \times 773)^{- 2}}$
Explanation
(d)
For the reversible reaction,
$N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)$
$\mathrm{\Delta}n_{g} = 2 - (1 + 3)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2 - 4 = - 2$
$R = 0.0821\ L\ atm\ K^{- 1}mol^{- 1}$
$T = 500 + 273 = 773$ K
$K_{p} = K_{c}(RT)^{\mathrm{\Delta}n_{g}}$
$K_{p} = K_{c} \times (RT)^{- 2}$
or $K_{c} = \frac{K_{p}}{(RT)^{- 2}}$
$= \frac{1.44 \times 10^{- 5}}{(0.082 \times 773)^{- 2}}$
or $\frac{1.44 \times 10^{- 5}}{(0.082 \times 773)^{- 2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.