Q. The reaction between $N_{2}$ and $H_{2}$ to form ammonia has $K_{c} = 6 \times 10^{- 2}$ at the temperature $500℃$. The numerical value of $K_{p}$ for this reaction is
A) $1.5 \times 10^{- 5}$
B) $1.5 \times 10^{- 6}$
C) $1.5 \times 10^{5}$
D) $1.5 \times 10^{6}$
Explanation
(a)
$K_{p} = K_{c}(RT)^{\mathrm{\Delta}n_{g}}$
Here, $\mathrm{\Delta}n_{g} = 2 - 4 = - 2$
$\therefore K_{p} = 6 \times 10^{- 2} \times (0.0821 \times 773)^{- 2}$
or, $K_{p} = \frac{6 \times 10^{- 2}}{(0.0821 \times 773)^{2}} = 1.5 \times 10^{- 5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.