For the reaction $\mathrm{Fe}_{2} \mathrm{~N}(\mathrm{~s})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g})=2 \mathrm{Fe}(\mathrm{s})+\mathrm{NH}_{3}(\mathrm{~g})$

(C) $\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{P}}(\mathrm{RT})^{1 / 2}$

(A) $\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}}(\mathrm{RT})$

(B) $\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{P}}(\mathrm{RT})^{-1 / 2}$

(D) $\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{P}}(\mathrm{RT})^{3 / 2}$

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Equilibrium

For the reaction

Fe_{2} N (s) + \frac{3}{2} H_{2} (g) = 2 Fe (s) + NH_{3} (g)

Answer: (C)

The two constants are linked by

K_{p} = K_{c} (R T)^{Δ n_{g}}

, where

Δ n_{g}

is the change in moles of gas. Here the gaseous products amount to

1

mole of

NH X_{3}

while the gaseous reactants are

\frac{3}{2}

moles of

H X_{2}

(the solids do not count), so

Δ n_{g} = 1 - \frac{3}{2} = - \frac{1}{2}

. Hence

K_{p} = K_{c} (R T)^{- 1/2}

, which rearranges to

K_{c} = K_{p} (R T)^{1/2}

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Single correct option