For the reaction, $\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{~g})$ the $\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{K}_{\mathrm{c}}}$ is equal to

JEE Main · 2004 · 24 Apr 2004 · Offline · Equilibrium

For the reaction,

CO (g) + Cl_{2} (g) ⇌ COCl_{2} (g)

the

\frac{K _{p}}{K _{c}}

is equal to

Answer: (A)

The two constants are tied together by

K_{p} = K_{c} (R T)^{Δ n}

, where

Δ n

is the change in moles of gas. For

CO (g) + Cl X_{2} (g) COCl X_{2} (g)

there are two moles of gas on the left and one on the right, so

Δ n = 1 - 2 = - 1

. That gives

\frac{K _{p}}{K _{c}} = (R T)^{- 1} = \frac{1}{R T}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option