At $-20^{\circ} \mathrm{C}$ and 1 atm pressure, a cylinder is filled with equal number of $\mathrm{H}_{2}$. $\mathrm{I}_{2}$ and HI molecules for the reaction $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$, the $\mathrm{K}_{\mathrm{P}}$ for the process is $\mathrm{x} \times 10^{-1} . \mathrm{x}=$ \_\_\_\_. [Given : $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Equilibrium

- 2 0^{\circ} C

and 1 atm pressure, a cylinder is filled with equal number of

H_{2}

I_{2}

and HI molecules for the reaction

H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)

, the

K_{P}

for the process is

x \times 1 0^{- 1} . x =

____.
[Given :

R = 0.082 L atm K^{- 1} mol^{- 1}

]

Answer: (C)

For

H_{2} + I_{2} ⇌ 2HI

Δ n_{g} = 0

, so

K_{P} = \frac{n _{H I}^{2}}{n _{H_{2}} n _{I_{2}}}

(the pressure/mole-number terms cancel).
With equal numbers of all three species,

K_{P} = \frac{n ^{2}}{n \cdot n} = 1

.
Given

K_{P} = x \times 1 0^{- 1}

1 = x \times 1 0^{- 1} \Rightarrow x = 10.

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Single correct option