At 1990 K and 1 atm pressure, there are equal number of \(\mathrm{Cl}_{2}\) molecules and Cl atoms in the reaction mixture. The value of \(\mathrm{K}_{\mathrm{p}}\) for the reaction \(\mathrm{Cl}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{Cl}_{(\mathrm{g})}\) under the above conditions is \(x \times 10^{-1}\). The value of x is ____. (Rounded off to the nearest integer)

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Equilibrium

At 1990 K and 1 atm pressure, there are equal number of

Cl_{2}

molecules and Cl atoms in the reaction mixture. The value of

K_{p}

for the reaction

Cl_{2 (g)} ⇌ 2 Cl_{(g)}

under the above conditions is

x \times 1 0^{- 1}

. The value of x is ____. (Rounded off to the nearest integer)

Answer: 5

Given that at equilibrium, the number of

Cl_{2}

molecules equals the number of Cl atoms.

Let the moles of

Cl_{2}

at equilibrium =

x

Let the moles of Cl atoms at equilibrium =

x

Total moles at equilibrium =

x + x = 2 x

Partial pressure of

Cl_{2}

P_{Cl_{2}} = \frac{x}{2 x} \times 1 = \frac{1}{2} atm

Partial pressure of Cl:

P_{Cl} = \frac{x}{2 x} \times 1 = \frac{1}{2} atm

The equilibrium constant

K_{p}

is:

K_{p} = \frac{( P _{Cl} ) ^{2}}{P _{Cl_{2}}} = \frac{( \frac{1}{2} ) ^{2}}{\frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} = 0.5 = 5 \times 1 0^{- 1}

Therefore,

x = 5

Answer: 5

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Integer answer