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JEE Main 202124 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 24 February
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Equilibrium

JEE Main 2021 · 24 February · Morning Shift · Chemistry · Chemical Equilibrium

At 1990K and 1atm pressure, there are equal number of Cl _ 2...

At

1990K

and

1atm

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\left(g\right)}

= 2{Cl}_{

\left(g\right)}

under the above conditions is

x \times {10}^{- 1} .

The value of

x

is______(Rounded off to the nearest integer)

Correct answer

0

Step-by-step solution

${Cl}_{2} ⇌ 2Cl$

Let mol of both of ${Cl}_{2}$ and $Cl$ is $x$

${P}_{Cl} = \dfrac{x}{2x} \times 1 = \dfrac{1}{2}atm$

${P}_{{Cl}_{2}} = \dfrac{x}{2x} \times 1 = \dfrac{1}{2}atm$

${K}_{p} = \dfrac{{\left(\dfrac{1}{2}\right)}^{2}}{\dfrac{1}{2}} = \dfrac{1}{2} = 0 . 5 ⇒ 5 \times {10}^{- 1}$

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