If for the reaction, 2 ICl I _2+Cl_2 ; K_c=0.14 and initial concentration of ICl is 0.6 M then equilibrium concentration of I_2 is

Correct answer: B. 0.128 M 2 ICl leftharpoons I _2+Cl_2 Initial 0.6 0 0 At eqm : 0.6-2 x x x aligned & K_c=0.14= [I_2][Cl_2] [ICl]^2 =(x^2)/((0.6-2 x)^2) & 0.37=(x)/(0.6-2 x) 0.222-0.74 x=x & 1.74 x=0.222 x=0.128 M aligned

AIIMS 201826 May

Exam: AIIMS
Year: 2018
Date: 26 May
Subject: Chemistry
Chapter: Chemical Equilibrium

AIIMS 2018 · 26 May · Chemistry · Chemical Equilibrium

If for the reaction, 2 ICl I _2+Cl_2 ; K_c=0.14 and initial...

If for the reaction,

2 \mathrm{ICl} \longrightarrow

\mathrm{I}

_2+\mathrm{Cl}_2 ; K_c=0.14

and initial concentration of

\mathrm{ICl}

0.6 \mathrm{M}

then equilibrium concentration of

\mathrm{I}_2

Options

A $0.37 \mathrm{M}$
B $0.128 \mathrm{M}$
C $0.224 \mathrm{M}$
D $0.748 \mathrm{M}$

Correct answer

B. $0.128 \mathrm{M}$

Step-by-step solution

2 \mathrm{ICl} \rightleftharpoons

\mathrm{I}

_2+\mathrm{Cl}_2

Initial

0.6 \quad 0 \quad 0

At eqm :

0.6-2 x \quad x \quad x

\begin{aligned} & K_c=0.14=\frac{\left[\mathrm{I}_2\right]\left[\mathrm{Cl}_2\right]}{[\mathrm{ICl}]^2}=\frac{x^2}{(0.6-2 x)^2} \\ & 0.37=\frac{x}{0.6-2 x} \Rightarrow 0.222-0.74 x=x \\ & \Rightarrow 1.74 x=0.222 \therefore x=0.128 \mathrm{M}\end{aligned}

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