If for the reaction, 2ICl→ I2+CI2; Kc=0.14 and initial concentration of ICl is 0.6 M then equilibrium concentration of I2 is

Question

If for the reaction, 2ICl→ I2+CI2; Kc=0.14 and initial concentration of ICl is 0.6 M then equilibrium concentration of I2 is (A) 0.37 M (B) 0.128 M (C) 0.224 M (D) 0.748 M

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/16f8fd77052d32bc4679adbf87095f57-.png" /> Kc=0.14=([I2][Cl2]/[ICI]2)=(x2/(0.6-2x)2) 0.37=(x/0.6-2x) ⇒ 0.222-0.74x=x ⇒ 1.74x=0.222 ∴ x=0.128 M

Q. If for the reaction, $2 I Cl \to I_{2} + C I_{2}; K_{c} = 0.14$ and initial concentration of $I Cl$ is 0.6 M then equilibrium concentration of $I_{2}$ is

0.37 M

0.128 M

0.224 M

0.748 M

$K_{c} = 0.14 = \frac{[ I _{2} ] [ C l _{2} ]}{[ I C I ] ^{2}} = \frac{x ^{2}}{( 0.6 - 2 x ) ^{2}}$
$0.37 = \frac{x}{0.6 - 2 x} \Rightarrow 0.222 - 0.74 x = x$
$\Rightarrow 1.74 x = 0.222 ∴ x = 0.128 M$

Q. If for the reaction, 2ICl→I2​+CI2​;Kc​=0.14 and initial concentration of ICl is 0.6 M then equilibrium concentration of I2​ is