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  4. 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

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Q. If for the reaction, $2ICl\to I_{2}+CI_{2}; K_{c}=0.14$ and initial concentration of $ICl$ is 0.6 M then equilibrium concentration of $I_2$ is

AIIMSAIIMS 2018Equilibrium

Solution:

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$K_{c}=0.14=\frac{\left[I_{2}\right]\left[Cl_{2}\right]}{\left[ICI\right]^{2}}=\frac{x^{2}}{\left(0.6-2x\right)^{2}}$
$0.37=\frac{x}{0.6-2x}\,\,\Rightarrow \,\,0.222-0.74x=x$
$\Rightarrow \,\,\,1.74x=0.222\,\,\therefore \,\,x=0.128\,M$