Answer
The $ICl$ concentration on this equilibrium is equal to 0.119M
Work Step by Step
1. Write the equilibrium constant expression:
$K_{eq} = \frac{[Products]}{[Reactants} = \frac{[ICl]^2}{[I_2][Cl_2]}$
2. Substitute the known values into the expression:
$81.9 = \frac{[ICl]^2}{0.0112 \times 0.0155}$
$81.9 = \frac{[ICl]^2}{1.74 \times 10^{-4}}$
$81.9 \times 1.74 \times 10^{-4} = [ICl]^2$
$0.01425 = [ICl]^2$
$[ICl] = \sqrt {0.01425} = 0.119M$
