Answer
$ [ ICl ] = 0.0601 \space M$
$ [ I_2 ] = 0.0199 \space M$
$ [ Cl_2 ] = 0.0199 \space M$
Work Step by Step
- Calculate the molarity:
$$Molarity = \frac{0.500 \space mol \space ICl}{5.00 \space L} = 0.100 \space mol \space ICl$$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ ICl ]& [ I_2 ]& [ Cl_2 ]\\
Initial& 0.100 & 0 & 0 \\
Change& -2 x& +x& +x\\
Equilibrium& 0.100 -2 x& 0 +x& 0 +x\\
\end{vmatrix}$$
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ I_2 ][ Cl_2 ]}{[ ICl ] ^{ 2 }}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ ICl ] = 0.100 \space M - 2x$
$ [ I_2 ] = 0 \space M + x$
$ [ Cl_2 ] = 0 \space M + x$
$$0.110 = \frac{(0 + x)(0 + x)}{(0.100-2x)^2}$$
3. Solve for x:
x = $0.0199\underline 4$
$ [ ICl ] = 0.100 \space M - 2(0.0199\underline 4) = 0.0601 \space M$
$ [ I_2 ] = 0.0199 \space M$
$ [ Cl_2 ] = 0.0199 \space M$
