Answer
$$K_C = 0.106$$
Work Step by Step
1. Calculate both concentrations:
$$[PCl_5]_{initial} = \frac{1.00 \times 10^{-3} \space mol}{250.0 \space mL} \times \frac{1000 \space mL}{1 \space L} = 4.00 \times 10^{-3} \space M$$
$$[Cl_2]_{eq} = \frac{9.65 \times 10^{-4} \space mol}{250.0 \space mL} \times \frac{1000 \space mL}{1 \space L} = 3.86 \times 10^{-3} \space M$$
2. At equilibrium, these are the concentrations of each compound:
$ [ PCl_5 ] = 4.00 \times 10^{-3} \space M - x$
$ [ PCl_3 ] = 0 \space M + x$
$ [ Cl_2 ] = 0 \space M + x$
3. Using the concentration of $ Cl_2 $ at equilibrium, find x:
$ 0 + x = 3.86 \times 10^{-3} $
$ x = 3.86 \times 10^{-3} - 0 $
$x = 3.86 \times 10^{-3} $
$ [ PCl_5 ] = 4.00 \times 10^{-3} \space M - 3.86 \times 10^{-3} =1.40 \times 10^{-4} $
$ [ PCl_3 ] = 0 \space M + 3.86 \times 10^{-3} =3.86 \times 10^{-3} $
$ [ Cl_2 ] = 0 \space M + 3.86 \times 10^{-3} =3.86 \times 10^{-3} $
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ PCl_3 ][ Cl_2 ]}{[ PCl_5 ]}$$
4. Substitute the values and calculate the constant value:
$$K_C = \frac{( 3.86 \times 10^{-3} )( 3.86 \times 10^{-3} )}{( 1.40 \times 10^{-4} )} = 0.106$$
