Answer
$ [ B ] = 4.50 \times 10^{-4} \space M$
$ [ C ] =2.50 \times 10^{-4} \space M$
Work Step by Step
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ B ] ^{ 2 }[ C ]}{[ A ]}$$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ A ]& [ B ]& [ C ]\\
Initial& 1.75 \times 10^{-3} & 1.25 \times 10^{-3} & 6.50 \times 10^{-4} \\
Change& -x& +2 x& +x\\
Equilibrium& 1.75 \times 10^{-3} -x& 1.25 \times 10^{-3} + 2 x& 6.50 \times 10^{-4} +x\\
\end{vmatrix}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ A ] = 1.75 \times 10^{-3} \space M - x$
$ [ B ] = 1.25 \times 10^{-3} \space M + 2x$
$ [ C ] = 6.50 \times 10^{-4} \space M + x$
3. Using the concentration of $ A $ at equilibrium, find x:
$ 1.75 \times 10^{-3} - x = 2.15 \times 10^{-3} $
$ - x = 2.15 \times 10^{-3} - 1.75 \times 10^{-3} $
$ x = 1.75 \times 10^{-3} - 2.15 \times 10^{-3} $
$x = -4.00 \times 10^{-4} $
$ [ B ] = 1.25 \times 10^{-3} \space M + 2*( -4.00 \times 10^{-4} )=4.50 \times 10^{-4} $
$ [ C ] = 6.50 \times 10^{-4} \space M + (-4.00 \times 10^{-4}) =2.50 \times 10^{-4} $
