Answer
$[HBr] \approx 0.267 M$
$[H_2] = [Br_2] = 1.80 \times 10^{-4} M$
Work Step by Step
- Calculate all the concentrations:
$$[HBr] = ( 3.20 )/(12.0) = 0.267 M$$
1. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[ H_2 ][ Br_2 ]}{[ HBr ]^{ 2 }}$$
2. At equilibrium, these are the concentrations of each compound:
$[ HBr ] = 0.267 \space M - 2x$
$[ H_2 ] = x$
$[ Br_2 ] = x$
$$4.59 \times 10^{-7} = \frac{( x)( x)}{(0.267-2x)^2}$$
x = $1.80 \times 10^{-4}$
$[HBr] = 0.267 - 2(1.80 \times 10^{-4}) \approx 0.267 M$
$[H_2] = [Br_2] = 1.80 \times 10^{-4} M$
