Answer
$n_{N_2O_4} = 0.945 \space mol$
$n_{NO_2} = 0.114 \space mol$
Work Step by Step
1. Calculate the new concentrations.
New volume = 0.750 L + 2.25 L = 3.00 L
$[N_2O_4] = \frac{0.971 \space mol}{3.00 \space L} = 0.324 \space M$
$[NO_2] = \frac{0.0580 \space mol}{3.00 \space L} = 0.0193 \space M$
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ NO_2 ] ^{ 2 }}{[ N_2O_4 ]}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ N_2O_4 ] = 0.324 \space M - x$
$ [ NO_2 ] = 0.0193 \space M + 2x$
$$4.61 \times 10^{-3} = \frac{(0.0193 + 2x)^2}{(0.324 - x)}$$
3. Solve for x:
In order to not have negative concentrations:
x = 0.00939
$ [ N_2O_4 ] = 0.324 \space M - x = 0.315 \space M$
$ [ NO_2 ] = 0.0193 \space M + 2x = 0.0381 \space M$
$n_{N_2O_4} = 0.945 \space mol$
$n_{NO_2} = 0.114 \space mol$
