## General Chemistry: Principles and Modern Applications (10th Edition)

$n_{N_2O_4} = 0.945 \space mol$ $n_{NO_2} = 0.114 \space mol$
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 $[ N_2O_4 ] = 0.324 \space M - x = 0.315 \space M$ $[ NO_2 ] = 0.0193 \space M + 2x = 0.0381 \space M$ $n_{N_2O_4} = 0.945 \space mol$ $n_{NO_2} = 0.114 \space mol$