Chemistry: Molecular Approach (4th Edition)

$Q_c \gt K_c$, therefore, the reaction is not at equilibrium, and will proceed to shift the equilibrium to the left, producing more reactants.
1. Determine each molarity. $[NO_2] = \frac{0.055 \space mol}{2.25 \space L} = 0.024$ $[N_2O_4] = \frac{0.082 \space mol}{2.25 \space L} = 0.036$ 2. Calculate the $Q_c$ value: - The exponent of each concentration is equal to its balance coefficient. $$Q_C = \frac{[Products]}{[Reactants]} = \frac{[ N_2O_4 ]}{[ NO_2 ] ^{ 2 }}$$ - Substitute the values and calculate the constant value: $$Q_C = \frac{( 0.036 )}{( 0.024 )^{ 2 }} = 63$$ $Q_c \gt K_c$, therefore, the reaction is not at equilibrium, and will proceed to shift the equilibrium to the left, producing more reactants.