Answer
$Q_c \gt K_c$, therefore, the reaction is not at equilibrium, and will proceed to shift the equilibrium to the left, producing more reactants.
Work Step by Step
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.
