Answer
$K_c \neq Q_c$, therefore, the system is not at equilibrium.
$Q_c \gt K_c$ therefore, the concentration of $NO_2$ will increase as the system proceeds to equilibrium.
Work Step by Step
- Calculate all the concentrations:
$$[NO_2] = ( 2.0 \times 10^{-3} )/(10.0) = 2.0 \times 10^{-4} M$$
$$[N_2O_4] = ( 1.5 \times 10^{-3} )/(10.0) = 1.5 \times 10^{-4} M$$
- The exponent of each concentration is equal to its balance coefficient.
$$Q_C = \frac{[Products]}{[Reactants]} = \frac{[ N_2O_4 ]}{[ NO_2 ] ^{ 2 }}$$
2. Substitute the values and calculate the constant value:
$$Q_C = \frac{( 1.5 \times 10^{-4} )}{( 2.0 \times 10^{-4} )^{ 2 }} = 4.0 \times 10^{3}$$
$K_c \neq Q_c$, therefore, the system is not at equilibrium.
$Q_c \gt K_c$ therefore, there is too much product, so the concentration of $NO_2$ will increase as the system proceeds to equilibrium.
