Answer
(a) Since $Q_c \lt K_c$, the mixture is not at equilbrium;
(b) And the net change will occur in the products direction, in order to increase the $Q_c$.
Work Step by Step
1. Calculate all the concentrations:
$$[SO_2] = ( 0.455 )/(1.90) = 0.239 M$$
$$[O_2] = ( 0.183 )/(1.90) = 0.0963 M$$
$$[SO_3] = ( 0.568 )/(1.90) = 0.299 M$$
- The exponent of each concentration is equal to its balance coefficient.
$$Q_c = \frac{[Products]}{[Reactants]} = \frac{[ SO_3 ] ^{ 2 }}{[ SO_2 ] ^{ 2 }[ O_2 ]}$$
2. Substitute the values and calculate the quocient value:
$$Q_c = \frac{( 0.299 )^{ 2 }}{( 0.239 )^{ 2 }( 0.0963 )} = 16.3$$
Since $Q_c \lt K_c$, the mixture is not at equilbrium, and the net change will occur in the products direction, in order to increase the $Q_c$.
