# Question #595b6

May 29, 2015

The equilibrium will shift to the left.

To determine the direction in which the equilibrium will shift once you mix the three gases together, you need to calculate the reaction quotient, ${Q}_{c}$.

The reaction quotient expresses the ratio between the concentrations of the products and the concentration of the reactant at a given moment during the reaction.

In your case, you know that you start with 0.100 M of each gas. The reaction quotient for these initial values will be

${Q}_{c} = \frac{{\left[S {O}_{2}\right]}_{0} \cdot {\left[C {L}_{2}\right]}_{0}}{{\left[S {O}_{2} C {l}_{2}\right]}_{0}}$

${Q}_{c} = \frac{\cancel{0.100} \cdot 0.100}{\cancel{0.100}} = 0.100$

Now compare this value to the value of the equilibrium constant, ${K}_{c}$. Since ${Q}_{c} > {K}_{c}$, the mixture contains more products than reactant.

As a result, the equilibrium will shift to the left, favoring the formation of more reactant, in your case sulfuryl chloride, $S {O}_{2} C {l}_{2}$.

The reverse reaction will thus proceed in order for the equilibrium to be established.