# When volume decreases and the product has less moles than reactants, why is there no increase in the reaction quotient?

Assume a reaction:

$$\ce{CO + Cl2 <=> COCl2}$$

My textbook says that a decrease in volume will result in the reaction quotient being less than the equilibrium constant. Why is that?

With a decrease in the volume, a forward reaction proceeds. Thus, more product and less reactant means there should be a higher reaction quotient.

Keep in mind that for reaction

$$\ce{r_{1}R_{1} + r_{2}R_{2} + \ldots + r_{n}R_{n} <=> p_{1}P_{1} +\ldots + p_{m}P_{m}}$$

the reaction quotient is:

$$Q = \frac{\prod_{i = 1}^{m} [P_{i}]^{p_{i}}}{\prod_{j = 1}^{n} [R_{j}]^{r_{j}}}$$

Specifically, here: $$Q = \frac{[\ce{COCl2}]}{[\ce{CO}][\ce{Cl2}]}$$

For your specific reaction, the reaction quotient is getting smaller because the denominator has a quadratic dependence on the pressure where as the numerator only has a linear dependence on pressure.

The goal as usual is to make $Q = K_{\mathrm{eq}}$, which is equilibrium.

Since now, $Q < K_{\mathrm{eq}}$, you want to push the reaction forwards. That will increase the concentrations of product and decrease the concentrations of reactant, making the reaction quotient bigger.

Intuitively, that makes sense in your example. There are two molecules combining to form one molecule. By decreasing the volume and increasing the pressure, you're squeezing them together into the product.