Effect of Pressure on the Position of Equilibrium

Question

The following general reaction proceeds with decrease in amount of substance, so increasing pressure should shift the position of equilibrium to the right.

$$\ce{2 A(g) <=> B(g)}$$

But let's say the forward reaction is endothermic, and both $\ce{A}$ and $\ce{B}$ are gases. This means if the position shifts to the right, there will be an increase in temperature and consequently an increase in gas pressure. Then why will the equilibrium shift towards right side upon increasing the pressure?

Does the effect of pressure on the position of equilibrium depend on whether we have an endothermic or exothermic reaction?

Answer 1

The equilibrium constant is increasing with T for endothermic reactions.

$$ \frac{\text{d} (\ln {K}) }{ \text{d} T} = \dfrac {\Delta_r H^{\circ}}{RT^2}$$

For a small enough temperature interval, where $\Delta_r H$ can be approximated as being constant:

$$ \ln \left({\frac{K_{T_2}}{K_{T_1}}}\right) = \dfrac {\Delta_r H^{\circ}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

Its value is pressure independent, at least for ideal gases, if expressed in partial pressures.

But the reaction quotient, (formally the same expression as for the equilibrium constant, but for any reaction state), decreases with increasing pressure, if volume of reactants is bigger than of products.

$$K = \frac{p_\text{B, eq} }{ p_\text{A, eq}^2}$$

$$Q = \frac{p_\text{B} }{ p_\text{A}^2}$$

For doubled pressure:

$$Q_2 = \frac{2 \cdot p_\text{B} }{ (2 \cdot p_\text{A})^2} = \frac{Q}{2}$$

Simultaneous changes of both temperature and pressure changes both values $K$ and $Q$.

• If $Q \lt K$ then the forward net reaction is ongoing.
• If $Q \gt K$ then the backward net reaction is ongoing.