Question

How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant or product, (b) decrease in the volume, (c) decrease in the temperature, (d) addition of a catalyst?

Short answer

For a gas-phase exothermic reaction, (a) removal of a reactant or product, and (d) addition of a catalyst do not affect the equilibrium constant (K) value. (b) Decrease in volume does not affect the K value but shifts the equilibrium towards reactants. And (c) decrease in temperature increases the K value for an exothermic reaction, as per the Van't Hoff equation.

Step-by-step solution

Restate Le Chatelier's Principle

Le Chatelier's principle states that when a system at equilibrium experiences a stress or change (like a change in concentration, volume, pressure or temperature), it will shift its position to counteract that stress and re-establish equilibrium.

Explain the Van't Hoff Equation

The Van't Hoff equation is used to determine how the change in temperature affects the equilibrium constant (K). The equation is given as, \( \frac{-d(lnK)}{dT} = \frac{\Delta H^o}{RT^2} \) where d(lnK)/dT is the temperature coefficient of the equilibrium constant, ΔH° is the standard enthalpy change of the reaction, R is the gas constant, and T is the absolute temperature. Now, let's analyze each change and its effect on the equilibrium constant:

(a) Removal of a Reactant or Product

When a reactant or product is removed from the system, the equilibrium will shift to counteract the stress by reforming more of the substance that was removed, as per Le Chatelier's principle. However, the value of the equilibrium constant (K) remains the same since it is only dependent on temperature. The concentrations of the reactants and products will change, but the ratio of their concentrations at equilibrium still remains constant.

(b) Decrease in Volume

Decreasing the volume of the reaction increases the pressure, and according to Le Chatelier's principle, the system will shift towards the direction where there are fewer moles of gas to reduce the pressure. For an exothermic reaction, the side with more moles of gas is the reactants side. Hence, the equilibrium will shift towards the reactants. However, as stated earlier, the value of the equilibrium constant (K) is not affected by the change in volume or pressure, but only by temperature.

(c) Decrease in Temperature

Since the given reaction is exothermic, it releases heat. Decreasing the temperature is equivalent to removing heat from the system, which is treated as a product in an exothermic reaction. According to Le Chatelier's principle, the reaction will shift to counteract the stress, which in this case, will favor the forward reaction, producing more heat. In terms of the Van't Hoff equation, a decrease in temperature results in an increase in the ln(K) value, which indicates an increase in the equilibrium constant for exothermic reactions (\(\Delta H^o < 0 \)).

(d) Addition of a Catalyst

Adding a catalyst changes the speed at which the equilibrium is achieved by lowering the activation energy for both forward and reverse reactions. However, it does not change the position of equilibrium at all. Therefore, the value of the equilibrium constant (K) will remain unchanged upon the addition of a catalyst. In conclusion, for an exothermic reaction, (a) removal of a reactant or product, and (d) addition of a catalyst do not affect the equilibrium constant value. (b) Decrease in volume does not affect the equilibrium constant value but shifts the equilibrium towards reactants. (c) Decrease in temperature increases the value of the equilibrium constant for an exothermic reaction.