Consider the reaction: 2SO2 (g) + O2 (g) ⇌ 2SO3 (g) ΔH = -198.2 kJ/mol. What would be the effect on the reaction if we were to:a. increase the temperatureb. increase the pressurec. increase [SO2]d. add a catalyst

Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

a. increase the temperature:

∵ ΔH has a negative value, the reaction is exothermic.

The heat can be represented as a part of the products.

Increase the T will increase the concentration of the products "heat".

So, the reaction will be shifted to the left side to suppress the increase in the concentration of products by increasing T.

So, the reaction will be shifted to the left side.

b. increase the pressure:

When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.

The reactants side (left) has 3.0 moles of gases and the products side (right) has 2.0 moles of gases.

Thus, increasing the pressure will shift the reaction to the side with lower moles of gas (right side).

So, the reaction will be shifted to the right side.

c. increase [SO₂]:

Increasing [SO₂] will increase the concentration of the reactants side, so the reaction will be shifted to the right side to suppress the increase in the concentration of SO₂.

So, the reaction will be shifted to the right side.

d. add a catalyst:

Catalyst increases the rate of the reaction without affecting the equilibrium position.

Catalyst increases the rate via lowering the activation energy of the reaction.

This can occur via passing the reaction in alternative pathway (changing the mechanism).

The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).

in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.

with adding a catalyst, both the forward and reverse reaction rates will speed up equally, which allowing the system to reach equilibrium faster.

So, it has no effect.