Which two changes would make this reaction favor the formation of products?

[tex]\[
2 \text{SO}_2 + \text{O}_2 \rightleftharpoons 2 \text{SO}_3 + \text{energy}
\][/tex]

A. Decreasing the temperature
B. Raising the pressure
C. Decreasing the pressure
D. Raising the temperature



Answer :

To determine which changes would favor the formation of products in the given reaction:

[tex]$ 2 \ \text{SO}_2 + \text{O}_2 \rightleftarrows 2 \ \text{SO}_3 + \text{energy} $[/tex]

we need to consider the principles of Le Chatelier's Principle. This principle tells us how a system at equilibrium will respond to changes in temperature, pressure, and concentration.

### Step-by-Step Analysis:

1. Identifying the Reaction Type:
- The given reaction is exothermic because it releases energy (heat) as a product.

2. Effect of Temperature:
- In an exothermic reaction (where energy is released as a product), decreasing the temperature favors the formation of products. This is because the system will adjust to produce more heat and counteract the temperature decrease.
- Raising the temperature would favor the reactants to absorb the extra heat and shift the equilibrium to the left (towards reactants).

3. Effect of Pressure:
- The number of moles of gas on the reactants side is 3 (2 moles of SO₂ and 1 mole of O₂).
- The number of moles of gas on the products side is 2 (2 moles of SO₃).
- Increasing the pressure on the system will favor the side with fewer moles of gas. In this case, increasing pressure will shift the equilibrium towards the products, which have fewer moles of gas (2 moles) compared to the reactants (3 moles).
- Decreasing the pressure would shift the equilibrium toward the side with more moles of gas (reactants side).

### Conclusion:
The changes that favor the formation of products are:
- A. Decreasing the temperature.
- B. Raising the pressure.

Thus, the choices that would make this reaction favor the formation of products are:
- (1) Decreasing the temperature.
- (2) Raising the pressure.

The corresponding choices are A and B.