Certainly! Let's analyze the chemical reaction and its shifts to understand the behavior of the components involved:
Given the reaction:
[tex]\[ 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) + 198 \text{kJ} \][/tex]
This is an exothermic reaction (releases heat).
### Understanding Reaction Shifts:
The reaction is described with a double arrow [tex]\(\rightleftharpoons\)[/tex], indicating it is in a dynamic equilibrium where it can proceed in both forward (to the right) and reverse (to the left) directions:
- Forward reaction: [tex]\(2 \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \text{SO}_3(g) + 198 \text{kJ}\)[/tex]
- Reverse reaction: [tex]\(2 \text{SO}_3(g) + 198 \text{kJ} \rightarrow 2 \text{SO}_2(g) + \text{O}_2(g)\)[/tex]
### Shift to the Left:
If the reaction shifts to the left, it means the equilibrium is moving towards forming more reactants and fewer products.
#### Changes When Reaction Shifts Left:
1. Products (SO[tex]\(_3\)[/tex]): The concentration of sulfur trioxide ([tex]\(\text{SO}_3\)[/tex]) will decrease because it is being converted back into reactants.
2. Reactants (SO[tex]\(_2\)[/tex] and O[tex]\(_2\)[/tex]): The concentrations of sulfur dioxide ([tex]\(\text{SO}_2\)[/tex]) and oxygen ([tex]\(\text{O}_2\)[/tex]) will increase as they are being formed from the product ([tex]\(\text{SO}_3\)[/tex]).
[tex]\[ 2 \text{SO}_3(g) \rightarrow 2 \text{SO}_2(g) + \text{O}_2(g) \][/tex]
Since [tex]\(\text{O}_2\)[/tex] is a reactant in the reverse reaction, as the reaction shifts to the left, more [tex]\(\text{O}_2\)[/tex] will be produced.
### Answer:
As the reaction shifts to the left:
- Amount of O[tex]\(_2\)[/tex]: The amount of oxygen ([tex]\(\text{O}_2\)[/tex]) increases because it is a product of the reverse reaction.
Hence, the correct answer is:
A. The amount of O[tex]\(_2\)[/tex] goes up.
