Answer :
To determine which stress will shift the given equilibrium system to the left, we need to apply Le Chatelier's principle. Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. The given reaction is:
[tex]\[ 2 \text{SO}_{2(g)} + \text{O}_{2(g)} \rightleftharpoons 2 \text{SO}_{3(g)} \][/tex]
with
[tex]\[ \Delta H = -98.8 \ \text{kJ/mol} \][/tex]
This means the reaction is exothermic (it releases heat).
Let's consider each stress described and determine how it would affect the equilibrium:
1. Increasing Temperature:
- For an exothermic reaction (negative [tex]\(\Delta H\)[/tex]), increasing the temperature adds heat to the system. According to Le Chatelier's principle, the equilibrium will shift to the left to absorb some of this added heat, counteracting the change.
2. Decreasing Concentration of [tex]\( \text{SO}_3 \)[/tex]:
- Removing [tex]\( \text{SO}_3 \)[/tex] from the system reduces the concentration of one of the products. The equilibrium will shift to the right to produce more [tex]\( \text{SO}_3 \)[/tex], counteracting the decrease in its concentration.
3. Increasing Concentration of [tex]\( \text{SO}_2 \)[/tex]:
- Adding more [tex]\( \text{SO}_2 \)[/tex] increases the concentration of one of the reactants. The equilibrium will shift to the right to consume the added [tex]\( \text{SO}_2 \)[/tex], counteracting the increase in its concentration.
4. Decreasing Volume:
- Decreasing the volume of the system increases the pressure. The equilibrium will shift towards the side with fewer moles of gas to decrease the pressure. In this reaction, both sides have the same total number of moles of gas (3 moles on each side: 2 moles of [tex]\( \text{SO}_2 \)[/tex] and 1 mole of [tex]\( \text{O}_2 \)[/tex] on the left, 2 moles of [tex]\( \text{SO}_3 \)[/tex] on the right). Therefore, decreasing volume will have no effect on the direction of the shift in this particular case.
Given these analyses, the correct stress that will shift the equilibrium system to the left is:
Increasing temperature.
[tex]\[ 2 \text{SO}_{2(g)} + \text{O}_{2(g)} \rightleftharpoons 2 \text{SO}_{3(g)} \][/tex]
with
[tex]\[ \Delta H = -98.8 \ \text{kJ/mol} \][/tex]
This means the reaction is exothermic (it releases heat).
Let's consider each stress described and determine how it would affect the equilibrium:
1. Increasing Temperature:
- For an exothermic reaction (negative [tex]\(\Delta H\)[/tex]), increasing the temperature adds heat to the system. According to Le Chatelier's principle, the equilibrium will shift to the left to absorb some of this added heat, counteracting the change.
2. Decreasing Concentration of [tex]\( \text{SO}_3 \)[/tex]:
- Removing [tex]\( \text{SO}_3 \)[/tex] from the system reduces the concentration of one of the products. The equilibrium will shift to the right to produce more [tex]\( \text{SO}_3 \)[/tex], counteracting the decrease in its concentration.
3. Increasing Concentration of [tex]\( \text{SO}_2 \)[/tex]:
- Adding more [tex]\( \text{SO}_2 \)[/tex] increases the concentration of one of the reactants. The equilibrium will shift to the right to consume the added [tex]\( \text{SO}_2 \)[/tex], counteracting the increase in its concentration.
4. Decreasing Volume:
- Decreasing the volume of the system increases the pressure. The equilibrium will shift towards the side with fewer moles of gas to decrease the pressure. In this reaction, both sides have the same total number of moles of gas (3 moles on each side: 2 moles of [tex]\( \text{SO}_2 \)[/tex] and 1 mole of [tex]\( \text{O}_2 \)[/tex] on the left, 2 moles of [tex]\( \text{SO}_3 \)[/tex] on the right). Therefore, decreasing volume will have no effect on the direction of the shift in this particular case.
Given these analyses, the correct stress that will shift the equilibrium system to the left is:
Increasing temperature.
