To determine which stress will shift the equilibrium of the given reaction to the left, let's analyze each option based on Le Chatelier's principle. The principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
The given equilibrium 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]. Since [tex]\(\Delta H\)[/tex] is negative, the reaction is exothermic (releases heat).
Now, let's analyze each type of stress:
1. Decreasing the concentration of SO[tex]\(_3\)[/tex]:
- Decreasing the concentration of a product will shift the equilibrium to the right, towards the products, to produce more SO[tex]\(_3\)[/tex].
2. Decreasing volume:
- Decreasing the volume of the reaction vessel increases the pressure. According to Le Chatelier's principle, the system will shift to the side with fewer moles of gas to counteract this increase in pressure.
- In this reaction, there are 3 moles of gas on the reactant side (2 SO[tex]\(_2\)[/tex] + 1 O[tex]\(_2\)[/tex]) and 2 moles of gas on the product side (2 SO[tex]\(_3\)[/tex]). Therefore, decreasing the volume (increasing pressure) will shift the equilibrium to the right, towards the side with fewer moles of gas.
3. Increasing the concentration of SO[tex]\(_2\)[/tex]:
- Increasing the concentration of a reactant will shift the equilibrium to the right, towards the products, to consume the excess reactant.
4. Increasing temperature:
- For an exothermic reaction, increasing the temperature adds heat to the system. According to Le Chatelier's principle, the system will shift to counteract this by absorbing the extra heat, which means the equilibrium will shift to the left, towards the reactants, to favor the endothermic direction (absorbing heat).
Given these analyses, the stress that will shift the equilibrium to the left is:
Increasing temperature.
