Consider the following reaction at equilibrium. What effect will reducing the pressure of the reaction mixture have on the system?

Xe(g) + 2 F2(g) XeF4(g)

Question

Consider the following reaction at equilibrium. What effect will reducing the pressure of the reaction mixture have on the system?

Xe(g) + 2 F2(g)  XeF4(g)

BlackTom AI · Accepted Answer

The question asks about the effect of reducing the pressure on an equilibrium involving gaseous species: Xe(g) + 2 F2(g) ⇌ XeF4(g).

First, recall Le Chatelier's principle: when a system at equilibrium is subjected to a change in pressure (or volume), it will shift in a direction that tends to counteract that change. For gaseous reactions, the pressure change is effectively related to the total number of moles of gas on each side.

Next, determine the number of moles of gas on each side of the equilibrium:
- Left side (reactants): Xe(g) + 2 F2(g) → total = 1 + 2 = 3 moles of gas.
- Right side (products): XeF4(g) → total = 1 mole of gas.

When the pressure is reduced, the system will favor the side with more gas moles, because increasing the number of gas molecules helps compensate for the reduction in pressure by occupying more volume. In this case, the left side has 3 moles, which is greater than the right side's 1 mole.

Therefore, the equilibrium will shift toward the left (toward the reactants) to increase the total number of gas molecules and partially counteract the pressure decrease.

Note: The provided answer option states that the reaction will shift to the left in the direction of reactants, which aligns with the reasoning above. Since the original set of answer choices is unavailable here, the analysis focuses on applying Le Chatelier’s principle and mole-number comparison to arrive at the expected directional shift.

Consider the following reaction at equilibrium. What effect will reducing the pressure of the reaction mixture have on the system? Xe(g) + 2 F2(g) XeF4(g)单项选择题

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