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For the reaction [tex]\(2 N_2O(g) = O_2(g) + 2 N_2(g)\)[/tex], what happens to the equilibrium position if the pressure increases?

A. Shifts to the right
B. Does nothing
C. Halves
D. Doubles
E. Shifts to the left

Sagot :

GinnyAnsweridn
Let's analyze the reaction:

[tex]\[ 2 \text{N}_2 \text{O} (g) \rightleftharpoons \text{O}_2 (g) + 2 \text{N}_2 (g) \][/tex]

This reaction involves gases, so according to Le Chatelier's Principle, we need to examine the effect of pressure changes on this equilibrium.

1. Count the moles of gas on each side of the reaction:
- On the left side (reactants), we have 2 moles of [tex]\( \text{N}_2 \text{O} \)[/tex].
- On the right side (products), we have 1 mole of [tex]\( \text{O}_2 \)[/tex] and 2 moles of [tex]\( \text{N}_2 \)[/tex].

So, there are:
- 2 moles of gas on the left side.
- 3 moles of gas on the right side.

2. Applying Le Chatelier's Principle:
- When the pressure increases, the equilibrium will shift towards the side of the reaction that has fewer moles of gas. This is because increasing pressure favors the direction that decreases the number of gas molecules, thus relieving the pressure.

3. Determine the direction of the shift:
- Since the left side has 2 moles of gas and the right side has 3 moles of gas, an increase in pressure will shift the equilibrium towards the left side (the side with fewer moles of gas).

Therefore, when the pressure increases, the equilibrium position will shift to the left.

So, the correct answer is:
- shifts to the left
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