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2 votes
The system below was at equilibrium in a

2.0 L container. What change will occur
for the system when the container is
shrunk to 1.0 L?
2NO(g) + O₂(g) + 2NO₂(g) + 113.06 kJ
Hint: How many moles of gas are on each side?
A. There is no change because there are the same
number of moles of gas on both sides.
B. The reactions shifts to the right (products) to produce
fewer moles of gas.
C. The reactions shifts to the left (reactants) to produce
more moles of gas.

The system below was at equilibrium in a 2.0 L container. What change will occur for-example-1

1 Answer

4 votes
To determine the change that will occur when the container is shrunk from 2.0 L to 1.0 L, we need to consider the effect on the equilibrium position of the given reaction: 2NO(g) + O₂(g) ⇌ 2NO₂(g) + 113.06 kJ.

The hint provides a clue about the number of moles of gas on each side of the equation. Let's analyze the equation:

On the left side, we have:
2 moles of NO
1 mole of O₂

On the right side, we have:
2 moles of NO₂

Comparing the number of moles of gas on each side, we find that there are 3 moles of gas on the left side and 2 moles of gas on the right side.

When the container is shrunk from 2.0 L to 1.0 L, the volume is reduced, which increases the pressure. According to Le Chatelier's principle, an increase in pressure will cause the equilibrium to shift in the direction that produces fewer moles of gas.

In this case, since there are fewer moles of gas on the right side of the equation (2 moles) compared to the left side (3 moles), the reaction will shift to the right (products) to produce fewer moles of gas.

Therefore, the correct answer is B. The reaction shifts to the right (products) to produce fewer moles of gas.

I hope this helps! :)
User Vasaka
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