The system below was at equilibrium in a 7.0 L container. What change will occur for the system when the container is shrunk to 2.5 L? 2SO2(g) + O₂(g) = 2SO3(g) + 198 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 left (reactants) to produce more moles of gas. C. The reactions shifts to the right (products) to produce fewer moles of gas.

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The system below was at equilibrium in a
7.0 L container. What change will occur
for the system when the container is
shrunk to 2.5 L?
2SO₂(g) + O₂(g) = 2SO3(g) + 198 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 left (reactants) to produce
more moles of gas.
C. The reactions shifts to the right (products) to produce
fewer moles of gas.
Transcribed Image Text:The system below was at equilibrium in a 7.0 L container. What change will occur for the system when the container is shrunk to 2.5 L? 2SO₂(g) + O₂(g) = 2SO3(g) + 198 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 left (reactants) to produce more moles of gas. C. The reactions shifts to the right (products) to produce fewer moles of gas.
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