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Consider the following reaction: 2SO₂(g) + O₂(g) =2SO₃(g) where delta H_rxn = -198 kJ. The amount of SO₂(g) at equilibrium increases when...

a-the temperature is decreased
b-more oxygen is added
c-SO₃ is removed
d-the volume is increased.

User JohnUS
by
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1 Answer

6 votes

Final answer:

The amount of SO₂ at equilibrium in the reaction 2SO₂(g) + O₂(g) = 2SO₃(g) increases when the volume of the system is increased, due to the shift in equilibrium towards the reactants side to counter the change.

Step-by-step explanation:

Considering the reaction:

2SO₂(g) + O₂(g) = 2SO₃(g) with a delta H_rxn of -198 kJ, we can analyze the effect of various changes on the equilibrium concentration of SO₂(g). Given that the reaction is exothermic (negative delta H_rxn), applying Le Chatelier's Principle allows us to predict the response of the system to different disturbances:

  • (a) Temperature Decrease: Lowering the temperature favors the exothermic reaction, hence increasing the production of SO₃ and reducing the concentration of SO₂.
  • (b) Adding Oxygen: More SO₃ will be formed, consuming more SO₂.
  • (c) Removing SO₃: This shifts the equilibrium to the right, promoting the consumption of SO₂ and O₂ to form more SO₃.
  • (d) Increasing Volume: This results in a decrease in the concentration of gases, favoring the direction with more gas molecules, which is the reverse reaction in this case, thus increasing the amount of SO₂.

Hence, the amount of SO₂ at equilibrium increases when the volume is increased.

User Stuckedunderflow
by
8.2k points
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