Which of the following equilibrium reactions is shifted to the right by increasing the pressure?A) C(s) + O2(g) CO2(g)B) 2 NO (g) + O2 (g) 2 NO2 (g)C) CO2(g) + NO(g) CO(g) + NO2(g)D) 2 SO3 (g) 2 SO2 (g) + O2 (g)E) H2(g) + I2(g) 2 HI(g)
Question
Which of the following equilibrium reactions is shifted to the right by increasing the pressure?A) C(s) + O2(g) CO2(g)B) 2 NO (g) + O2 (g) 2 NO2 (g)C) CO2(g) + NO(g) CO(g) + NO2(g)D) 2 SO3 (g) 2 SO2 (g) + O2 (g)E) H2(g) + I2(g) 2 HI(g)
Solution
The equilibrium of a reaction is affected by pressure changes according to Le Chatelier's principle. This principle states that if a system at equilibrium is subjected to a change in pressure, temperature, or concentration, the system will shift its equilibrium position to counteract the effect of the disturbance.
In the context of pressure, the system will shift towards the side with fewer moles of gas to decrease the pressure, or towards the side with more moles of gas to increase the pressure.
Let's analyze each reaction:
A) C(s) + O2(g) ⇌ CO2(g) - The number of moles of gas doesn't change, so increasing the pressure doesn't shift the equilibrium.
B) 2 NO (g) + O2 (g) ⇌ 2 NO2 (g) - The number of moles of gas doesn't change, so increasing the pressure doesn't shift the equilibrium.
C) CO2(g) + NO(g) ⇌ CO(g) + NO2(g) - The number of moles of gas doesn't change, so increasing the pressure doesn't shift the equilibrium.
D) 2 SO3 (g) ⇌ 2 SO2 (g) + O2 (g) - The number of moles of gas increases from 2 to 3. Therefore, increasing the pressure will shift the equilibrium to the left, not the right.
E) H2(g) + I2(g) ⇌ 2 HI(g) - The number of moles of gas decreases from 2 to 1. Therefore, increasing the pressure will shift the equilibrium to the right.
So, the answer is E) H2(g) + I2(g) ⇌ 2 HI(g).
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