Question

a mixture of reactants and products for the reaction shown below is at equilibrium in a 5.0 l container. what would most likely happen to the equilibrium if the volume of the container were reduced to 3.0 l?
n₂(g) + 3h₂(g) ⇌ 2nh₃(g)
a. more n₂ and h₂ would be produced until a new equilibrium position was reached.
b. more nh₃ would be produced until a new equilibrium position was reached.
c. more n₂, h₂, and nh₃ would be produced until a new equilibrium position was reached.
d. the concentrations of n₂, h₂, and nh₃ would decrease until equilibrium was reached again.

Explanation:

To solve this, we use Le Chatelier's principle. The reaction is \( \ce{N2(g) + 3H2(g) <=> 2NH3(g)} \). When the volume of the container is reduced, the pressure increases. The system will shift to the side with fewer moles of gas to counteract the pressure increase. On the reactant side, there are \( 1 + 3 = 4 \) moles of gas, and on the product side, there are 2 moles of gas. So the equilibrium will shift to the right (towards \( \ce{NH3} \)) to produce more \( \ce{NH3} \) until a new equilibrium is reached.

• Option A is incorrect because producing more \( \ce{N2} \) and \( \ce{H2} \) would be a shift to the left, which is opposite of what is needed.
• Option C is incorrect as all three can't be produced more; the shift is towards the side with fewer moles.
• Option D is incorrect because reducing volume increases concentration initially, and the system shifts to re - establish equilibrium, not decrease all concentrations.

Answer:

B. More \( \ce{NH3} \) would be produced until a new equilibrium position was reached.