In which of the following reactions would the yield of product be increased by lowering the pressure? A) H$_2$(g) + I$_2$(g) ⇌ 2HI(g) B) N$_2$(g) + 3H$_2$(g) ⇌ 2NH$_3$(g) C) N$_2$O(g) ⇌ 2NO(g) D) CO(g) + 2H$_2$(g) ⇌ CH$_3$OH(g) - Scottish Highers Chemistry - Question 23 - 2019

To determine which reaction's product yield increases by lowering the pressure, we apply Le Chatelier's principle. This principle states that if a system at equilibrium is subjected to a change in pressure, the equilibrium will shift towards the side with fewer moles of gas to minimize the change.

Let's analyze each option:

A) H$_2$(g) + I$_2$(g) ⇌ 2HI(g): The number of moles of gas on both sides is equal (2 moles versus 2 moles), so changing the pressure will not affect the yield.

B) N$_2$(g) + 3H$_2$(g) ⇌ 2NH$_3$(g): There are 4 moles of gas on the reactant side and 2 moles on the product side, hence a decrease in pressure would shift the equilibrium toward the reactants, decreasing the yield of NH$_3$.

C) N$_2$O(g) ⇌ 2NO(g): There is 1 mole of gas on the reactant side and 2 moles of gas on the product side. Lowering the pressure would shift the equilibrium to the side with fewer moles of gas, which means it would favor the reactants, again reducing the yield of NO.

D) CO(g) + 2H$_2$(g) ⇌ CH$_3$OH(g): The reactants involve 3 moles of gas while the product is a single molecule (1 mole). Lowering the pressure would favor the reaction forming methanol, thus increasing its yield.

Therefore, the correct answer is C) N$_2$O(g) ⇌ 2NO(g) because it is the only reaction that would show a change in equilibrium if the pressure decreases.