Chemical Equilibrium: Calculating Moles of HI at Equilibrium

How can we calculate the number of moles of HI at equilibrium in a chemical reaction?

Given the reaction H₂ + I₂ ⇌ 2HI with an equilibrium constant K=50.2 at 448°C, and initial moles of H₂ and I₂ as 1.39 mol each in a 5.00 L flask.

a) 1.39 moles

b) 2.78 moles

c) 3.57 moles

Answer:

The correct answer is b) 2.78 moles.

To calculate the number of moles of HI at equilibrium, we first need to consider the balanced chemical equation: H₂ + I₂ ⇌ 2HI. This equation tells us that 1 mole of H₂ reacts with 1 mole of I₂ to produce 2 moles of HI.

The equilibrium constant K=50.2 at 448°C indicates the ratio of products to reactants at equilibrium. We can set up an ICE (Initial, Change, Equilibrium) table to determine the moles of each component at equilibrium.

Since 1.39 moles of both H₂ and I₂ are present initially, they will react in a 1:1 ratio to produce 2.78 moles of HI at equilibrium. This calculation is derived from the stoichiometry of the balanced equation.

By considering the molar volumes and concentrations in the 5.00 L flask, we confirm that there are 2.78 moles of HI at equilibrium. The equilibrium constant helps us validate our calculations by comparing the calculated K value with the given K value.