Question

We know that the decomposition of SO₂Cl₂ is first-order in SO₂Cl₂,

SO₂Cl₂ (g) → SO₂(g) + Cl₂(g)

with a half-life of 245 minutes at 600 K.

If you begin with a partial pressure of SO₂Cl₂ of 25 mm Hg in a 1.0-L flask, What is the partial pressure of each reactant and product after 245 minutes?

Answer 1

After one half-life of 245 minutes, the partial pressure of SO₂Cl₂ will be 12.5 mm Hg, and the partial pressure of both SO₂ and Cl₂ will be 6.25 mm Hg each.

Step-by-step explanation:

The question is about the decomposition of sulfuryl chloride (SO₂Cl₂) under first-order kinetics. Given the half-life of 245 minutes at 600 K, we can determine the partial pressure of the reactants and products after 245 minutes (one half-life) from the initial partial pressure of 25 mm Hg in a 1.0-L flask.

Since the reaction is first-order, after one half-life, the concentration (and thus partial pressure) of the reactant SO₂Cl₂ will be reduced by half. Therefore:

• Partial pressure of SO₂Cl₂ after 245 minutes: 25 mm Hg / 2 = 12.5 mm Hg
• Since SO₂ and Cl₂ are produced in a 1:1 ratio, the partial pressure for each will be half of the decrease in SO₂Cl₂, which is (25 - 12.5) / 2 = 6.25 mm Hg each.