Question

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a

75.0 L tank with 3.8 mol of sulfur dioxide gas and 7.0 mol of oxygen gas, and when the mixture has come to equilibrium measures the amount of sulfur trioxide
gas to be 1.5 mol
Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2
significant digits.

Answer 1

`\large \boxed{5.1}`

Step-by-step explanation:

1. Initial concentrations of reactants

[SO₂] = (3.8 mol)/(75 L) = 0.0507 mol·L⁻¹

[O₂] = (7.0 mol)/(75 L) = 0.0933 mol·L⁻¹

2. Equilibrium concentration of SO₃

[SO₃] = (1.5 mol)/(75 L) = 0.0200 mol·L⁻¹

3. Set up an ICE table

2SO₂ + O₂ ⇌ 2SO₃

I/mol·L⁻¹: 0.0507 0.0933 0

C/mol·L⁻¹: -2x -x +2x

E/mol·L⁻¹: 0.0507-2x 0.0933-x 2x

4. Calculate x

We know the final concentration of SO₃ is 0.0200 mol·L⁻¹, so

2x = 0.0200

x = 0.0100

5. Find the final concentrations of the reactants

Insert the numbers into the ICE table.

2SO₂ + O₂ ⇌ 2SO₃

I/mol·L⁻¹: 0.0507 0.0933 0

C/mol·L⁻¹: -0.0200 -0.0100 +0.0200

E/mol·L⁻¹: 0.0307 0.0833 0.0200

6. Calculate K

`K_{\text{eq}} = \frac{\text{[SO$_(3)$]}^(2)}{\text{[SO}_(2)]^(2)\text{[O$_(2)$]}} = (0.0200^(2))/(0.0307^(2)*0.0833) =\mathbf{5.1}\\\\\text{The value of the equilibrium constant is $\large \boxed{\mathbf{5.1}}$}`