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 1.5 L flask with 0.59 atm of sulfur dioxide gas and 2.9 atm of oxygen gas at 35.0 °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be 0.53 atm.
Calculate the pressure equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.
Kp=_______.

Answer 1

Answer:

P SO₂ = 0.06atm

P O₂ = 2.635atm

P SO₃ = 0.53atm

Kp = 29.6

Explanation:

The reaction of Sulfur dioxide and oxygen react to form sulfur trioxide is as follows:

2SO₂(g) + O₂(g) ⇄ 2SO₃(g)

And Kp is defined as:

$Kp = \frac{P_{SO_3}^2}{P_{SO_2}^2P_{O_2}}$

Where P represents the pressure at equilibrium of each reactant.

If you add, in the first, 0.59atm of SO₂ and 2.9atm of O₂, the equilibrium pressures will be:

P SO₂ = 0.59atm - 2X

P O₂ = 2.9atm - X

P SO₃ = 2X

Where X represents the reaction coordiante.

As equilibrium pressure of SO₃ is 0.53atm:

0.53atm = 2X

0.265atm = X

Replacing, equilibrium pressures of each species will be:

P SO₂ = 0.59atm - 2×0.265atm

P O₂ = 2.9atm - 0.265atm

P SO₃ = 2×0.265atm

P SO₂ = 0.06atm

P O₂ = 2.635atm

P SO₃ = 0.53atm

And Kp will be:

$Kp = \frac{P_{SO_3}^2}{P_{SO_2}^2P_{O_2}}$

$Kp = \frac{0.53^2}{{0.06}^2*{2.635}}$

Kp = 29.6