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 500. mL flask with 3.7 atm of sulfur dioxide gas and 2.3 atm of oxygen gas, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be 2.2 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.

Answer 1

The pressure equilibrium constant is
`K_p = 323`

Step-by-step explanation:

From the question we are told that

The volume of the flask is
`V = 50 mL = 50 *10^(-3) L`

The pressure of sulfur dioxide is
`P_s = 3.7 \ atm`

The pressure of oxygen gas
`P_o = 2.3 \ atm`

The pressure of sulfur trioxide at equilibrium is
`P_t = 2.2 \ atm`

The chemical equation for this reaction is

`2 SO_2_((g)) + O_2_((g))` ⇄
`2SO_3_((g))`

The partial pressure of oxygen at equilibrium is mathematically evaluated as

`P_p__(O)} = P_o - P_t`

Substituting values

`P_p__(O)} = 2.3 -2.2`

`P_p__(O)} = 0.1 \ atm`

The partial pressure of sulfur dioxide at equilibrium is mathematically evaluated as

`P_p__(s)} = P_s - P_t`

Substituting values

`P_p__(S)} = 3.7 -2.2`

`P_p__(O)} = 1.5 \ atm`

From the chemical equation pressure constant is mathematically represented as

`K_p = ([P_t]^2)/([P_p__(o)) ]^2 [P_p__(s)}]}`

Substituting values

`K_p = \frac{[2.2]^2}{[ 0.1 ]^2 [{ 1.5}]}`

`K_p = 323`