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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.

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Answer:

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

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