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 50.0 l tank with 4.1 mol of sulfur dioxide gas and 4.0 mol of oxygen gas, and when the mixture has come to equilibrium measures the amount of sulfur trioxide gas to be 2.1 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.

Explanation:

Step1: Write the balanced chemical equation

$$2SO_{2}(g)+O_{2}(g) ightleftharpoons 2SO_{3}(g)$$

Step2: Calculate the initial concentrations

The volume of the tank \(V = 50.0\space L\).
The initial concentration of \(SO_{2}\), \(C_{SO_{2},initial}=\frac{4.1\space mol}{50.0\space L}=0.082\space M\)
The initial concentration of \(O_{2}\), \(C_{O_{2},initial}=\frac{4.0\space mol}{50.0\space L}=0.080\space M\)
The initial concentration of \(SO_{3}\), \(C_{SO_{3},initial} = 0\space M\)

Step3: Calculate the equilibrium concentrations

The equilibrium amount of \(SO_{3}\) is \(n_{SO_{3},equilibrium}=2.1\space mol\).
The equilibrium concentration of \(SO_{3}\), \(C_{SO_{3},equilibrium}=\frac{2.1\space mol}{50.0\space L}=0.042\space M\)
From the stoichiometry of the reaction, for every \(2\) moles of \(SO_{3}\) formed, \(2\) moles of \(SO_{2}\) are consumed and \(1\) mole of \(O_{2}\) is consumed.
The change in concentration of \(SO_{2}\), \(\Delta C_{SO_{2}}=- 0.042\space M\)
The equilibrium concentration of \(SO_{2}\), \(C_{SO_{2},equilibrium}=0.082 - 0.042=0.040\space M\)
The change in concentration of \(O_{2}\), \(\Delta C_{O_{2}}=-\frac{0.042}{2}=- 0.021\space M\)
The equilibrium concentration of \(O_{2}\), \(C_{O_{2},equilibrium}=0.080-0.021 = 0.059\space M\)

Step4: Write the expression for \(K_{c}\)

$$K_{c}=\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}$$
Substitute the equilibrium concentrations:
$$K_{c}=\frac{(0.042)^{2}}{(0.040)^{2}\times0.059}$$
$$K_{c}=\frac{0.001764}{0.0016\times0.059}$$
$$K_{c}=\frac{0.001764}{9.44\times10^{-5}}$$
$$K_{c}\approx19$$

Answer:

\(19\)