Question

An engineer examining the oxidation of \(\mathrm{SO}_{2}\) in the manufacture of sulfuric acid determines that \(K_{\mathrm{c}}=1.7 \times 10^{8}\) at \(600 . \mathrm{K}:\) $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ (a) At equilibrium, \(P_{\mathrm{SO}_{3}}=300 .\) atm and \(P_{\mathrm{O}_{2}}=100 .\) atm. Calculate \(P_{\mathrm{sO}_{2}}\). (b) The engineer places a mixture of \(0.0040 \mathrm{~mol}\) of \(\mathrm{SO}_{2}(g)\) and \(0.0028 \mathrm{~mol}\) of \(\mathrm{O}_{2}(g)\) in a 1.0 -L container and raises the temperature to \(1000 \mathrm{~K}\). At equilibrium, \(0.0020 \mathrm{~mol}\) of \(\mathrm{SO}_{3}(g)\) is present. Calculate \(K_{c}\) and \(P_{\mathrm{SO}_{2}}\) for this reaction at \(1000 . \mathrm{K}\).

Short answer

For part (a), \(P_{\mathrm{SO_{2}}} \approx 0.073 \) atm. For part (b), \(K_{c} \approx 555.6\)

Step-by-step solution

Write the Equilibrium Expression

For the reaction \[2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\] the equilibrium constant expression in terms of partial pressures is given by: \[K_{p} = \frac{{(P_{\mathrm{SO}_{3}})^{2}}}{{(P_{\mathrm{SO}_{2}})^{2} \cdot P_{\mathrm{O}_{2}}}}\]

Solve for \(P_{\mathrm{SO}_{2}}\) in Part (a)

Given:\(K_{p} = 1.7 \times 10^{8}\)\(P_{\mathrm{SO}_{3}} = 300\) atm\(P_{\mathrm{O}_{2}} = 100\) atmSubstitute these into the equilibrium expression:\[1.7 \times 10^{8} = \frac{{(300)^{2}}}{{(P_{\mathrm{SO}_{2}})^{2} \cdot 100}}\]Solve for \(P_{\mathrm{SO}_{2}}\):\[(1.7 \times 10^{8}) \cdot 100 = \frac{{90000}}{{(P_{\mathrm{SO}_{2}})^{2}}}\]\[1.7 \times 10^{10} = \frac{{90000}}{{(P_{\mathrm{SO}_{2}})^{2}}}\]\[P_{\mathrm{SO}_{2}} = \sqrt{\frac{{90000}}{{1.7 \times 10^{10}}}}\]\[P_{\mathrm{SO}_{2}} \approx 0.073\] atm

Initial Conditions for Part (b)

Place 0.0040 mol of \(\mathrm{SO}_{2}(g)\) and 0.0028 mol of \(\mathrm{O}_{2}(g)\) in a 1.0 L container:\([\mathrm{SO}_{2}]_{0} = 0.0040\) M\([\mathrm{O}_{2}]_{0} = 0.0028\) MAt equilibrium, 0.0020 mol of \(\mathrm{SO}_{3}(g)\) is present:\([\mathrm{SO}_{3}]_{eq} = 0.0020\) M

Changes in Concentration

Since 2 moles of SO\(\mathrm{SO}_{3}\) are produced from 2 moles of \(\mathrm{SO}_{2}\) and 1 mole of \(\mathrm{O}_{2}\), the changes in concentration are:\(\mathrm{SO}_{2}: 0.0040 - 2x\)\(\mathrm{O}_{2}: 0.0028 - x\)\(\mathrm{SO}_{3}: 0.0020\)From stoichiometry: \(x = 0.0010\)

Calculate Equilibrium Concentrations

Substitute \(x = 0.0010\) M:\([\mathrm{SO}_{2}]_{eq} = 0.0040 - 2 \times 0.0010 = 0.0020\) M\([\mathrm{O}_{2}]_{eq} = 0.0028 - 0.0010 = 0.0018\) M\([\mathrm{SO}_{3}]_{eq} = 0.0020\) M

Calculate \(K_{c}\)

Substitute equilibrium concentrations into the equilibrium expression:\[K_{c} = \frac{{[\mathrm{SO}_{3}]^{2}}}{{[\mathrm{SO}_{2}]^{2} [\mathrm{O}_{2}]}}\]\[K_{c} = \frac{{(0.0020)^{2}}}{{(0.0020)^{2} (0.0018)}} = \frac{{0.000004}}{{0.000000004\times 0.0018}} \approx 555.6\]

Calculate \(P_{\mathrm{SO}_{2}}\) at 1000 K

Use the partial pressure formula in the same format as earlier. For the equilibrium partial pressures:Given in consistency with the concentration, calculate the partial pressure using:\[P_{\text{SO}2} = (R \cdot T \cdot P)/n \]:All concentration units apply conformity.

Key concepts

chemical equilibrium

To understand how chemical equilibrium works, imagine a reversible chemical reaction where the reactants convert into products and the products can also convert back into the reactants. At equilibrium, the rate at which reactants turn into products equals the rate at which products turn back into reactants. This balance means the concentrations of all species remain constant over time.

Equilibrium does not mean the reactants and products are in equal amounts; it just indicates that their concentrations do not change. This dynamic balance is key to solving equilibrium problems, like in the oxidation of \(\text{SO}_2\) for sulfuric acid production mentioned in the exercise.

partial pressure

Partial pressure defines the pressure exerted by a single type of gas in a mixture of gases. It's an important concept in calculating equilibrium constants for reactions involving gases.

For example, in the reaction given, we have partial pressures for each gas: \(\text{SO}_2\), \(\text{O}_2\), and \(\text{SO}_3\). Each gas's partial pressure adds up to the total pressure in the container. Using these values is crucial for solving the equilibrium problem.

It's helpful to remember that the partial pressure of a gas is directly proportional to its mole fraction in the mixture and the total pressure.

reaction stoichiometry

Reaction stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. This relationship helps us determine how the concentration of one substance changes relative to another during the reaction.

In the given example, the reaction \(\text{2 SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons \text{2 SO}_3(\text{g})\) shows that 2 moles of \(\text{SO}_2\) react with 1 mole of \(\text{O}_2\) to produce 2 moles of \(\text{SO}_3\). Knowing these ratios is essential for solving for equilibrium concentrations.

For instance, if you have initial amounts of \(\text{SO}_2\) and \(\text{O}_2\), you can use stoichiometry to find the changes in their concentrations as the reaction proceeds.

equilibrium constant (Kc)

The equilibrium constant \(\text{Kc}\) expresses the ratio of the concentrations of products to reactants at equilibrium for a reaction occurring in a closed system. Each concentration is raised to the power of the coefficient from the balanced chemical equation.

For the given reaction, \(\text{Kc} = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} \), which allows you to determine the equilibrium concentrations if you know \(\text{Kc}\) and starting concentrations.

In the exercise, we calculated \(\text{Kc}\) at higher temperatures, which is crucial for understanding how equilibrium shifts with temperature changes.

equilibrium constant (Kp)

The equilibrium constant \(\text{Kp}\) is similar to \(\text{Kc}\) but used for gaseous reactions, involving partial pressures instead of concentrations. The relation between \(\text{Kc}\) and \(\text{Kp}\) involves the ideal gas constant and the temperature, given by the formula: \(\text{Kp} = \text{Kc}(RT)^{\text{Δn}}\), where \(\text{Δn}\) is the change in moles of gas.

In the problem, \(\text{Kp}\) is given as \(\text{1.7 × 10}^8\) for the reaction at 600 K. Using \(\text{Kp}\), you can solve for unknown partial pressures when you know the equilibrium partial pressures of other gases.

Understanding how \(\text{Kp}\) works helps you analyze gas-phase reactions' equilibrium conditions.