Problem 39 - Chemistry: The Central Science

Question

At \(100^{\circ} \mathrm{C}, K_{c}=0.078\) for the reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ In an equilibrium mixture of the three gases, the concentrations of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and \(\mathrm{SO}_{2}\) are \(0.108 \mathrm{M}\) and \(0.052 \mathrm{M}, \mathrm{re}\) - spectively. What is the partial pressure of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

Step-by-Step Solution

Answer

The partial pressure of \(\mathrm{Cl}_2\) in the equilibrium mixture is approximately \(0.162 \, atm\).

Step 1: Write the equilibrium constant expression

The equilibrium expression for the given reaction is: \[K_c = \frac{[\mathrm{SO}_2][\mathrm{Cl}_2]}{[\mathrm{SO}_2\mathrm{Cl}_2]}\]

Step 2: Identifying the given values and converting them into partial pressures:

We are given the following concentrations: \[ [\mathrm{SO_2\,Cl_2}] = 0.108 \, M\] \[ [\mathrm{SO_2}] = 0.052 \, M\] We are also given the value of Kc as 0.078: \[ K_c = 0.078 \] Now, let the partial pressure of Cl2 at the equilibrium be P. Therefore, \[ [\mathrm{Cl_2}] = P\]

Step 3: Substitute the given values into the equilibrium constant expression and solve for P:

Plugging in the given values into the Kc expression, we have: \[0.078 = \frac{(0.052)(P)}{(0.108)}\] Now, we need to solve for P. To do this, we can multiply both sides of the equation by 0.108: \[(0.078)(0.108) = (0.052)(P)\] Now, divide both sides by 0.052 to isolate P: \[P = \frac{(0.078)(0.108)}{(0.052)}\] Now, calculate P: \[P \approx 0.162 \]

Step 4: Conclusion

The partial pressure of Cl2 in the equilibrium mixture is approximately 0.162 atm.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_c\) for reactions in solution, is a crucial concept in chemistry. It provides a snapshot of a chemical reaction at equilibrium. At this point, the rate of the forward reaction equals the rate of the reverse reaction, leading to stable concentrations of reactants and products. For the reaction \( \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) \), the equilibrium constant is defined as:\[K_c = \frac{[\mathrm{SO}_2][\mathrm{Cl}_2]}{[\mathrm{SO}_2\mathrm{Cl}_2]}\]This expression is derived from the law of mass action, where the concentrations of the products are divided by the concentrations of the reactants, each raised to the power of their coefficients in the balanced equation. A high \(K_c\) value indicates a reaction that favors the production of products at equilibrium, while a low \(K_c\) suggests that reactants are favored. For the given reaction, \(K_c = 0.078\), meaning that at equilibrium, the reaction slightly favors reactants.

Partial Pressure

In gas-phase reactions, partial pressure plays a role similar to concentration in solutions. It refers to the pressure that a single gas component in a mixture would exert if it occupied the entire volume alone. For the reaction involving \( \mathrm{SO}_{2}, \mathrm{Cl}_{2}, \) and \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \), partial pressures are used in calculations to determine the state of equilibrium.When you know the equilibrium constants and concentrations, it's possible to calculate unknown partial pressures using the equilibrium constant expression. In this context, if the concentration of \( \mathrm{Cl}_2 \) is not directly known, we can represent it as \( P \). Substituting into the equilibrium expression allows us to solve for this partial pressure, which reveals how much \( \mathrm{Cl}_2 \) is produced or utilized at equilibrium. The ability to calculate partial pressures underscores how understanding equilibrium constants and reaction conditions can predict the behavior of gas mixtures.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of reactants and products in a reaction mixture when the rates of the forward and reverse reactions are equal. At this point, the system reaches a state where no further changes in concentration occur over time. For the reaction \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \rightleftharpoons \mathrm{SO}_{2} + \mathrm{Cl}_{2} \), we initially have concentrations provided for \( \mathrm{SO}_{2\mathrm{Cl}_2} \) and \( \mathrm{SO}_{2} \). These concentrations allow us to apply the equilibrium constant expression and find unknowns, such as the concentration of \( \mathrm{Cl}_{2} \) if needed.Calculating equilibrium concentration involves substituting known values into the equilibrium expression and solving for unknowns. This helps chemists understand the proportion of reactants and products at equilibrium under given conditions. Such knowledge allows for the prediction and manipulation of reaction yields in industrial and laboratory settings. The calculated concentrations confirm how the reaction parameters, such as temperature, affect the system's equilibrium state and finally, the output of desired products.