Question

At \(25^{\circ} \mathrm{C},\) gaseous \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}(g)\) and \(\mathrm{Cl}_{2}(g)\) to the extent that 12.5\(\%\) of the original \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (by moles) has decomposed to reach equilibrium. The total pressure (at equilibrium) is 0.900 atm. Calculate the value of \(K_{\mathrm{p}}\) for this system.

Short answer

The value of \(K_p\) for the decomposition of \(SO_2Cl_2\) at \(25^{\circ}C\) and 0.900 atm is 0.0159.

Step-by-step solution

Write the balanced chemical equation

The decomposition of SO2Cl2 can be represented by the following balanced chemical equation: \(SO_2Cl_2(g) \rightleftharpoons SO_2(g) + Cl_2(g)\)

Determine initial concentrations of reactants and products

We are told that initially, 12.5% of the SO2Cl2 has decomposed. Let's assume that we started with 1 mole of SO2Cl2. This means that at the time that the reaction reaches equilibrium, there will be: - 0.875 moles of SO2Cl2 (since 12.5% has decomposed) - 0.125 moles of SO2 and Cl2 (since 12.5% of the initial SO2Cl2 has decomposed into these products)

Calculate the change in concentrations at equilibrium

Since we are given the total pressure at equilibrium (0.900 atm), we need to determine the volume of the container in order to calculate the concentrations. We can use the Ideal Gas Law to find the volume: \(PV = nRT\) Where P is pressure, V is volume, n is the number of moles, R is the gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin. First, we need to convert the temperature to Kelvin: \(T = 25^\circ C + 273.15 = 298.15 K\) We also know that at equilibrium, there are a total of 1 mole of gas (0.875 moles of SO2Cl2 + 0.125 moles of SO2 + 0.125 moles of Cl2). Thus: \(V = \frac{nRT}{P} = \frac{(1 \; mole)(0.0821 \; L \; atm/mol \; K)(298.15 \; K)}{0.900 \; atm} = 27.297 \; L\) Now we can find the equilibrium concentrations: - [SO2Cl2] = 0.875 moles / 27.297 L = 0.0320 M - [SO2] = 0.125 moles / 27.297 L = 0.00458 M - [Cl2] = 0.125 moles / 27.297 L = 0.00458 M

Find the partial pressures of each species

To find the partial pressures, we can use the mole fractions and the total pressure: - Partial pressure of SO2Cl2: \((0.875/1) \times 0.900 \; atm = 0.788 \; atm\) - Partial pressure of SO2: \((0.125/1) \times 0.900 \; atm = 0.112 \; atm\) - Partial pressure of Cl2: \((0.125/1) \times 0.900 \; atm = 0.112 \; atm\)

Calculate the value of K_p

To find K_p, we use the balanced chemical equation and the partial pressures at equilibrium: \(K_p = \frac{P_{SO_2} \times P_{Cl_2}}{P_{SO_2Cl_2}} = \frac{(0.112 \; atm)(0.112 \; atm)}{0.788 \; atm} = 0.0159\) Therefore, the value of K_p for this system is 0.0159.

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs in a closed system when the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of reactants and products remain constant over time. It does not mean the reactants and products are equal in concentration or pressure, but they remain stable.

In the decomposition of \(\mathrm{SO}_2\mathrm{Cl}_2\), equilibrium is reached when the formation of \(\mathrm{SO}_2\) and \(\mathrm{Cl}_2\) from \(\mathrm{SO}_2\mathrm{Cl}_2\) happens at the same rate as the recombination of \(\mathrm{SO}_2\) and \(\mathrm{Cl}_2\) back into \(\mathrm{SO}_2\mathrm{Cl}_2\). Thus, the composition of the system does not change over time when equilibrium is reached.

A critical factor in chemical equilibrium is the equilibrium constant, \(K_p\), which is determined by the ratio of the partial pressures of the products to the reactants, each raised to the power of their stoichiometric coefficients, at equilibrium. The magnitude of \(K_p\) gives insight into the position of the equilibrium:

• A smaller \(K_p\) indicates the equilibrium position favors reactants.
• A larger \(K_p\) suggests a shift towards products.

Ideal Gas Law

The Ideal Gas Law is an equation of state for an ideal gas which describes the relationships among the four variables: pressure (P), volume (V), temperature (T), and the amount of substance in moles (n). This fundamental equation is given by:

\(PV = nRT\)

where R is the ideal gas constant (0.0821 L atm/mol K). This law helps us understand how gases behave under different conditions.

In the given exercise, the Ideal Gas Law is used to calculate the volume of the system when the total pressure and temperature are known. By rearranging the equation, we can find the volume needed to reach equilibrium:\[V = \frac{nRT}{P}\]This calculation enables us to determine concentrations of different species in the reaction, which are critical for finding the equilibrium constant.

Partial Pressure

Partial pressure is the pressure exerted by a single gas in a mixture of gases. It is crucial for finding equilibrium constants involving gases, like \(K_p\). Each gas in a mixture behaves as if it occupies the whole volume of the container by itself.

To find the partial pressure:

• Use the mole fraction of the gas and multiply it by the total pressure of the mixture.

In the context of our \(\mathrm{SO}_2\mathrm{Cl}_2\) decomposition reaction, the partial pressures are calculated based on the mole fractions of each gas:

• The mole fraction of \(\mathrm{SO}_2\mathrm{Cl}_2\) is \(0.875\), leading to a partial pressure of \(0.788\ \, atm\).
• The mole fractions for \(\mathrm{SO}_2\) and \(\mathrm{Cl}_2\) are both \(0.125\), resulting in partial pressures of \(0.112\ \, atm\) each.

Recognizing the partial pressures helps to plug the values into the equilibrium constant expression.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where a single compound breaks down into two or more simpler substances. It often requires energy input and produces gaseous products from a solid or liquid reactant.

In this exercise, the decomposition of \(\mathrm{SO}_2\mathrm{Cl}_2\) into \(\mathrm{SO}_2\) and \(\mathrm{Cl}_2\) represents the decomposition reaction itself:\[\mathrm{SO}_2\mathrm{Cl}_2(g) \rightleftharpoons \mathrm{SO}_2(g) + \mathrm{Cl}_2(g)\]This reaction involves spontaneous splitting at a specific temperature till equilibrium is reached.

Decomposition reactions are significant both in laboratories and natural processes, showcasing the breakdown of complex substances into simpler ones, often driven by external agents like heat or light. In our specific exercise, it illustrates a real-world application of finding measurable quantities like \(K_p\) to assess reaction extent in equilibrium scenarios.