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.

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

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)

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

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: \((0.125/1) \times 0.900 \; atm = 0.112 \; atm\)

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.