When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 5.00 -Lflaskat \(310 \mathrm{~K}\), \(40 \%\) of the $\mathrm{SO}_{2} \mathrm{Cl}_{2}\( decomposes to \)\mathrm{SO}_{2}\( and \)\mathrm{Cl}_{2}$ : $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{P}\) for this reaction at \(310 \mathrm{~K}\). (c) According to Le Châtelier's principle, would the percent of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes increase, decrease or stay the same if the mixture was transferred to a 1.00 -L vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes when 2.00 mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(1.00-\mathrm{L}\) vessel at \(310 \mathrm{~K}\).

The balanced reaction equation is already given in the prompt: \(\mathrm{SO}_{2}\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) \)

First, let us find the initial concentration of SO2Cl2 in the 5.0 L flask: Initial concentration of SO2Cl2 = \(\frac{2.00\, \text{mol}}{5.00\, \text{L}} = 0.40\, \text{M}\) Then, we find the change in concentrations at equilibrium by utilizing the percentage of decomposition provided. Since \(40 \%\) of the SO2Cl2 decomposes, the change in concentration would be 0.4 times the initial concentration. Concentration change of SO2Cl2 = \(0.40 \times 0.40 = 0.16\, \text{M}\) Therefore, at equilibrium: [SO2Cl2] = \(0.40\, \text{M} - 0.16\, \text{M} = 0.24\, \text{M}\) [SO2] = [Cl2] = \(0.16\, \text{M}\)

Now, calculate Kc with the equilibrium concentrations: \(K_c = \frac{\text{[SO2]}\times\text{[Cl2]}}{\text{[SO2Cl2]}} = \frac{0.16\, \text{M} \times 0.16\, \text{M}}{0.24\, \text{M}} = 0.107\) To find conversion factor to calculate Kp, we need the given temperature, which is 310 K. \(K_p = K_c(RT)^{\Delta n}\) where R is the ideal gas constant (\(8.314\, \frac{\text{J}}{\text{mol K}}\)). In this case, the reaction involves 1 mole of reactant and 2 moles of products, so ∆n = 2 - 1 = 1. Now calculate Kp: \(K_p = 0.107 (8.314\, \text{J/mol K} \times 310\, \text{K})^1 = 274.699\) Thus, Kc=0.107 and Kp=274.699.

Le Châtelier's Principle states that if a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the reactants or products, the system will adjust to counteract the change and restore equilibrium. In this case, we are decreasing the volume of the flask, causing an increase in pressure. Since the forward reaction produces more moles of gas (2 moles) than the reverse reaction (1 mole), the system will tend to move in the direction of the reverse reaction (consuming the products and producing more reactants) to counteract the increased pressure. Therefore, the percentage of SO2Cl2 decomposition would decrease.

Initial concentration of SO2Cl2 in a 1 L flask: \(\frac{2.00\, \text{mol}}{1.00\, \text{L}} = 2.00\, \text{M}\) Let x represent the change in concentration of SO2Cl2 due to decomposition: Equilibrium concentrations: [SO2Cl2] = \(2.00 \, \text{M} - x \, \text{M}\) [SO2] = [Cl2] = \(x\, \text{M}\) Now, we will use the calculated Kc value (0.107) to determine x: \(0.107 = \frac{x^2}{(2.00 - x)}\) Solving for x (using a quadratic equation solver), we get two possible values: ~0.803 M (rounded) or ~0.289 M. Since 0.803 M decomposition is not feasible and x should be lower than 1, we choose x = 0.289 M. The percentage of decomposition in a 1.00 L vessel can be calculated with the formula: \(\text{Percentage of decomposition} = \frac{\text{Decomposed SO2Cl2 concentration}}{\text{Initial SO2Cl2 concentration}} \times 100\%\) Inserting the respective values, we get: \(\text{Percentage of decomposition} = \frac{0.289\, \text{M}}{2.00\, \text{M}} \times 100\% = 14.45\% \) (rounded to 2 decimals)