Question

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 2.00-L flask at \(303 \mathrm{~K}, 56 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ Calculate \(K_{c}\) for this reaction at this temperature.

Short answer

The equilibrium constant, Kc, at 303 K for the given reaction is 0.7136.

Step-by-step solution

Calculate the Initial Molar Concentration of SO2Cl2

The initial moles of SO2Cl2 are given as 2.00 moles. The volume of the flask is given as 2.00 L. To find the initial molar concentration of SO2Cl2, use the formula: Molar concentration = Moles / Volume \[C_{SO_{2}Cl_2} = 2.00\ mol / 2.00\ L = 1.00\ M\]

Calculate the Change in Moles

We are given that 56% of SO2Cl2 is decomposed. So let's calculate the change in moles of SO2, Cl2 and SO2Cl2. Change in moles of SO2Cl2 = 56% of 2.00 moles \[\Delta n_{SO_{2}Cl_2} = 0.56 \times 2.00\ mol = 1.12\ mol\] Since 1 mole of SO2 and 1 mole of Cl2 is formed when 1 mole of SO2Cl2 is decomposed, the change in moles for SO2 and Cl2 will be the same as that of SO2Cl2. \[\Delta n_{SO_2} = \Delta n_{Cl_2} = 1.12\ mol\]

Calculate the Equilibrium Moles

Now, let's find the equilibrium moles for SO2, Cl2, and SO2Cl2. Equilibrium moles of SO2Cl2 = Initial moles of SO2Cl2 - Change in moles of SO2Cl2 \[n_{SO_{2}Cl_2(eq)} = 2.00\ mol - 1.12\ mol = 0.88\ mol\] Equilibrium moles of SO2 = Change in moles of SO2 (since initially there is no SO2) \[n_{SO_{2}(eq)} = 1.12\ mol\] Equilibrium moles of Cl2 = Change in moles of Cl2 (since initially there is no Cl2) \[n_{Cl_{2}(eq)} = 1.12\ mol\]

Calculate the Equilibrium Concentrations

Now, we will calculate equilibrium concentrations using the equilibrium moles calculated in step 3 and the volume of the flask. \[C_{SO_{2}Cl_2(eq)} = n_{SO_{2}Cl_2(eq)} / 2.00\ L = 0.88\ mol / 2.00\ L = 0.44\ M\] \[C_{SO_{2}(eq)} = n_{SO_{2}(eq)} / 2.00\ L = 1.12\ mol / 2.00\ L = 0.56\ M\] \[C_{Cl_{2}(eq)} = n_{Cl_{2}(eq)} / 2.00\ L = 1.12\ mol / 2.00\ L = 0.56\ M\]

Calculate the Equilibrium Constant, Kc

Now, we will use the equilibrium concentrations of SO2, Cl2, and SO2Cl2 to calculate the equilibrium constant, Kc, using the equation: \[K_{c} = \frac{[SO_{2}][Cl_{2}]}{[SO_{2}Cl_{2}]}\] Plug in the equilibrium concentrations we calculated in step 4: \[K_{c} = \frac{(0.56\ M)(0.56\ M)}{(0.44\ M)} = \frac{0.3136\ M^2}{0.44\ M} = 0.7136\] So, the equilibrium constant, Kc, at 303 K for this reaction is 0.7136.

Key concepts

Molar Concentration

Understanding molar concentration is critical for solving chemical equilibrium problems. Molar concentration, often termed molarity, is a way to express the concentration of a solution. It measures the number of moles of a solute present in one liter of solution. This is calculated using the formula:\[ \text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Liters of solution}} \]For example, in the chemical equilibrium exercise given, we start with 2.00 moles of \(\text{SO}_2\text{Cl}_2\) in a 2.00-liter flask. The initial molarity was calculated as follows:- \(\text{Molarity of } \text{SO}_2\text{Cl}_2 = \frac{2.00 \text{ moles}}{2.00 \text{ L}} = 1.00 \text{ M}\).Understanding this value helps in determining how much of the substance reacts and how much is left at equilibrium.

Chemical Equilibrium

Chemical equilibrium is a fundamental concept where a chemical reaction and its reverse reaction occur at the same rate, resulting in no net change in concentrations of the reactants and products. At equilibrium:- The forward and reverse reaction rates are equal.- Concentrations of reactants and products remain constant over time.For the decomposition of \(\text{SO}_2\text{Cl}_2\), equilibrium is achieved when the rate at which \(\text{SO}_2\text{Cl}_2\) decomposes into \(\text{SO}_2\) and \(\text{Cl}_2\) equals the rate of their recombination.This concept is explored by calculating the equilibrium concentrations of the substances involved, further emphasizing the dynamic nature of chemical equilibrium.

Reaction Decomposition

Decomposition reactions are processes where one compound breaks down into two or more simpler substances. In the given problem:- \(\text{SO}_2\text{Cl}_2(g)\) decomposes into \(\text{SO}_2(g)\) and \(\text{Cl}_2(g)\).The given decomposition percentage indicates how much of the original compound breaks down. From the exercise, 56% of \(\text{SO}_2\text{Cl}_2\) decomposes, allowing us to calculate the change in moles for each substance involved:- \(\Delta n_{\text{SO}_2\text{Cl}_2} = 0.56 \times 2.00 \text{ moles} = 1.12 \text{ moles}\)The decomposition directly results in the creation of \(\text{SO}_2\) and \(\text{Cl}_2\) as shown. It's important to grasp this stoichiometric relationship when calculating the equilibrium concentrations.

SO2Cl2 Decomposition

The decomposition of sulfuryl chloride, \(\text{SO}_2\text{Cl}_2\), into sulfur dioxide \(\text{SO}_2\) and chlorine \(\text{Cl}_2\) is a typical example of a reaction achieving equilibrium. In this process:- Initially, we have pure \(\text{SO}_2\text{Cl}_2\) with no \(\text{SO}_2\) or \(\text{Cl}_2\).- As the reaction proceeds, \(\text{SO}_2\text{Cl}_2\) decomposes, forming \(\text{SO}_2\) and \(\text{Cl}_2\).The equilibrium concentrations calculated inform us about the proportion of each substance at equilibrium:- \(C_{\text{SO}_2\text{Cl}_2(eq)} = 0.44 \text{ M}\)- \(C_{\text{SO}_2(eq)} = C_{\text{Cl}_2(eq)} = 0.56 \text{ M}\)Finally, using these values in the equilibrium expression \(K_{c} = \frac{[\text{SO}_2][\text{Cl}_2]}{[\text{SO}_2\text{Cl}_2]}\), we appreciate how this specific equilibrium reflects quantitative changes directly.