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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

Expert verified
At 25°C, the decomposition reaction of SO2Cl2 can be represented as \(SO_2Cl_2(g) \rightleftharpoons SO_2(g) + Cl_2(g)\). For a given initial moles of SO2Cl2, we calculated the remaining moles, and the moles of SO2 and Cl2 formed at equilibrium. Then, we determined the mole fractions of each gas and used the total pressure at equilibrium (0.900 atm) to find the partial pressures. Finally, we used the expression \(K_p=\frac{P_{SO_2}P_{Cl_2}}{P_{SO_2Cl_2}}\) to calculate the value of Kp for the system.

Step by step solution

01

Find the initial moles of SO2Cl2

Let's assume there is initially 1 mole of SO2Cl2. Since 12.5% of the SO2Cl2 decomposes at equilibrium, we can calculate the moles of SO2Cl2 remaining (0.875 moles) and the moles of SO2 and Cl2 formed (0.125 moles each).
02

Calculate the mole fractions of each gas at equilibrium

Now that we know the moles of each gas at equilibrium, we can calculate their mole fractions (χ). Mole fraction = moles of component / total moles in the system. Mole fraction of SO2Cl2 (χ_SO2Cl2) = 0.875 / (0.875 + 0.125 + 0.125) = 0.875 / 1.125 Mole fraction of SO2 (χ_SO2) = 0.125 / 1.125 Mole fraction of Cl2 (χ_Cl2) = 0.125 / 1.125
03

Calculate the partial pressures of each gas at equilibrium

Now that we have the mole fractions, we can find the partial pressures using the total pressure (0.900 atm): Partial pressure of SO2Cl2 (P_SO2Cl2) = 0.875 / 1.125 * 0.900 atm Partial pressure of SO2 (P_SO2) = 0.125 / 1.125 * 0.900 atm Partial pressure of Cl2 (P_Cl2) = 0.125 / 1.125 * 0.900 atm
04

Calculate Kp for the system

Finally, use the partial pressures in the expression for Kp: \(K_p = \frac{P_{SO_2}P_{Cl_2}}{P_{SO_2Cl_2}}\) Plug in the values calculated above for partial pressures and calculate Kp.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant (Kp)
Understanding the equilibrium constant, noted as Kp, is crucial when studying chemical reactions that reach a state where the reactants and products are formed at the same rate. Kp represents the ratio of the product of partial pressures of the products to that of the reactants, raised to the power of their stoichiometric coefficients, at a constant temperature. For a reaction like \(aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\), the equilibrium constant is given by:\[ K_p = \frac{P_C^c \times P_D^d}{P_A^a \times P_B^b} \]where P represents the partial pressure of each gas. Kp is dimensionless and provides insight into the position of equilibrium; a higher Kp means more products at equilibrium, while a lower Kp suggests a greater concentration of reactants. It's important for students to note that Kp is only applicable for reactions involving gases, as it is based on partial pressures.

Partial Pressure
Partial pressure is a fundamental concept in gas law, which reveals the pressure a single gas in a mixture would exert if it alone occupied the entire volume. When multiple gases are present, like in the chemical equilibrium of our example, each contributes to the total pressure proportionally to its amount. The formula for calculating the partial pressure of any gas is:\[ P_i = \frac{n_iRT}{V} \]However, when dealing with a mixture at known total pressure, we use mole fractions to determine each gas's partial pressure:\[ P_i = \text{mole fraction of } i \times \text{total pressure} \]Understanding this will help students recognize how changes in pressure and mole fraction affect the partial pressure of each gas component.

Mole Fractions
Mole fractions are a way to express the composition of a mixture of substances. The mole fraction, represented by \( \text{χ} \), is the ratio of the number of moles of one component to the total number of moles of all components in the mixture. For instance:\[ \text{χ}_i = \frac{\text{moles of component } i}{\text{total moles in the mixture}} \]Understanding mole fractions is indispensable when calculating the partial pressures in a gas mixture, as seen in the step-by-step solution to our exercise. By using mole fractions, students can predict the behavior of individual gases in a mixture, and calculate variables like density and partial pressure more effectively.

Chemical Equilibrium
Chemical equilibrium occurs when the rates of the forward and reverse chemical reactions are equal, leading to no net change in the concentrations of reactants and products over time. In a closed system, reactants convert to products and vice versa at the same rate. At this point, the system is said to be at equilibrium. It’s important to note that equilibrium does not mean the reactants and products are present in equal amounts, but rather that their ratios remain constant. Students should be aware that variables like temperature, pressure, and concentration can disturb the equilibrium, potentially shifting it in favor of the reactants or products according to Le Chatelier's principle. Learning to calculate the equilibrium constant helps in predicting the extent of reaction and direction in which the equation will proceed on changing these variables.

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Most popular questions from this chapter

The equilibrium constant is 0.0900 at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$$.For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a. A 1.0 -L flask contains 1.0 mole of HOCI, 0.10 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.10 mole of \(\mathrm{H}_{2} \mathrm{O}\) b. A \(2.0-\) L flask contains 0.084 mole of HOCI, 0.080 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.98 mole of \(\mathrm{H}_{2} \mathrm{O}\) c. A 3.0 -L flask contains 0.25 mole of HOCI, 0.0010 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.56 mole of \(\mathrm{H}_{2} \mathrm{O}\).

An \(8.00-\mathrm{g}\) sample of \(\mathrm{SO}_{3}\) was placed in an evacuated container, where it decomposed at \(600^{\circ} \mathrm{C}\) according to the following reaction: $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$$ At equilibrium the total pressure and the density of the gaseous mixtures were 1.80 atm and \(1.60 \mathrm{g} / \mathrm{L},\) respectively. Calculate \(K_{\mathrm{p}}\) for this reaction.

Consider the following reactions:\(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g) \quad\) and \(\quad \mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g)\).List two property differences between these two reactions that relate to equilibrium.

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g),\) consider two possibilities: (a) you mix 0.5 mole of each reactant, allow the system to come to equilibrium, and then add another mole of \(\mathrm{H}_{2}\) and allow the system to reach equilibrium again, or (b) you mix 1.5 moles of \(\mathrm{H}_{2}\) and 0.5 mole of \(\mathrm{I}_{2}\) and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$.equilibrium constants \(\mathbf{r}_{\mathbf{p}}\).\(300^{\circ} \mathrm{C}, \quad 4.34 \times 10^{-3}\) \(500^{\circ} \mathrm{C}, \quad 1.45 \times 10^{-5}\) \(600^{\circ} \mathrm{C}, \quad 2.25 \times 10^{-6}\) Is the reaction exothermic or endothermic?

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