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Chapter 15: Problem 45

At \(120^{\circ} \mathrm{C}, K_{c}=0.090\) for the reaction $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ In an equilibrium mixture of the three gases, the concentrations of \(\mathrm{SO}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{SO}_{2}\) are $0.100 \mathrm{M}\( and \)0.075 \mathrm{M}$, respectively. What is the partial pressure of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

Short Answer

Expert verified
The partial pressure of Cl2 in the equilibrium mixture is 0.0434 atm.

Step by step solution

01

Write the Balanced Chemical Equation

The balanced chemical equation is already given: \(\mathrm{SO}_{2}\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\)
02

Write the Expression for the Equilibrium Constant, Kc

For the reaction, the expression for Kc is: \(K_c = \frac{[\mathrm{SO}_{2}][\mathrm{Cl}_{2}]}{[\mathrm{SO}_{2}\mathrm{Cl}_{2}]}\) We are given that \(K_c = 0.090\) at 120°C.
03

Find the Change in Concentrations during the Reaction

Let the change in the concentration of reactants and products during the reaction be x. Since the balanced equation has a 1:1:1 stoichiometric ratio, we can find the change in concentration for each species. At equilibrium: \([SO_2Cl_2] = 0.100 - x\) \([SO_2] = 0.075 + x\) \([Cl_2] = x\)
04

Substitute the Concentrations back into the Kc Expression

Substitute the concentrations into the Kc expression: \(0.090 = \frac{(0.075 + x)(x)}{(0.100 - x)}\)
05

Solve for x

Solve the above equation for x: \(0.090 = \frac{0.075x + x^2}{0.100 - x}\) Multiply both sides by \((0.100 - x)\): \(0.009 - 0.090x = 0.075x + x^2\) Rearrange the equation: \(x^2 + 0.165x - 0.009 = 0\) Now, solve the quadratic equation for x. We can use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) In this case, \(a=1\), \(b=0.165\), and \(c=-0.009\). Upon solving for x, we find that there are two possible values: \(x = -0.996\) and \(x = 0.00906\). Since the negative value of x doesn't make sense in the context of concentrations, we choose the positive value for x: \(x = 0.00906 \, \mathrm{M}\) This means that the equilibrium concentration of Cl2 is 0.00906 M.
06

Calculate the Partial Pressure of Cl2

To calculate the partial pressure of Cl2, first, we can assume that the total pressure (P) is 1 atm since it is not given. Then, we can use the mole fraction of Cl2 in the gas mixture to find its partial pressure. Mole fraction of Cl2, \(X_{Cl2} = \frac{[Cl_2]}{([SO_2] + [SO_2Cl_2] + [Cl_2])} = \frac{0.00906}{(0.075 + 0.100 - 0.00906 + 0.00906)}\) Calculate \(X_{Cl2}\): \(X_{Cl_2} = 0.0434\) Now, calculate the partial pressure of Cl2 using the total pressure and the mole fraction: \(P_{Cl_2} = X_{Cl_2} \times P\) \(P_{Cl_2} = 0.0434 \times 1 \, \mathrm{atm}\) \(P_{Cl_2} = 0.0434 \, \mathrm{atm}\) So, the partial pressure of Cl2 in the equilibrium mixture is 0.0434 atm.

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Most popular questions from this chapter

Suppose that the gas-phase reactions \(A \longrightarrow B\) and $B \longrightarrow A\( are both elementary processes with rate constants of \)2.5 \times 10^{-2} \mathrm{~min}^{-1}\( and \)2.5 \times 10^{-1} \mathrm{~min}^{-1}$, respectively. (a) What is the value of the equilibrium constant for the equilibrium $\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?(\mathbf{b})$ Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?

At \(800 \mathrm{~K},\) the equilibrium constant for the reaction \(\mathrm{A}_{2}(g) \rightleftharpoons 2 \mathrm{~A}(g)\) is $K_{c}=3.1 \times 10^{-4}$. (a) Assuming both forward and reverse reactions are elementary reactions, which rate constant do you expect to be larger, \(k_{f}\) or \(k_{r} ?\) (b) If the value of $k_{f}=0.27 \mathrm{~s}^{-1}\(, what is the value of \)k_{r}\( at \)800 \mathrm{~K} ?$ (c) Based on the nature of the reaction, do you expect the forward reaction to be endothermic or exothermic? (d) If the temperature is raised to $1000 \mathrm{~K}\(, will the reverse rate constant \)k_{r}$ increase or decrease? Will the change in \(k_{r}\) be larger or smaller than the change in \(k_{f}\) ?

A mixture of \(1.374 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and \(70.31 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) is heated in a 2.00-L vessel at \(700 \mathrm{~K}\). These substances react according to $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ At equilibrium, the vessel is found to contain \(0.566 \mathrm{~g}\) of \(\mathrm{H}_{2}\). (a) Calculate the equilibrium concentrations of $\mathrm{H}_{2}, \mathrm{Br}_{2},\( and \)\mathrm{HBr} .$ (b) Calculate \(K_{c}\)

(a) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

Consider the following equilibrium between oxides of nitrogen $$3 \mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g)$$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? (c) At constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture?
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