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Chapter 15: Problem 87

At \(700 \mathrm{~K}\) the equilibrium constant for the reaction $$\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)$$ is \(K_{p}=0.76 .\) A flask is charged with 2.00 atm of \(\mathrm{CCl}_{4}\), which then reaches equilibrium at \(700 \mathrm{~K}\). (a) What fraction of the \(\mathrm{CCl}_{4}\) is converted into \(\mathrm{C}\) and \(\mathrm{Cl}_{2} ?\) (b) What are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

Short Answer

Expert verified
(a) The fraction of the CCl4 converted into C and Cl2 is approximately 0.0556 or 5.56%. (b) The partial pressures of CCl4 and Cl2 at equilibrium are approximately 1.89 atm and 0.2224 atm, respectively.

Step by step solution

01

1. Expression for equilibrium constant Kp in terms of partial pressures

For the reaction: \(\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s) + 2 \mathrm{Cl}_{2}(g)\) Let's initially assign the partial pressure of CCl4 as \(P_{\mathrm{CCl}_4}\) and that of Cl2 as \(P_{\mathrm{Cl}_2}\). As stated in the problem, the equilibrium constant, Kp, is given as 0.76. The expression for Kp is \(K_{p} = \frac{P_{\mathrm{Cl}_2}^2}{P_{\mathrm{CCl}_4}}\)
02

2. Relationship between change in partial pressures

At the beginning, partial pressure of CCl4 is 2.00 atm and there is no Cl2. Let x be the pressure of CCl4 that reacts and converts to Cl2. Thus, at equilibrium, the new partial pressures are: \(P_{\mathrm{CCl}_4} = 2.00 - x\) \(P_{\mathrm{Cl}_2} = 2x\)
03

3. Substitute pressures in the Kp expression and solve for x

Plug the partial pressures into the Kp expression and solve for x: \(0.76 = \frac{(2x)^2}{(2.00 - x)}\) Solve the quadratic equation for x. Multiplying both sides by (2.00-x): \(1.52x - 0.76x^2 = 4x^2\) Rearrange terms: \(5.28x^2 - 1.52x = 0\) Now we can solve for x: \(x = \frac{-1.52 \pm \sqrt{1.52^2-4(5.28)(0)}}{2(5.28)}\) At this point, we have a positive and a negative value of x. However, the negative value has no physical meaning in this problem since it would imply an increase in the partial pressure of CCl4, contradicting the reaction. Thus, we must choose the positive value: \(x = \frac{-1.52 + \sqrt{1.52^2}}{2(5.28)} \approx 0.1112\)
04

4. Calculate the fraction of CCl4 converted

The fraction of CCl4 converted is given by the ratio of x to the initial partial pressure of CCl4: Fraction converted = \(\frac{x}{2.00}\) Fraction converted = \(\frac{0.1112}{2.00} \approx 0.0556\)
05

Part (a) Answer

(a) The fraction of the CCl4 converted into C and Cl2 is approximately 0.0556 or 5.56%.
06

5. Determine partial pressures of CCl4 and Cl2 at equilibrium

Now we will calculate the equilibrium partial pressures of CCl4 and Cl2 using the value of x: \(P_{\mathrm{CCl}_4} = 2.00 - x = 2.00 - 0.1112 \approx 1.89 \mathrm{~atm} \) \(P_{\mathrm{Cl}_2} = 2x = 2(0.1112) \approx 0.2224\mathrm{~atm} \)
07

Part (b) Answer

(b) The partial pressures of CCl4 and Cl2 at equilibrium are approximately 1.89 atm and 0.2224 atm, respectively.

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Most popular questions from this chapter

Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? (a) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) ; K_{p}=5.0 \times 10^{12}\) (b) \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) ; K_{c}=5.8 \times 10^{-18}\)

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