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Chapter 15: Problem 89

At \(700 \mathrm{~K}\), the equilibrium constant for the reaction $$\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)$$ is \(K_{p}=77\). A flask is charged with \(202.7 \mathrm{kPa}\) 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
The fraction of conversion of \(\mathrm{CCl}_4\) into \(\mathrm{C}\) and \(\mathrm{Cl}_2\) is approximately \(0.24\), and the equilibrium partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) are \(154.1\mathrm{kPa}\) and \(97.2\mathrm{kPa}\), respectively.

Step by step solution

01

Write the equilibrium expressions

Let \(x\) be the moles of \(\mathrm{CCl}_4\) that decompose. The initial pressure of \(\mathrm{CCl}_4\) is \(202.7\mathrm{kPa}\). So, the equilibrium pressure of \(\mathrm{CCl}_4\) will be \((202.7-x)\mathrm{kPa}\). Since the molecular coefficient of \(\mathrm{Cl}_2\) in the balanced chemical reaction is 2, each mole of decomposed \(\mathrm{CCl}_4\) will produce 2 moles of \(\mathrm{Cl}_2\). Therefore, the equilibrium pressure of \(\mathrm{Cl}_2\) will be \((2x)\mathrm{kPa}\). Note that solid carbon does not contribute to the equilibrium expression.
02

Write the expression for \(K_p\)

The equilibrium constant for a given reaction is defined as the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients to the product of the partial pressures of the reactants raised to their stoichiometric coefficients. Thus, for the given reaction, the expression for \(K_p\) can be written as: $$K_p=\frac{P_{\mathrm{Cl}_2}^2}{P_{\mathrm{CCl}_4}}$$ Substitute the equilibrium expressions of the partial pressures in terms of \(x\): $$77=\frac{(2x)^2}{202.7-x}$$
03

Solve for \(x\)

Solve the equation for \(x\): $$77(202.7-x)=(2x)^2$$ $$30.5(202.7-x)=2x^2$$ $$6163.5-30.5x=2x^2$$ Rearrange the equation: $$2x^2+30.5x-6163.5=0$$ Solve this quadratic equation using the quadratic formula: $$x = \frac{-30.5\pm\sqrt{(-30.5)^2-4(2)(-6163.5)}}{2(2)}$$ After solving, we get two possible values for \(x\), but only one value is physically meaningful (i.e., \(x < 202.7\)): $$x\approx48.6$$
04

Calculate the fraction of conversion and equilibrium pressures

(a) Calculate the fraction of conversion of \(\mathrm{CCl}_4\) into \(\mathrm{C}\) and \(\mathrm{Cl}_2\): Fraction of conversion \(=\frac{x}{202.7}=\frac{48.6}{202.7}\approx0.24\) (b) Calculate the equilibrium partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\): \(\mathrm{P_{CCl}_4}=202.7-x=202.7-48.6\approx154.1\mathrm{kPa}\) \(\mathrm{P_{Cl}_2}=2x=2(48.6)\approx97.2\mathrm{kPa}\) Therefore, the fraction of conversion of \(\mathrm{CCl}_4\) into \(\mathrm{C}\) and \(2\mathrm{Cl}_2\) is approximately \(0.24\), and the equilibrium partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) are \(154.1\mathrm{kPa}\) and \(97.2\mathrm{kPa}\), respectively.

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Most popular questions from this chapter

At \(1285^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) is $K_{c}=1.04 \times 10^{-3} .\( A \)1.00-\mathrm{L}$ vessel containing an equilibrium mixture of the gases has \(1.50 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\mathrm{Br}(g)\) in the vessel?

Write the expressions for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}(g)\) (b) $\mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SiCl}_{4}(g)$ (c) $\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)$ (d) $\mathrm{O}_{2}(g)+2 \mathrm{CO}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)$ (e) $\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}^{+}(a q)$ (f) $\mathrm{Fe}^{2+}(a q)+\mathrm{Ce}^{4+}(a q) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)$ (g) $\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$

Assume that the equilibrium constant for the dissociation of molecular bromine, \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\), at 800 \(\mathrm{K}\) is \(K_{c}=5.4 \times 10^{-3}\). (a) Which species predominates at equilibrium, \(\mathrm{Br}_{2}\) or Br, assuming that the concentration of \(\mathrm{Br}_{2}\) is larger than $5.4 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?$ (b) Assuming both forward and reverse reactions are elementary processes, which reaction has the larger numeric value of the rate constant, the forward or the reverse reaction?

For the equilibrium $$\mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \operatorname{BrCl}(g) $$ at \(400 \mathrm{~K}, K_{c}=7.0 .\) If $0.25 \mathrm{~mol}\( of \)\mathrm{Br}_{2}\( and \)0.55 \mathrm{~mol}$ of \(\mathrm{Cl}_{2}\) are introduced into a 3.0-L container at \(400 \mathrm{~K}\), what will be the equilibrium concentrations of $\mathrm{Br}_{2}, \mathrm{Cl}_{2}\(, and \)\mathrm{BrCl}$ ?

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}(g)+\mathrm{Cl}(g)$$ At \(25^{\circ} \mathrm{C}\), the rate constants for the forward and reverse reactions are $1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{~s}^{-1}\( and \)9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1}$, respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?(\mathbf{b})\) Are reactants or products more plentiful at equilibrium?
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