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Chapter 12: Problem 83

At a certain temperature, the reaction $$\mathrm{Xe}(g)+2 \mathrm{~F}_{2}(g) \rightleftharpoons \mathrm{XeF}_{4}(g)$$ gives a \(50.0 \%\) yield of \(\mathrm{XeF}_{4}\), starting with \(\mathrm{Xe}\left(P_{\mathrm{X}_{e}}=0.20 \mathrm{~atm}\right)\) and \(\mathrm{F}_{2}\) \(\left(P_{\mathrm{F}_{2}}=0.40 \mathrm{~atm}\right)\). Calculate \(K\) at this temperature. What must the initial pressure of \(\mathrm{F}_{2}\) be to convert \(75.0 \%\) of the xenon to \(\mathrm{XeF}_{4} ?\)

Short Answer

Expert verified
Answer: The initial pressure of \(F_2\) needed to convert \(75 \%\) of the xenon to \(XeF_4\) is \(0.60\, \text{atm}\).

Step by step solution

01

Write the equilibrium constant expression

The equilibrium constant \(K_p\) is given by the expression: $$K_p = \frac{P_{XeF_4}}{P_{Xe}\times P_{F_2}^2}$$
02

Define the pressure changes for the reaction

Let the initial pressures be given as \(P_{Xe}=0.20\,\text{atm}\) and \(P_{F2}=0.40\,\text{atm}\). Let \(x\) represent the change in pressure of the reactants and products at equilibrium. Then the changes in pressure are: - Xenon: \(0.20 - x\) - Fluorine: \(0.40 - 2x\) - Xenon hexafluoride: \(0 + x\)
03

Use the reaction yield to find the equilibrium pressures

Since the reaction yield is \(50.0\%\), we know that only half of the initial pressure of xenon is actually converted. Therefore, the equilibrium pressure of \(Xe\), \(XeF_4\), and \(F_2\) are: $$P_{Xe} = 0.20 - 0.10 = 0.10\,\text{atm}$$ $$P_{XeF_4} = 0.10\,\text{atm}$$ $$P_{F_2} = 0.40 - 2(0.10) = 0.20\,\text{atm}$$
04

Calculate the equilibrium constant \(K_p\)

Substitute the equilibrium pressures into the equilibrium constant expression: $$K_p = \frac{0.10}{(0.10)(0.20)^2} = 25$$
05

Set up the new pressure changes for 75% conversion

Now we have to find the initial pressure of \(F_2\), denoted as \(P'_{F_2}\), required to convert \(75.0\%\) of xenon to \(XeF_4\). Let \(x'\) represent the new change in pressure of the reactants and products at equilibrium: - Xenon: \(0.20 - x'\) - Fluorine: \(P'_{F_2} - 2x'\) - Xenon hexafluoride: \(0 + x'\) The new equilibrium pressures for 75.0% conversion are: $$P'_{Xe} = 0.20\,\text{atm} - \frac{3}{4} (0.20\,\text{atm}) = 0.05\,\text{atm}$$ $$P'_{XeF_4} = \frac{3}{4} (0.20\,\text{atm}) = 0.15\,\text{atm}$$ $$P'_{F_2} = P'_{F_2} - 2x'$$
06

Calculate the initial pressure of \(F_2\) for 75% conversion

Substitute the new equilibrium pressures into the equilibrium constant expression and solve for the initial pressure of \(F_2\): $$K_p = \frac{0.15}{(0.05)(P'_{F_2} - 2(0.15))^2} = 25$$ Solving for \(P'_{F_2}\), we get: $$P'_{F_2} = 0.60\,\text{atm}$$ The initial pressure of \(F_2\) needed to convert \(75 \%\) of the xenon to \(XeF_4\) is \(0.60\, \text{atm}\).

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Most popular questions from this chapter

For the following reactions, predict whether the pressure of the reactants or products increases or remains the same when the volume of the reaction vessel is increased. (a) \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\)

Calculate \(K\) for the formation of methyl alcohol at \(100^{\circ} \mathrm{C}\) : $$\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ given that at equilibrium, the partial pressures of the gases are \(P_{\mathrm{CO}}=0.814 \mathrm{~atm}, P_{\mathrm{H}_{2}}=0.274 \mathrm{~atm}\), and \(P_{\mathrm{CH}_{3} \mathrm{OH}}=0.0512 \mathrm{~atm} .\)

Write a chemical equation for an equilibrium system that would lead to the following expressions (a-d) for \(K\). (a) \(K=\frac{\left(P_{\mathrm{H}_{2}}\right)^{2}\left(P_{\mathrm{O}_{2}}\right)^{3}}{\left(P_{\mathrm{SO}_{2}}\right)^{2}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{2}}\) (b) \(K=\frac{\left(P_{\mathrm{F}_{1}}\right)^{1 / 2}\left(P_{\mathrm{I}_{2}}\right)^{1 / 2}}{P_{\mathrm{IF}}}\) (c) \(K=\frac{\left[\mathrm{Cl}^{-}\right]^{2}}{\left(P_{\mathrm{C}_{2}}\right)\left[\mathrm{Br}^{-}\right]^{2}}\) (d) \(K=\frac{\left(P_{\mathrm{NO}}\right)^{2}\left(P_{\mathrm{H}_{3} \mathrm{O}}\right)^{4}\left[\mathrm{Cu}^{2+}\right]^{3}}{\left[\mathrm{NO}_{3}^{-}\right]^{2}\left[\mathrm{H}^{+}\right]^{8}}\)

A compound, \(\mathrm{X}\), decomposes at \(131^{\circ} \mathrm{C}\) according to the following equation: $$2 \mathrm{X}(g) \rightleftharpoons \mathrm{A}(g)+3 \mathrm{C}(g) \quad K=1.1 \times 10^{-3}$$ If a flask initially contains \(\mathrm{X}, \mathrm{A}\), and \(\mathrm{C}\), all at partial pressures of \(0.250 \mathrm{~atm}\), in which direction will the reaction proceed?

Hemoglobin (Hb) binds to both oxygen and carbon monoxide. When the carbon monoxide replaces the oxygen in an organism, the following reaction occurs: $$\mathrm{HbO}_{2}(a q)+\mathrm{CO}(g) \rightleftharpoons \mathrm{HbCO}(a q)+\mathrm{O}_{2}(g)$$ At \(37^{\circ} \mathrm{C}, K\) is about 200 . When equal concentrations of \(\mathrm{HbO}_{2}\) and \(\mathrm{HbCO}\) are present, the effect of CO inhalation is fatal. Assuming \(\mathrm{P}_{\mathrm{O}_{2}}=0.21 \mathrm{~atm}\), what is \(\mathrm{P}_{\mathrm{CO}}\) when \(\left[\mathrm{HbO}_{2}\right]=[\mathrm{HbCO}] ?\)
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