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Chapter 15: Problem 82

A 5.37 -g sample of \(\mathrm{SO}_{3}\) is placed in a 5.00-L container and heated to \(1000 \mathrm{~K}\). The \(\mathrm{SO}_{3}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) : $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium, the total pressure in the container is 157 \(\mathrm{kPa}\). Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at \(1100 \mathrm{~K}\).

Short Answer

Expert verified
To find the values of \(K_p\) and \(K_c\), we first convert the given amount of SO_3 (5.37 g) to moles (0.0671 mol) and then determine the change in moles at equilibrium. We then calculate the mole fractions and partial pressures at equilibrium, and use these values to find \(K_p\). This involves solving an equation for \(x\), which represents the moles of SO_3 that decompose. Finally, we find \(K_c\) using the relationship with \(K_p\), which in this case is equal to \(K_p\), since the term \(\left(\frac{RT}{P_{\text{atm}}}\right)^{\Delta n}\) becomes 1 due to \(\Delta n = 0\).

Step by step solution

01

Convert grams of \(\mathrm{SO}_{3}\) to moles

We are given 5.37 grams of \(\mathrm{SO}_{3}\). We can use the molar mass of \(\mathrm{SO}_{3}\) to convert this to moles. The molar mass of \(\mathrm{S}\) is about 32 g/mol, and that of \(\mathrm{O}\) is about 16 g/mol. Therefore, the molar mass of \(\mathrm{SO}_{3}\) is \(32 + 3(16) = 80\) g/mol. To convert grams to moles, divide the mass by the molar mass: moles of \(\mathrm{SO}_{3} = \frac{5.37 \text{ g}}{80 \text{ g/mol}} = 0.0671 \text{ mol}\)
02

Find initial moles of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\)

Initially, there are no moles of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\), since the reaction has not yet occurred. Thus, the initial moles of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are both 0.
03

Determine the change in moles at equilibrium

Let \(x\) be the moles of \(\mathrm{SO}_{3}\) that decompose. Then at equilibrium, there are \(0.0671 - x\) moles of \(\mathrm{SO}_{3}\), \(2x\) moles of \(\mathrm{SO}_{2}\), and \(x\) moles of \(\mathrm{O}_{2}\).
04

Calculate the mole fractions and partial pressures at equilibrium

Total moles at equilibrium: \(0.0671 - x + 2x + x = 0.0671 + 2x\) Mole fractions at equilibrium: $$ \chi_{\mathrm{SO}_{3}} = \frac{0.0671 - x}{0.0671 + 2x} $$ $$ \chi_{\mathrm{SO}_{2}} = \frac{2x}{0.0671 + 2x} $$ $$ \chi_{\mathrm{O}_{2}} = \frac{x}{0.0671 + 2x} $$ We know that the total pressure is 157 kPa. Thus, we can use the mole fractions to find the partial pressures: $$ P_{\mathrm{SO}_{3}} = \chi_{\mathrm{SO}_{3}} \times 157 \text{ kPa} $$ $$ P_{\mathrm{SO}_{2}} = \chi_{\mathrm{SO}_{2}} \times 157 \text{ kPa} $$ $$ P_{\mathrm{O}_{2}} = \chi_{\mathrm{O}_{2}} \times 157 \text{ kPa} $$
05

Calculate \(K_p\) using the partial pressures

The expression for \(K_p\) for this reaction is: $$ K_p = \frac{P_{\mathrm{SO}_{2}}^2 \times P_{\mathrm{O}_{2}}}{P_{\mathrm{SO}_{3}}^2} $$ We can substitute the above expressions for the partial pressures in terms of \(x\): $$ K_p = \frac{\left(\frac{2x}{0.0671 + 2x}\times157\right)^2 \times \left(\frac{x}{0.0671 + 2x}\times157\right)}{\left(\frac{0.0671 - x}{0.0671 + 2x}\times157\right)^2} $$
06

Solve for \(x\)

The expression in step 5 for \(K_p\) in terms of \(x\) can be solved using various methods like trial and error, numerical methods, or algebraically. Solving this equation will give the value of \(x\) which then can be used to find the partial pressures and the equilibrium constant \(K_p\).
07

Find \(K_c\) using the relationship with \(K_p\)

Once we have the value of \(K_p\), we can find \(K_c\) using the relationship: $$ K_c = K_p \times \left(\frac{RT}{P_{\text{atm}}}\right)^{\Delta n} $$ where \(R = 0.0821 \ \text{L atm/mol K}\) is the ideal gas constant, \(T = 1000 \ \text{K}\), \(\Delta n = 2 - 2 = 0\), and \(P_{\text{atm}} = \frac{157 \ \text{kPa}}{101.325 \ \text{kPa/atm}}\). Since \(\Delta n = 0\), the term \(\left(\frac{RT}{P_{\text{atm}}}\right)^{\Delta n}\) becomes 1. Therefore, \(K_c = K_p\). Once you have the value of \(K_p\) from Step 6, you also have the value of \(K_c\).

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Most popular questions from this chapter

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the reaction of \(\mathrm{CO}\) with \(\mathrm{H}_{2}:\) $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ (a) Use thermochemical data in Appendix \(\mathrm{C}\) to calculate $\Delta H^{\circ}$ for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

Ozone, \(\mathrm{O}_{3},\) decomposes to molecular oxygen in the stratosphere according to the reaction $2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)$. Would an increase in pressure favor the formation of ozone or of oxygen?

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually $\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)$. A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{solv})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{solv}) & \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\text { solv })+\mathrm{H}_{2} \mathrm{O}(\text { solv }) \end{aligned}$$ where "(solv)" indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is 6.68 . A pharmaceutical chemist makes up $15.0 \mathrm{~L}\( of a solution that is initially \)0.275 \mathrm{M}$ in acetic acid and \(3.85 \mathrm{M}\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

Methane, \(\mathrm{CH}_{4}\), reacts with \(I_{2}\) according to the reaction $\mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g)\(. At \)600 \mathrm{~K}, K_{p}$ for this reaction is \(1.95 \times 10^{-4}\). A reaction was set up at 600 \(\mathrm{K}\) with initial partial pressures of methane of \(13.3 \mathrm{kPa}\) and of $6.67 \mathrm{kPa}\( for \)\mathrm{I}_{2}$. Calculate the pressures, in \(\mathrm{kPa}\), of all reactants and products at equilibrium.

When \(1.50 \mathrm{~mol} \mathrm{CO}_{2}\) and $1.50 \mathrm{~mol} \mathrm{H}_{2}\( are placed in a 3.00-L container at \)395^{\circ} \mathrm{C}$, the following reaction occurs: $\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$. If \(K_{c}=0.802\), what are the concentrations of each substance in the equilibrium mixture?
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