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Chapter 15: Problem 111

An equilibrium mixture of \(\mathrm{SO}_{2}, \mathrm{SO}_{3},\) and \(\mathrm{O}_{2}\) gases is maintained in a \(2.05 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=35.5\) for the reaction $$2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})$$ (a) If the numbers of moles of \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\) in the flask are equal, how many moles of \(\mathrm{O}_{2}\) are present? (b) If the number of moles of \(\mathrm{SO}_{3}\) in the flask is twice the number of moles of \(\mathrm{SO}_{2}\), how many moles of \(\mathrm{O}_{2}\) are present?

Short Answer

Expert verified
The number of moles of \(O_2\) are approximately 0.028 for case (a) and approximately 0.113 for case (b).

Step by step solution

01

Analyzing Case (a)

In case (a), the number of moles of \(\mathrm{SO_{2}}\) and \(\mathrm{SO_{3}}\) are equal. This implies that \(n(\mathrm{SO_{2}}) = n(\mathrm{SO_{3}})\). Let's assume this to be 'x' moles. Additionally, let's assume the number of moles of \(O_2\) to be 'y'. The expression for the equilibrium constant for the given reaction is \[K_c = \frac{(n(\mathrm{SO_{3}})/V)^2}{(n(\mathrm{SO_{2}})/V)^2\times (n(\mathrm{O_{2}})/V)}\] Since the volume, V, is the same for all the gases, it cancels out from the equation and we have: \[K_c = \frac{(n(\mathrm{SO_{3}}))^2}{(n(\mathrm{SO_{2}}))^2\times n(\mathrm{O_{2}})}\] Using this formula, we insert the known values and solve for y.
02

Calculating moles of \(O_2\) for Case (a)

For case (a), since \(n(\mathrm{SO_{2}}) = n(\mathrm{SO_{3}}) =x\), the formula becomes: \[35.5 = \frac{x^2}{x^2\times y} = \frac{1}{y}\] Solving for y, we get \[y = \frac{1}{35.5} = 0.028 moles\] Thus, there are approximately 0.028 moles of \(O_2\) in the flask.
03

Analyzing Case (b)

In case (b), the number of moles of \(\mathrm{SO_{3}}\) is twice the number of moles of \(\mathrm{SO_{2}}\). This implies that \(n(\mathrm{SO_{2}}) =x\) and \(n(\mathrm{SO_{3}}) =2x\). We use the same expression for equilibrium and adjust the values according to these quantities to find \(y\) the moles of \(O_2\).
04

Calculating moles of \(O_2\) for Case (b)

To solve for the moles of \(O_2\) in case (b), we insert the given values into the equilibrium formula: \[35.5 = \frac{(2x)^2}{x^2 \times y}\] Solving for y, we get \[y = \frac{4}{35.5} = 0.113 moles\] Thus, there are approximately 0.113 moles of \(O_2\) when the number of moles of \(\mathrm{SO_{3}}\) is twice the number of moles of \(\mathrm{SO_{2}}\).

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Most popular questions from this chapter

A classic experiment in equilibrium studies dating from 1862 involved the reaction in solution of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) to produce ethyl acetate and water. $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O}$$ The reaction can be followed by analyzing the equilibrium mixture for its acetic acid content. $$\begin{array}{r} 2 \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightleftharpoons \\ \mathrm{Ba}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In one experiment, a mixture of 1.000 mol acetic acid and 0.5000 mol ethanol is brought to equilibrium. A sample containing exactly one-hundredth of the equilibrium mixture requires \(28.85 \mathrm{mL} 0.1000 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. Calculate the equilibrium constant, \(K_{c}\), for the ethanol-acetic acid reaction based on this experiment.

For the reaction \(\mathrm{SO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{aq}), K=1.25 \mathrm{at}\) \(25^{\circ} \mathrm{C} .\) Will the amount of \(\mathrm{SO}_{2}(\mathrm{g})\) be greater than or less than the amount of \(\mathrm{SO}_{2}(\mathrm{aq}) ?\)

Explain why the percent of molecules that dissociate into atoms in reactions of the type \(\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) always increases with an increase in temperature.

Determine \(K_{c}\) for the reaction $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NOBr}(\mathrm{g})$$ from the following information (at \(298 \mathrm{K}\) ). $$\begin{aligned} 2 \mathrm{NO}(\mathrm{g}) & \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) K_{\mathrm{c}}=2.1 \times 10^{30} \\ \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{NOBr}(\mathrm{g}) \quad K_{\mathrm{c}}=1.4 \end{aligned}$$

Formamide, used in the manufacture of pharmaceuticals, dyes, and agricultural chemicals, decomposes at high temperatures. $$\begin{array}{r} \mathrm{HCONH}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \\ K_{\mathrm{c}}=4.84 \text { at } 400 \mathrm{K} \end{array}$$ If \(0.186 \mathrm{mol} \mathrm{HCONH}_{2}(\mathrm{g})\) dissociates in a 2.16 Lflask at 400 K, what will be the total pressure at equilibrium?
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