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Chapter 15: Problem 29

Can a mixture of \(2.2 \mathrm{mol} \mathrm{O}_{2}, 3.6 \mathrm{mol} \mathrm{SO}_{2},\) and \(1.8 \mathrm{mol}\) \(\mathrm{SO}_{3}\) be maintained indefinitely in a \(7.2 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=100\) in this reaction? Explain. $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) $$

Short Answer

Expert verified
No, the given mixture cannot be maintained indefinitely in the flask because the initial conditions do not correspond to the state of equilibrium for this reaction at the given temperature. The reaction will shift towards the products to reach equilibrium.

Step by step solution

01

Understand the Reaction

The reaction under consideration is: \[2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\] At equilibrium, the concentrations of the constituents obey the law of mass action, which is given by \(K_c\), the equilibrium constant.
02

Calculate the Initial Molar Concentration of Reactants and Products

Before any reaction occurs, the molar concentration of each substance can be calculated by dividing the number of moles by the volume of the flask. For \(SO_2\), \(O_2\), and \(SO_3\), the initial molar concentrations are: \[ [SO_2] = \frac{3.6 \mathrm{mol}}{7.2 \mathrm{L}} = 0.5 \mathrm{M} \] \[ [O_2] = \frac{2.2 \mathrm{mol}}{7.2 \mathrm{L}} = 0.306 \mathrm{M} \] \[ [SO_3] = \frac{1.8 \mathrm{mol}}{7.2 \mathrm{L}} = 0.25 \mathrm{M} \]
03

Calculate the Reaction Quotient, Q

The next step is determining the reaction quotient \(Q\), which is similar to the equilibrium constant but involves the initial concentrations, before equilibrium is reached. Its calculation is based on the law of mass action as well: \[ Q = \frac{([SO_3]^2)}{([SO_2]^2 [O_2])} = \frac{(0.25 \mathrm{M})^2}{(0.5 \mathrm{M})^2 \times 0.306 \mathrm{M}} = 0.326 \]
04

Comparing Q and Kc to Predict the Reaction Shift

Let's compare \(Q\) with \(K_c\). If \(Q < K_c\), the reaction will shift to the right, i.e., towards the products, to achieve equilibrium. If \(Q > K_c\), it will shift to the left, i.e., towards the reactants. Here, \(Q = 0.326\) and \(K_c = 100\). Since \(Q < K_c\), the reaction will shift to the right to attain equilibrium, converting some \(SO_2\) and \(O_2\) into \(SO_3\). Thus, the given mixture cannot be maintained indefinitely, as the reaction will proceed until equilibrium is reached.

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Most popular questions from this chapter

Write an equilibrium constant, \(K_{\mathrm{p}},\) for the formation from its gaseous elements of (a) 1 mol \(\mathrm{NOCl}(\mathrm{g})\) (b) \(2 \mathrm{mol} \mathrm{ClNO}_{2}(\mathrm{g}) ;\) (c) \(1 \mathrm{mol} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{g}) ;\) (d) \(1 \mathrm{mol}\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\)

In the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}), 0.455\) \(\mathrm{mol} \mathrm{SO}_{2}, 0.183 \mathrm{mol} \mathrm{O}_{2},\) and \(0.568 \mathrm{mol} \mathrm{SO}_{3}\) are introduced simultaneously into a 1.90 L vessel at \(1000 \mathrm{K}\). (a) If \(K_{c}=2.8 \times 10^{2},\) is this mixture at equilibrium? (b) If not, in which direction will a net change occur?

A mixture of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) and \(\mathrm{CH}_{4}(\mathrm{g})\) in the mole ratio 2: 1 was brought to equilibrium at \(700^{\circ} \mathrm{C}\) and a total pressure of 1 atm. On analysis, the equilibrium mixture was found to contain \(9.54 \times 10^{-3} \mathrm{mol} \mathrm{H}_{2} \mathrm{S} .\) The \(\mathrm{CS}_{2}\) pre- sent at equilibrium was converted successively to \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and then to \(\mathrm{BaSO}_{4} ; 1.42 \times 10^{-3} \mathrm{mol} \mathrm{BaSO}_{4}\) was obtained. Use these data to determine \(K_{\mathrm{p}}\) at \(700^{\circ} \mathrm{C}\) for the reaction $$\begin{aligned} 2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{CH}_{4}(\mathrm{g}) \rightleftharpoons \mathrm{CS}_{2}(\mathrm{g})+& 4 \mathrm{H}_{2}(\mathrm{g}) \\\ & K_{\mathrm{p}} \text { at } 700^{\circ} \mathrm{C}=? \end{aligned}$$

For the reaction \(2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\) \(K_{\mathrm{c}}=1.8 \times 10^{-6}\) at \(184^{\circ} \mathrm{C} . \mathrm{At} 184^{\circ} \mathrm{C},\) the value of \(K_{\mathrm{c}}\) for the reaction \(\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g})\) is (a) \(0.9 \times 10^{6}\) (b) \(7.5 \times 10^{2}\) (c) \(5.6 \times 10^{5}\) (d) \(2.8 \times 10^{5}\)

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