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Chapter 15: Problem 31

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?

Short Answer

Expert verified
a) The mixture is not in equilibrium. b) The reaction will occur in the direction of forming more SO3, that is to the right.

Step by step solution

01

Calculate Concentrations

Calculating concentrations, n(V), for SO2, O2, and SO3 to initiate the process. Concentration = Amount (in mol) / Volume (in L). Concentration of SO2 = 0.455 mol/1.9 L = 0.239 M. Concentration of O2 = 0.183 mol/1.9 L = 0.096 M. Concentration of SO3 = 0.568 mol/1.9 L = 0.299 M.
02

Calculate Reaction Quotient

To determine whether a reaction is at equilibrium, a comparison between the Reaction Quotient, Qc, and the Equilibrium constant, Kc, has to be made. First, calculate the Reaction Quotient (Qc) using the formula Q = [Products] / [Reactants]. For the reaction, Q = [SO3]^2 /[SO2]^2 [O2] = (0.299)^2 / (0.239)^2*0.096 = 1.837.
03

Compare Qc and Kc to Determine Equilibrium

Now we compare Qc and Kc. Here, Qc (1.837) is less than Kc (280), the reaction will therefore proceed to the right. This means the reaction isn't at equilibrium, hence more product (SO3) needs to be formed from the reactants (SO2 and O2).
04

Determine the Reaction Direction

If a reaction is not at equilibrium, it will proceed in the forward or reverse direction in order to reach equilibrium. As our Qc < Kc, then our reaction will proceed to the right: SO2 and O2 will be used up, and more SO3 will be formed, until equilibrium is reached.

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Most popular questions from this chapter

The following reaction represents the binding of oxygen by the protein hemoglobin (Hb): $$\mathrm{Hb}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{aq}) \rightleftharpoons \mathrm{Hb}: \mathrm{O}_{2}(\mathrm{aq}) \quad \Delta H<0$$ Explain how each of the following affects the amount of \(\mathrm{Hb}: \mathrm{O}_{2}:\) (a) increasing the temperature; (b) decreasing the pressure of \(\mathrm{O}_{2} ;\) (c) increasing the amount of 6 hemoglobin.

Continuous removal of one of the products of a chemical reaction has the effect of causing the reaction to go to completion. Explain this fact in terms of Le Châtelier's principle.

Starting with \(0.3500 \mathrm{mol} \mathrm{CO}(\mathrm{g})\) and \(0.05500 \mathrm{mol}\) \(\mathrm{COCl}_{2}(\mathrm{g})\) in a \(3.050 \mathrm{L}\) flask at \(668 \mathrm{K},\) how many moles of \(\mathrm{Cl}_{2}(\mathrm{g})\) will be present at equilibrium? $$\begin{aligned} \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{g}) & \\ K_{\mathrm{c}} &=1.2 \times 10^{3} \mathrm{at} \ 668 \mathrm{K} \end{aligned}$$

Equilibrium is established at \(1000 \mathrm{K},\) where \(K_{\mathrm{c}}=281\) for the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) .\) The equilibrium amount of \(\mathrm{O}_{2}(\mathrm{g})\) in a \(0.185 \mathrm{L}\) flask is 0.00247 mol. What is the ratio of \(\left[\mathrm{SO}_{2}\right]\) to \(\left[\mathrm{SO}_{3}\right]\) in this equilibrium mixture?

In the equilibrium described in Example \(15-12,\) the percent dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) can be expressed as $$\frac{3.00 \times 10^{-3} \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4}}{0.0240 \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4} \text { initially }} \times 100 \%=12.5 \%$$ What must be the total pressure of the gaseous mixture if \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\) is to be \(10.0 \%\) dissociated at \(298 \mathrm{K} ?\) $$ \mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=0.113 \text { at } 298 \mathrm{K} $$
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