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Chapter 15: Problem 23

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?

Short Answer

Expert verified
The ratio of the concentrations of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) in the equilibrium mixture is \(\frac{1}{\sqrt{281 * 0.0134}}\). Simply put, if you want the numerical value for the ratio, just evaluate this expression.

Step by step solution

01

Calculate the concentration of \(\mathrm{O}_{2}\)

Given that the number of moles of \(\mathrm{O}_{2}\) is 0.00247 mol and the volume is 0.185 L, the concentration of \(\mathrm{O}_{2}\) can be calculated using the formula \(c = n/V\), where \(n\) is the number of moles and \(V\) is the volume. Thus, \([O_2] = \frac{0.00247}{0.185} = 0.0134 \, \text{M}\).
02

Substitute the values into the \(K_c\) expression

The \(K_c\) for the reaction is given as 281. The general form of the equilibrium expression for the reaction is \(K_c = \frac{[\mathrm{SO}_{3}]^2}{[\mathrm{SO}_{2}]^2[\mathrm{O}_{2}]}\). Substitute \(K_c = 281\) and \([O_2] = 0.0134\) into the equation. We get \(281 = \frac{[\mathrm{SO}_{3}]^2}{[\mathrm{SO}_{2}]^2 * 0.0134}\). The target of this exercise is to find the ratio \(\frac{[\mathrm{SO}_{2}]}{[\mathrm{SO}_{3}]}\), so it's better to rewrite this expression as \(\frac{[\mathrm{SO}_{3}]}{[\mathrm{SO}_{2}]} = \sqrt{\frac{281 * 0.0134}{[\mathrm{SO}_{2}]^2}\). From this expression, it is evident that the ratio requested in the exercise is the reciprocal of the square root of \(\frac{281 * 0.0134}{[\mathrm{SO}_{2}]^2}\).
03

Simplify the expression for the ratio

Take the reciprocal of the square root from the previous step, which gives \(\frac{[\mathrm{SO}_{2}]}{[\mathrm{SO}_{3}]} = \frac{1}{\sqrt{281 * 0.0134}}\). This will be the final expression for the ratio. Algebraically, this cannot be further simplified without additional information.

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Most popular questions from this chapter

For the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\) \(\mathrm{H}_{2}(\mathrm{g}), K_{\mathrm{p}}=23.2\) at \(600 \mathrm{K} .\) Explain which of the fol- lowing situations might equilibrium: (a) \(\quad P_{\mathrm{CO}}=P_{\mathrm{H}_{2} \mathrm{O}}=P_{\mathrm{CO}_{2}}=P_{\mathrm{H}_{2}} ; \quad\) (b) \(\quad P_{\mathrm{H}_{2}} / P_{\mathrm{H}_{2} \mathrm{O}}=\) \(P_{\mathrm{CO}_{2}} / P_{\mathrm{CO}} ; \quad(\mathrm{c}) \quad\left(P_{\mathrm{CO}_{2}}\right)\left(P_{\mathrm{H}_{2}}\right)=\left(P_{\mathrm{CO}}\right)\left(P_{\mathrm{H}_{2} \mathrm{O}}^{2}\right)\) (d) \(P_{\mathrm{CO}_{2}} / P_{\mathrm{H}_{2} \mathrm{O}}=P_{\mathrm{H}_{2}} / P_{\mathrm{CO}}\)

Determine \(K_{c}\) for the reaction \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})+\) \(\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g}),\) given the following data at \(298 \mathrm{K}\) $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=1.0 \times 10^{-9}$$ $$\operatorname{NOCl}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2} \mathrm{Cl}(\mathrm{g}) \quad K_{\mathrm{p}}=1.1 \times 10^{2}$$ $$\mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2} \mathrm{Cl}(\mathrm{g}) \quad K_{\mathrm{p}}=0.3$$

The Deacon process for producing chlorine gas from hydrogen chloride is used in situations where \(\mathrm{HCl}\) is available as a by-product from other chemical processes. $$\begin{aligned} 4 \mathrm{HCl}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{Cl}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&-114 \mathrm{kJ} \end{aligned}$$ A mixture of \(\mathrm{HCl}, \mathrm{O}_{2}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{Cl}_{2}\) is brought to equilibrium at \(400^{\circ} \mathrm{C}\). What is the effect on the equilibrium amount of \(\mathrm{Cl}_{2}(\mathrm{g})\) if (a) additional \(\mathrm{O}_{2}(\mathrm{g})\) is added to the mixture at constant volume? (b) \(\mathrm{HCl}(\mathrm{g})\) is removed from the mixture at constant volume? (c) the mixture is transferred to a vessel of twice the volume? (d) a catalyst is added to the reaction mixture? (e) the temperature is raised to \(500^{\circ} \mathrm{C} ?\)

The volume of the reaction vessel containing an equilibrium mixture in the reaction \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) is increased. When equilibrium is re-established, (a) the amount of \(\mathrm{Cl}_{2}\) will have increased; (b) the amount of \(\mathrm{SO}_{2}\) will have decreased; (c) the amounts of \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) will have remained the same; (d) the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) will have increased.

Recall the formation of methanol from synthesis gas, the reversible reaction at the heart of a process with great potential for the future production of automotive fuels (page 663 ). $$\begin{aligned} \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}) & \\ K_{\mathrm{c}}=& 14.5 \mathrm{at} 483 \mathrm{K} \end{aligned}$$ A particular synthesis gas consisting of 35.0 mole percent \(\mathrm{CO}(g)\) and 65.0 mole percent \(\mathrm{H}_{2}(\mathrm{g})\) at a total pressure of 100.0 atm at \(483 \mathrm{K}\) is allowed to come to equilibrium. Determine the partial pressure of \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) in the equilibrium mixture.
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