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Chapter 17: Problem 84

Consider the following reaction at 298 K: $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ An equilibrium mixture contains \(\mathrm{O}_{2}(g)\) and \(\mathrm{SO}_{3}(g)\) at partial pressures of 0.50 \(\mathrm{atm}\) and 2.0 \(\mathrm{atm}\) , respectively. Using data from Appendix \(4,\) determine the equilibrium partial pressure of \(\mathrm{SO}_{2}\) in the mixture. Will this reaction be most favored at a high or a low temperature, assuming standard conditions?

Short Answer

Expert verified
The equilibrium partial pressure of \(\mathrm{SO_2}\) in the mixture is \(2.28 \: \mathrm{atm}\), and the reaction is favored at low temperatures due to its exothermic nature.

Step by step solution

01

Write the balanced equation

The balanced equation is given in the problem: \(2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g)\)
02

Express Kp in terms of partial pressures

Let's express the equilibrium constant (Kp) for the given reaction in terms of the partial pressures of the gases: \(Kp = \frac{(P_{SO_3})^2}{(P_{O_2})(P_{SO_2})^2}\)
03

Get the equilibrium constant from the Appendix 4

We need to find the equilibrium constant for this reaction in Appendix 4. Look for the value Kp for the reaction at 298 K. In this case, \(Kp = 2.94 \times 10^{-4}\)
04

Solve for the pressure of SO2

We are given the pressures of O2 (0.50 atm) and SO3 (2.0 atm), and we have the equilibrium constant Kp. Now, we can plug these values into our Kp expression and solve for the pressure of SO2: \( \begin{align*} 2.94 \times 10^{-4} &= \frac{(2.0)^2}{(0.50)(P_{SO_2})^2} \\ P_{SO_2} &= \sqrt{\frac{(2.0)^2}{(0.50)(2.94 \times 10^{-4})}} \\ P_{SO_2} &= 2.28 \mathrm{atm} \end{align*} \) So the equilibrium partial pressure of SO2 is 2.28 atm.
05

Determine if the reaction is favored at high or low temperature

To determine if the reaction is favored at high or low temperature, we will consider the heat of reaction (ΔH). Look for the value of ΔH for the reaction of formation of SO3 in Appendix 4: ΔH for the formation of SO3(g): -197 kJ/mol Since the reaction is exothermic (ΔH < 0), the reaction will be favored at low temperatures. This is because, according to Le Chatelier's principle, lowering the temperature of an exothermic reaction will force the equilibrium to shift towards the products, favoring the formation of SO3. In conclusion, the equilibrium partial pressure of SO2 in the mixture is 2.28 atm, and the reaction is favored at low temperatures.

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