Consider the following reaction at \(298 \mathrm{K}:\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$$ An equilibrium mixture contains \(\mathrm{O}_{2}(g)\) and \(\mathrm{SO}_{3}(g)\) at partial pressures of 0.50 atm and 2.0 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 SO2 in the mixture is approximately 0.019 atm. The reaction is most favored at low temperatures, assuming standard conditions, as increasing the temperature will shift the equilibrium towards the reactants due to Le Châtelier's principle.

Step by step solution

01

Write the equilibrium expression and find Kp

Using the balanced chemical equation, we can write the equilibrium expression for Kp. \(K_p = \frac{[SO_3]^2}{[SO_2]^2[O_2]}\) Next, let's find the value of Kp at 298 K. According to the Appendix 4: - ΔGº(SO2) = -300.19 kJ/mol - ΔGº(O2) = 0 kJ/mol - ΔGº(SO3) = -370.40 kJ/mol We can calculate ΔGº for the reaction and use it to find the Kp value: ΔGº(reaction) = 2 x ΔGº(SO3) - 2 x ΔGº(SO2) - ΔGº(O2) ΔGº(reaction) = 2 x (-370.40) - 2 x (-300.19) - 0 = -140.42 kJ/mol Now, we can use the formula: \(ΔGº = -RT \ln{K_p}\) where R is the ideal gas constant, \(8.314 J/(mol \cdot K)\), and T is the temperature in Kelvin. Solving for Kp: \(K_p = e^{\frac{-ΔGº}{RT}}\) \(K_p = e^{\frac{-(-140.42 \times 10^3)}{(8.314)(298)}}\) \(K_p \approx 1.04 \times 10^5\)
02

Solve for the partial pressure of SO2

Now that we have the Kp value, we can plug in the given partial pressures of O2 and SO3 and solve for the partial pressure of SO2: \((1.04 \times 10^5) = \frac{(2.0)^2}{[SO_2]^2(0.50)}\) To solve for [SO2]: \([SO_2] = \sqrt{\frac{(2.0)^2}{(1.04 \times 10^5)(0.50)}}\) \([SO_2] \approx 0.019\,atm\) The equilibrium partial pressure of SO2 in the mixture is approximately 0.019 atm.
03

Check if the reaction is favored at high or low temperature

Looking at the given reaction, we see that there are two moles of gaseous reactants and three moles of gaseous products. According to Le Châtelier's principle, increasing the temperature of the reaction will shift the reaction toward the side with fewer moles of gas. In this case, increasing the temperature will shift the equilibrium towards the reactants. Thus, the reaction is most favored at low temperatures, assuming standard conditions.

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