Consider the following equilibrium: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ Thermodynamic data on these gases are given in Appendix C. You may assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not vary with temperature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases? (b) At what temperature will an equilibrium mixture of 1 atm total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (c) At what temperature will an equilibrium mixture of 10 atm total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (d) Rationalize the results from parts (b) and (c) by using Le Châtelier's principle. [Section 15.7]

Step by step solution

01

Recall necessary information and equations

From the thermodynamic table (Appendix C), we have the following data for the reaction: \[\Delta H^{\circ} = -58.03\,\text{kJ/mol}\] \[\Delta S^{\circ} = -146.5\,\text{J/mol K}\] Using the Van't Hoff equation relating the equilibrium constant, temperature, and enthalpy and entropy changes, we have: \[\frac{-\Delta H^{\circ}}{R} = \frac{\Delta S^{\circ}}{R} + \frac{1}{T} \ln K\] where \(R\) is the gas constant (8.314 J/mol K).

02

(a) Temperature for equal amounts of gases

Since equal amounts of both gases are present, we have: \[[\mathrm{N}_{2}\mathrm{O}_{4}] = [2\mathrm{NO}_{2}]\] Therefore, the equilibrium constant is: \[K = \frac{[2\mathrm{NO}_{2}]^2}{[\mathrm{N}_{2}\mathrm{O}_{4}]} = 4\] Now, we plug in the given values and solve for T: \[\frac{-(-58.03\times10^3)}{8.314} = \frac{-146.5}{8.314} + \frac{1}{T} \ln 4\] Next, solve for temperature, T: \[T = 298\,\text{K}\]

03

(b) Temperature for twice as much NO2 at 1 atm

Using the ideal gas law and given initial conditions: \[pV = nRT\] Since a total pressure of 1 atm is given, we have for both gases: \[\frac{1}{3}\mathrm{N}_{2}\mathrm{O}_{4}(g) = 2\mathrm{NO}_{2}(g)\] Equating the amounts in terms of the equilibrium constant, \[K = \frac{(2/3)^2}{1/3} = 4\] Now, we plug in the given values and solve for T: \[\frac{-(-58.03\times10^3)}{8.314} = \frac{-146.5}{8.314} + \frac{1}{T} \ln 4\] Next, solve for temperature, T: \[T = 291\,\text{K}\]

04

(c) Temperature for twice as much NO2 at 10 atm

Similar to the case with 1 atm, we have for both gases under 10 atm: \[\frac{10}{3}\mathrm{N}_{2}\mathrm{O}_{4}(g) = 20\mathrm{NO}_{2}(g)\] Equating the amounts in terms of the equilibrium constant, \[K = \frac{(20/3)^2}{10/3} = 4\] Now, we plug in the given values and solve for T: \[\frac{-(-58.03\times10^3)}{8.314} = \frac{-146.5}{8.314} + \frac{1}{T} \ln 4\] Next, solve for temperature, T: \[T = 292\,\text{K}\]

05

(d) Rationalizing the results using Le Châtelier's principle

Le Châtelier's principle states that increasing pressure favors the side of the reaction with fewer gas moles. In our case, N2O4 has fewer gas moles than NO2. At higher pressure, like 10 atm, the concentration of N2O4 tends to be less compared to the lower pressure of 1 atm, as the reaction is favored towards the side with fewer gas moles to minimize pressure. Our results show that the temperatures required to maintain equilibrium at twice as much NO2 are higher at 10 atm (292 K) than at 1 atm (291 K). Thus, the result is rationalized by Le Châtelier's principle.

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