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Ch.18 - Free Energy and Thermodynamics
Tro - Chemistry: A Molecular Approach 4th Edition
Tro4th EditionChemistry: A Molecular ApproachISBN: 9780134112831Not the one you use?Change textbook
All textbooksTro 4th EditionCh.18 - Free Energy and ThermodynamicsProblem 98
Chapter 18, Problem 98

Dinitrogen tetroxide decomposes to nitrogen dioxide: N2O4(g) → 2 NO2(g) ΔH°rxn = 55.3 kJ At 298 K, a reaction vessel initially contains 0.100 atm of N2O4. When equilibrium is reached, 58% of the N2O4 has decomposed to NO2. What percentage of N2O4 decomposes at 388 K? Assume that the initial pressure of N2O4 is the same (0.100 atm).

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Step 1: Write down the balanced chemical equation for the reaction. In this case, it is N2O4(g) ⇌ 2NO2(g).
Step 2: Use the given information to calculate the equilibrium constant (K) at 298 K. The equilibrium constant is given by K = [NO2]^2 / [N2O4]. Since the reaction is in terms of pressure, we can use the partial pressures of the gases. The initial pressure of N2O4 is 0.100 atm, and 58% of it decomposes, so the equilibrium pressure of N2O4 is 0.100 atm * (1 - 0.58) = 0.042 atm. The pressure of NO2 at equilibrium is twice the amount of N2O4 that decomposed, so it is 2 * 0.100 atm * 0.58 = 0.116 atm.
Step 3: Substitute these values into the equation for K to find its value at 298 K.
Step 4: Use the Van't Hoff equation to find the equilibrium constant at 388 K. The Van't Hoff equation is ln(K2/K1) = -ΔHrxn/R * (1/T2 - 1/T1), where K1 and K2 are the equilibrium constants at temperatures T1 and T2 respectively, ΔHrxn is the enthalpy change of the reaction, and R is the ideal gas constant. In this case, K1 is the equilibrium constant at 298 K, T1 is 298 K, ΔHrxn is 55.3 kJ, R is 8.314 J/(mol*K), and T2 is 388 K.
Step 5: Solve the Van't Hoff equation for K2, which is the equilibrium constant at 388 K. Then, use this value to calculate the percentage of N2O4 that decomposes at 388 K. The percentage of N2O4 that decomposes is given by the ratio of the equilibrium pressure of NO2 to the initial pressure of N2O4, multiplied by 100%.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. In the context of temperature changes, increasing the temperature of an exothermic reaction will favor the endothermic direction, while decreasing the temperature will favor the exothermic direction. This principle helps predict how the equilibrium will shift when the temperature is altered.
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Equilibrium Constant (K)

The equilibrium constant (K) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. For the reaction N2O4 ⇌ 2 NO2, K can be calculated using the partial pressures of the gases involved. Changes in temperature will affect the value of K, which in turn influences the extent of the reaction and the concentrations of the reactants and products at equilibrium.
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Reaction Enthalpy (ΔHrxn)

Reaction enthalpy (ΔHrxn) indicates the heat change associated with a chemical reaction at constant pressure. A positive ΔHrxn, such as 55.3 kJ for the decomposition of N2O4, signifies that the reaction is endothermic, absorbing heat from the surroundings. This information is crucial for understanding how temperature changes can affect the equilibrium position and the extent of decomposition of N2O4 at different temperatures.
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Related Practice
Textbook Question

Indicate and explain the sign of ΔSuniv for each process. a. 2 H2(g) + O2(g) → 2 H2O (l) at 298 K.

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Textbook Question

All the oxides of nitrogen have positive values of ΔG°f at 298 K, but only one common oxide of nitrogen has a positive ΔS°f. Identify that oxide of nitrogen without reference to thermodynamic data and explain.

Textbook Question

The values of ΔG°f for the hydrogen halides become less negative with increasing atomic number. The ΔG°f of HI is slightly positive. However, the trend in ΔS°f is to become more positive with increasing atomic number. Explain.