Problem 79 - Chemistry The Central Science

Question

Consider the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\). (a) Using data from Appendix \(\mathrm{C},\) calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are 0.40 atm and 1.60 atm, respectively.

Step-by-Step Solution

Answer

(a) To calculate the standard Gibbs free energy change (ΔG⁰), first find the standard enthalpy change (ΔH⁰) and standard entropy change (ΔS⁰) using the data in Appendix C: ΔH⁰(reaction) = Σ ΔH⁰(products) - Σ ΔH⁰(reactants) and ΔS⁰(reaction) = Σ ΔS⁰(products) - Σ ΔS⁰(reactants). Then, use the equation ΔG⁰ = ΔH⁰ - TΔS⁰ to find the standard Gibbs free energy change. (b) To calculate ΔG at 298 K given the partial pressures, first find the reaction quotient (Q) using the formula Q = (P(N₂O₄))/(P(NO₂)²) with the given partial pressures of NO₂ and N₂O₄ as 0.40 atm and 1.60 atm, respectively. Next, use the equation ΔG = ΔG⁰ + RT ln(Q) to find ΔG, where R is the gas constant (8.314 J/mol K) and T is the temperature (298 K).

Step 1: Find the standard enthalpy change and standard entropy change

Look up the standard enthalpy change (ΔH⁰) and the standard entropy change (ΔS⁰) for the reaction in Appendix C. We can find the values for the individual species and use the equation ΔH⁰(reaction) = Σ ΔH⁰(products) - Σ ΔH⁰(reactants) and ΔS⁰(reaction) = Σ ΔS⁰(products) - Σ ΔS⁰(reactants).

Step 2: Calculate the standard Gibbs free energy change

Use the equation ΔG⁰ = ΔH⁰ - TΔS⁰ to find the standard Gibbs free energy change. (b) Calculate ΔG at 298K given the partial pressures.

Step 3: Find the reaction quotient (Q)

The reaction quotient Q is defined as the ratio of the product of partial pressures of products to the product of partial pressures of reactants. In this case, Q = (P(N₂O₄))/(P(NO₂)²). Given the partial pressures of NO₂ and N₂O₄ as 0.40 atm and 1.60 atm respectively, calculate the value of Q.

Step 4: Calculate ΔG using the reaction quotient

Use the equation ΔG = ΔG⁰ + RT ln(Q) to find ΔG. R is the gas constant (8.314 J/mol K), T is the temperature (298 K), and Q is the reaction quotient calculated in step 3. Following these steps above will give you the answers for ΔG⁰ and ΔG at 298 K for the given reaction and partial pressures.

Key Concepts

Standard Enthalpy Change

In a chemical reaction, the standard enthalpy change, \( \Delta H^{\circ} \), is a way to express the heat absorbed or released under standard conditions (1 atm pressure and 298 K). It is calculated using the equation:

• \( \Delta H^{\circ}(\text{reaction}) = \Sigma \Delta H^{\circ}(\text{products}) - \Sigma \Delta H^{\circ}(\text{reactants}) \)

Standard enthalpy values are usually provided for individual substances in tables like Appendix C of textbooks.
By using these values, you can determine if the reaction releases heat (exothermic, \( \Delta H^{\circ} < 0 \)) or absorbs heat (endothermic, \( \Delta H^{\circ} > 0 \)). This helps in predicting how energy changes during the formation of products from reactants.
Remember to ensure your units match when making calculations, as enthalpy is often given in kilojoules per mole (kJ/mol). Understanding \( \Delta H^{\circ} \) is crucial for assessing the energy profile of chemical reactions.

Standard Entropy Change

Entropy, represented by \( \Delta S^{\circ} \), measures the degree of disorder or randomness in a system. The standard entropy change of a reaction indicates how the entropy of the system changes when reactants convert to products under standard conditions.

• \( \Delta S^{\circ}(\text{reaction}) = \Sigma \Delta S^{\circ}(\text{products}) - \Sigma \Delta S^{\circ}(\text{reactants}) \)

A positive \( \Delta S^{\circ} \) indicates an increase in disorder, implying more randomness in the products compared to reactants. Conversely, a negative \( \Delta S^{\circ} \) suggests that the reaction results in a more ordered system.
Changes in entropy are a crucial component when calculating Gibbs Free Energy, as they provide insight into the feasibility of a reaction proceeding under certain conditions. Thermodynamics tells us that reactions with increasing entropy are generally more favored, from an order-disorder perspective.

Reaction Quotient

The reaction quotient, \( Q \), is a snapshot of a reaction's position at any given point in time, distinct from the equilibrium state. It is calculated using the partial pressures or concentrations of the reactants and products at a particular moment:

• For the given reaction, \( Q = \frac{P_{\text{products}}}{P_{\text{reactants}}^{\text{coefficients}}} \)

In our specific case:

• \( Q = \frac{P(\mathrm{N}_{2} \mathrm{O}_{4})}{P(\mathrm{NO}_{2})^{2}} \)

Here, \( Q \) helps us compare the current reaction concentration to the equilibrium state, aiding in predicting which direction the reaction will shift to reach equilibrium.
If \( Q < K \), the reaction will move towards the products. If \( Q > K \), the reaction will favor the formation of reactants. Thus, \( Q \) plays a pivotal role in understanding real-time reaction dynamics and is integral in computing non-standard Gibbs Free Energy changes.

Partial Pressures

Partial pressures are a critical concept in understanding gas mixtures. Each gas in a mixture exerts pressure independently of other gases, known as its partial pressure. For a reaction involving gases, like the one in our exercise, understanding partial pressures is key to calculating reactions under non-standard conditions.
The total pressure is the sum of the partial pressures of all gases present, based on Dalton's Law:

• Total Pressure = sum of all individual partial pressures

In terms of chemical reactions, the partial pressures of reactants and products help determine the reaction quotient \( Q \).
For instance, in the reaction \(2 \mathrm{NO}_{2} \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}\), if \(P(\mathrm{NO}_{2})\) is 0.40 atm and \(P(\mathrm{N}_{2} \mathrm{O}_{4})\) is 1.60 atm, these values are plugged into the expression for \( Q \) to understand how far the reaction is from equilibrium. Partial pressures thus serve as a vital tool in thermodynamic calculations for gaseous systems, providing valuable insight into the state of chemical processes.