Problem 78 - Chemistry The Central Science

Question

Consider the reaction 2 \(\mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) .(\mathbf{a})\) 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

The standard Gibbs free energy change (ΔG°) for the reaction 2 NO₂(g) → N₂O₄(g) at 298 K is -4.71 kJ/mol. When the partial pressures of NO₂ and N₂O₄ are 0.40 atm and 1.60 atm, respectively, the Gibbs free energy change (ΔG) at 298 K is 2.61 kJ/mol.

Step 1: Calculation of Standard Gibbs Free Energy Change (ΔG°)

We'll use the provided standard Gibbs free energies of formation (ΔGf°) from Appendix C to calculate ΔG° for the reaction. For NO₂(g): \( \Delta G_{f}^{\circ} = 51.30 \thinspace \dfrac{kJ}{mol} \) For N₂O₄(g): \( \Delta G_{f}^{\circ} = 97.89 \thinspace \dfrac{kJ}{mol} \) Using the formula for ΔG°: \( \Delta G^{\circ} = \Sigma n_i \Delta G_f^{\circ}(products) - \Sigma n_j \Delta G_f^{\circ}(reactants) \) \( \Delta G^{\circ} = [\Delta G_f^{\circ}(N_{2}O_{4})] - 2[\Delta G_f^{\circ}(NO_{2})] \) Calculating ΔG° using the given values: \( \Delta G^{\circ} = 97.89 - 2 \times 51.30 = -4.71 \thinspace kJ/mol \)

Step 2: Calculation of Reaction Quotient (Q)

In order to calculate ΔG, we first need to find the reaction quotient Q, which is given by: \( Q = \dfrac{P_{(N_{2}O_{4})}}{P_{(NO_{2})}^2} \) Substituting the given partial pressures of NO₂ (0.40 atm) and N₂O₄ (1.60 atm) into the equation: \( Q = \dfrac{1.60}{(0.40)^2} = 10 \)

Step 3: Calculation of Gibbs Free Energy Change (ΔG)

Now that we have Q and ΔG°, we can use the following equation to calculate ΔG: \( \Delta G = \Delta G^{\circ} + RT \ln Q \) Where: R = 8.314 J/(mol·K) T = 298 K Substitute the calculated values: \( \Delta G = (-4.71 \times 10^3 J/mol) + (8.314 J/(mol·K)) \times 298 K \times \ln (10) \) Calculating ΔG: \( \Delta G = -4.71 \times 10^{3} + 8.314 \times 298 \times 2.303 = 2.61 \thinspace kJ/mol \) So, the Gibbs free energy change (ΔG) for the given reaction at 298 K with partial pressures of NO₂ and N₂O₄ as 0.40 atm and 1.60 atm, respectively, is 2.61 kJ/mol.

Key Concepts

Standard Free Energy Change

Standard Free Energy Change, denoted as \( \Delta G^{\circ} \), indicates the free energy difference between reactants and products under standard conditions. These conditions typically involve:

• 1 atm pressure for gases
• 1 M concentration for solutions
• Pure substances in their most stable form at 1 atm

For the reaction \( 2 \mathrm{NO}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) \), we use the standard Gibbs free energies of formation (\( \Delta G_f^{\circ} \)) from Appendix C to determine the \( \Delta G^{\circ} \) of the reaction. Using the formula:\[ \Delta G^{\circ} = \Sigma n_i \Delta G^{\circ}_f(\text{products}) - \Sigma n_j \Delta G^{\circ}_f(\text{reactants}) \]Substitute the \( \Delta G_f^{\circ} \) values of \( \mathrm{N}_{2} \mathrm{O}_{4}(g) \) and \( \mathrm{NO}_{2}(g) \), we find \( \Delta G^{\circ} = -4.71 \thinspace \text{kJ/mol} \). This tells us about the reaction's favorability under standard conditions.
If \( \Delta G^{\circ} < 0 \), the reaction is spontaneous under standard conditions. If \( \Delta G^{\circ} > 0 \), the reaction is non-spontaneous.

Reaction Quotient

The Reaction Quotient, \( Q \), is a dimensionless number that provides a snapshot of the system's status concerning equilibrium. It is expressed as the ratio of the product of the concentrations (or partial pressures) of products to reactants, each raised to the power of their stoichiometric coefficients. The formula for \( Q \) in our reaction is:\[ Q = \frac{P_{(\mathrm{N_{2}O_{4})}}}{P_{(\mathrm{NO_{2})}^2}} \]Using the partial pressures given in the problem (\( P_{\mathrm{NO}_{2}} = 0.40 \) atm and \( P_{\mathrm{N_{2}O}_{4}} = 1.60 \) atm), we calculate:\[ Q = \frac{1.60}{(0.40)^2} = 10 \]This \( Q \) is compared to the equilibrium constant \( K \) to understand the direction in which the reaction will proceed:

• If \( Q < K \), the reaction proceeds forward to reach equilibrium.
• If \( Q > K \), the reaction will move in reverse.

Here, \( Q = 10 \) suggests how the reaction quotient affects the free energy and drives the reaction if it is not at equilibrium.

Partial Pressure

Partial pressure is the pressure that a gas in a mixture would exert if it alone occupied the entire volume. In chemical reactions involving gases, understanding partial pressures is pivotal to predict reaction behavior. They relate to total pressure via Dalton's Law of Partial Pressures, which states:
\[ P_\text{total} = P_{1} + P_{2} + \ldots + P_{n} \]
where \( P_i \) is the partial pressure of a given component. In kinetics and gaseous equilibria, partial pressures are used to calculate the reaction quotient \( Q \).
In our specific example, the partial pressures of \( \mathrm{NO}_{2} \) (0.40 atm) and \( \mathrm{N}_{2} \mathrm{O}_{4} \) (1.60 atm) influenced \( Q \) and subsequently the expression for \( \Delta G \). Pressures significantly affect reaction pathways, particularly in the field of gas-phase reactions, helping chemists anticipate changes in reactivity and stability.

Thermodynamics

Thermodynamics is the branch of physical chemistry that deals with energy changes in chemical reactions. It primarily revolves around concepts such as energy, work, enthalpy, entropy, and Gibbs free energy. It helps us understand phenomena such as spontaneity and equilibrium.

• Spontaneity: Determined mainly by \( \Delta G \). A negative \( \Delta G \) means the reaction can proceed spontaneously.


• Equilibrium: At equilibrium, \( \Delta G = 0 \) and \( Q = K \).


• Temperature and Pressure Influence: Changes can shift equilibria and alter \( \Delta G \).

In our scenario, thermodynamic principles help us deduce how changes in reaction conditions, such as different partial pressures or temperatures, impact \( \Delta G \), thus indicating the reaction's position concerning equilibrium.
This particular exercise underscores these concepts by linking calculated \( \Delta G \) to being indicative of whether or not certain reaction conditions favor product formation at 298 K.