Question

Consider the reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g) $$ Given that the \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}\) is \(173.4 \mathrm{~kJ} / \mathrm{mol},\) calculate (a) the standard free energy of formation of \(\mathrm{NO},(\mathrm{b}) K_{P}\) for the reaction, and (c) \(K_{\mathrm{c}}\) for the reaction.

Short answer

The standard free energy of formation of NO is 86.7 kJ/mol, and both \(K_P\) and \(K_C\) are \(4.49 \times 10^{-31}\).

Step-by-step solution

Understanding the Reaction

Given the reaction \( \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g) \), we know that the standard Gibbs free energy change \( \Delta G^{\circ} \) is related to the free energy of formation of \( \mathrm{NO} \). The reaction shows that 2 moles of \( \mathrm{NO} \) are formed.

Calculating Standard Free Energy of Formation

The standard Gibbs free energy change \( \Delta G^{\circ} \) for a reaction is related to the standard Gibbs free energies of formation \( \Delta G_{f}^{\circ} \) of the products and reactants. Using \( \Delta G^{\circ} = \sum \Delta G_{f,\text{products}}^{\circ} - \sum \Delta G_{f,\text{reactants}}^{\circ} \), we get: \[ \Delta G^{\circ} = 2\Delta G_{f,\mathrm{NO}}^{\circ} - (\Delta G_{f,\mathrm{N}_2}^{\circ} + \Delta G_{f,\mathrm{O}_2}^{\circ}) \]. Given \( \Delta G^{\circ}(\mathrm{N}_2) = 0 \) and \( \Delta G^{\circ}(\mathrm{O}_2) = 0 \), \( 173.4 \text{ kJ/mol} = 2\Delta G_{f,\mathrm{NO}}^{\circ} \). Solving for \( \Delta G_{f,\mathrm{NO}}^{\circ} \) gives us \( \Delta G_{f,\mathrm{NO}}^{\circ} = 86.7 \text{ kJ/mol} \).

Relating ΔG° to Equilibrium Constant K_P

The equilibrium constant \( K_P \) can be found using the relation \( \Delta G^{\circ} = -RT \ln K_P \). Rearranging gives \( K_P = e^{-\frac{\Delta G^{\circ}}{RT}} \). Substitute \( \Delta G^{\circ} = 173.4 \times 10^3 \text{ J/mol} \), \( R = 8.314 \text{ J/mol K} \), and \( T = 298 \text{ K} \). Calculating gives \( K_P = e^{-\frac{173400}{(8.314)(298)}} \).

Solving for K_P

Calculate \( \frac{173400}{(8.314)(298)} \approx 70.16 \), so \( K_P = e^{-70.16} \approx 4.49 \times 10^{-31} \).

Relating K_P to K_C

The relationship between \( K_P \) and \( K_C \) is \( K_P = K_C(RT)^{\Delta n} \), where \( \Delta n \) is the change in moles of gas, \( \Delta n = 2 - (1 + 1) = 0 \). Thus, \( K_P = K_C \), so \( K_C = 4.49 \times 10^{-31} \).

Key concepts

Gibbs Free Energy

Gibbs Free Energy is a central concept in thermodynamics that helps determine whether a chemical reaction will occur spontaneously. It is represented by the symbol \( \Delta G \). If \( \Delta G \) is negative, the reaction will proceed spontaneously. If \( \Delta G \) is positive, the reaction is non-spontaneous. In a chemical reaction, the standard Gibbs free energy change, \( \Delta G^{\circ} \), reflects the energy difference when reactants transform into products under standard conditions. For the reaction \( \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g) \), the standard Gibbs free energy change \( \Delta G^{\circ} \) is given as 173.4 kJ/mol. This value indicates that under standard conditions, the reaction is non-spontaneous since it requires energy to proceed.

Equilibrium Constant

The equilibrium constant \( K \) represents the ratio of the concentration of products to reactants at equilibrium. It tells us about the favorability of a reaction proceeding towards products or staying at reactants. The equilibrium constant can be expressed in terms of pressure \( K_P \) or concentration \( K_C \). These constants provide insights into how far a reaction will proceed.To relate Gibbs free energy to the equilibrium constant, we use the equation:\[ \Delta G^{\circ} = -RT \ln K \]Where:

• \( R \) is the universal gas constant \( (8.314 \text{ J/mol K}) \)
• \( T \) is the temperature in Kelvin
• \( \ln \) stands for the natural logarithm

Using this equation, we can solve for \( K_P \) or \( K_C \). For the reaction provided, we calculated \( K_P \) to be approximately \( 4.49 \times 10^{-31} \), indicating a strong favor toward the reactants at equilibrium.

Chemical Reaction

Chemical reactions involve the transformation of substances through the breaking and forming of bonds. Each reaction has a balanced chemical equation that represents the stoichiometry, or the quantitative relationships of the reactants and products. Understanding the stoichiometry is crucial to calculating changes in energy and equilibrium positions.In the reaction \( \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g) \), the stoichiometry indicates that one mole of nitrogen gas reacts with one mole of oxygen gas to form two moles of nitrogen monoxide gas. The stoichiometric coefficients (the numbers in front of the chemical species) guide us in calculating the standard free energy of formation and the changes in mole numbers, \( \Delta n \), which are essential for translating between \( K_P \) and \( K_C \). Here, \( \Delta n = 0 \), leading to the conclusion that \( K_P = K_C \). This relationship helps in understanding how gas-phase equilibria respond to changes in conditions.