Question

Free Energy and Equilibrium (Section) Consider the reaction 2 \(\mathrm{NO} 2(g) \rightarrow \mathrm{N} 2 \mathrm{O} 4(g) .\) (a) Using data from Appendix 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.

Short answer

The standard Gibbs free energy change (ΔG°) for the reaction 2NO₂(g) → N₂O₄(g) at 298 K is 1.0 kJ/mol. Under the given partial pressures of NO₂ and N₂O₄ at 298 K, the Gibbs free energy change (ΔG) is 11.3 kJ/mol.

Step-by-step solution

Calculate ΔG° using standard Gibbs free energy of formation values

To calculate the standard Gibbs free energy change (ΔG°) for the reaction, we can use the following formula: ΔG° = ΣnGf°(products) - ΣmGf°(reactants) where n and m are the stoichiometric coefficients, and Gf° are the standard Gibbs free energy of formation values. From Appendix C, the standard Gibbs free energy of formation for NO₂(g) is 51.3 kJ/mol and for N₂O₄(g) is 103.6 kJ/mol. Now we can substitute the values into the formula: ΔG° = 1 × 103.6 - 2 × 51.3 ΔG° = \(103.6 - 102.6\) ΔG° = 1.0 kJ/mol

Calculate ΔG under given partial pressures

We can use the following equation to calculate the Gibbs free energy change (ΔG) under non-standard conditions: ΔG = ΔG° + RTlnQ where R is the gas constant (8.314 J/mol•K), T is the temperature in Kelvin (298 K), and Q is the reaction quotient. For this reaction, the reaction quotient (Q) is given by: Q = \(\frac{[N_{2}O_{4}]}{[NO_{2}]^{2}}\) We can convert partial pressures to the reaction quotient by replacing concentrations with partial pressures: Q = \(\frac{P_{N_{2}O_{4}}}{P_{NO_{2}}^{2}}\) We are given partial pressures: PNO₂ = 0.40 atm and PN₂O₄ = 1.60 atm. Substituting these values into the equation for Q, we get: Q = \(\frac{1.60}{(0.40)^{2}}\) Q = 10 Now that we have the value of Q, we can calculate ΔG: ΔG = ΔG° + RTlnQ ΔG = \(1.0 × 10^{3}\) J/mol + \((8.314 J/mol \cdot K)(298 K)\)ln(10) ΔG = \(1.0 × 10^{3}\) J/mol + (24710515.71 * 0.00004168) ΔG = \(1.0 × 10^{3}\) J/mol + 10.3 kJ/mol ΔG = 11.3 kJ/mol The Gibbs free energy change, ΔG, at 298 K and given partial pressures is 11.3 kJ/mol.

Key concepts

Chemical Equilibrium

Understanding the concept of chemical equilibrium is fundamental to grasping how reactions occur in nature and industry. Chemical equilibrium is achieved in a chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time, even though the reactions continue to proceed at the molecular level. The equilibrium state doesn't mean that the reactants and products are in equal concentrations, but rather that their ratios remain consistent.

In the reaction given in the exercise, \( 2 \mathrm{NO}_2(g) \rightleftharpoons \mathrm{N}_2\mathrm{O}_4(g) \), when it reaches equilibrium, the rate at which \(NO_2\) reacts to form \(N_2O_4\) is equal to the rate at which \(N_2O_4\) decomposes back into \(NO_2\). This condition is essential for predicting how changes in conditions, like temperature or pressure, affect the position of the equilibrium.

Reaction Quotient

The reaction quotient, designated as \(Q\), predicts the direction of reaction and whether a system is at equilibrium. It is calculated in a similar way to the equilibrium constant \(K\), but uses the current concentrations or partial pressures of reactants and products, not those at equilibrium.

For the generic reaction \(a\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D}\), \(Q\) is given by \(Q = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b} \), where the letters represent the stoichiometric coefficients. If \(Q < K\), the forward reaction is favored to reach equilibrium. If \(Q > K\), the reaction will shift towards the reactants to achieve equilibrium. Only when \(Q = K\), the system is at equilibrium and no net reaction occurs.

In our exercise, after calculating the reaction quotient with the partial pressures provided, we can compare this value with the equilibrium constant to determine which direction the reaction will proceed to reach equilibrium.

Standard Gibbs Free Energy of Formation

The standard Gibbs free energy of formation, \(\Delta G^\circ_f\), is the energy change that occurs when one mole of a compound is formed from its elements in their standard states. It's a key concept for understanding and predicting the spontaneity of reactions at standard conditions (1 atm pressure and 298 K temperature).

The value of \(\Delta G^\circ_f\) is particularly useful for calculating the overall standard Gibbs free energy change for a reaction, \(\Delta G^\circ\), which tells us whether a reaction is spontaneously favorable (negative \(\Delta G^\circ\)) or nonspontaneous (positive \(\Delta G^\circ\)) under standard conditions. This calculation is made using the formula \(\Delta G^\circ = \Sigma n\Delta G^\circ_f(\text{products}) - \Sigma m\Delta G^\circ_f(\text{reactants})\), where \(n\) and \(m\) are the stoichiometric coefficients.

In the exercise provided, by determining the \(\Delta G^\circ\) for the reaction and further adjusting for non-standard conditions using the reaction quotient, we can accurately predict the spontaneity and directionality of the reaction at any given set of conditions.