Question

Consider the following reaction at \(800 . \mathrm{K}\) : $$\mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g)$$ An equilibrium mixture contains the following partial pressures: \(P_{\mathrm{N}_{2}}=0.021 \mathrm{~atm}, P_{\mathrm{F}_{2}}=0.063 \mathrm{~atm}, P_{\mathrm{NF}_{3}}=0.48 \mathrm{~atm} .\) Calculate \(\Delta G^{\circ}\) for the reaction at \(800 . \mathrm{K}\).

Short answer

The standard Gibbs free energy change for the reaction at 800 K is approximately -8665 J/mol.

Step-by-step solution

Calculate the Reaction Quotient (Q)

We will use the equilibrium partial pressures to calculate the reaction quotient Q. The equation for Q in terms of partial pressures is: \(Q = \frac{(P_{NF_3})^2}{P_{N_2} \times (P_{F_2})^3}\) Plug in the given partial pressures: \(Q = \frac{(0.48 \space atm)^2}{(0.021 \space atm) \times (0.063 \space atm)^3}\)

Calculate the Gibbs Free Energy Change (ΔG)

Now that we have the reaction quotient (Q), we can find the Gibbs free energy change (ΔG) using the relation: ΔG = -RT·lnQ Given temperature T = 800 K and using the gas constant R = 8.314 J/mol·K: ΔG = - (8.314 J/mol·K) × (800 K) × ln(Q) Solve for ΔG: ΔG = -15321.6 J/mol

Calculate the Standard Gibbs Free Energy Change (ΔG°)

Now, we can use the equation relating ΔG, ΔG°, and Q to solve for ΔG°. Rearrange the equation to isolate ΔG°: ΔG° = ΔG - RT·lnQ ΔG° = -15321.6 J/mol - (8.314 J/mol·K) × (800 K) × ln(Q) Plug in the given values and solve for ΔG°: ΔG° ≈ -8665 J/mol So, the standard Gibbs free energy change for the reaction at 800 K is approximately -8665 J/mol.

Key concepts

Understanding Reaction Quotient

The reaction quotient, denoted by the symbol \( Q \), is a measure used to determine how far a reaction has proceeded towards equilibrium. It compares the current state of reaction to the state at equilibrium. To calculate it, you'll look at the ratio of the concentrations (or partial pressures in the case of gases) of the products to the reactants, much like an equilibrium constant \( K \), but for any moment during the reaction:\[Q = \frac{(P_{NF_3})^2}{P_{N_2} \times (P_{F_2})^3}\]In this expression, the partial pressures \( P_{NF_3} \), \( P_{N_2} \), and \( P_{F_2} \) represent the current conditions for nitrogen trifluoride, nitrogen, and fluorine, respectively. These pressures are used to calculate \( Q \), which can be compared to the equilibrium constant \( K \). If \( Q = K \), the system is at equilibrium; if \( Q < K \), the reaction will proceed forward, and if \( Q > K \), the reaction will proceed in reverse to reach equilibrium.

Role of Partial Pressure

Partial pressure is the pressure exerted by a single gas in a mixture of gases. It plays a fundamental role in chemical reactions involving gases, particularly when calculating reaction quotients and equilibrium constants. In the given exercise, understanding the partial pressures of \( N_2 \), \( F_2 \), and \( NF_3 \) is essential:

• \( P_{N_2} = 0.021 \, \text{atm} \)
• \( P_{F_2} = 0.063 \, \text{atm} \)
• \( P_{NF_3} = 0.48 \, \text{atm} \)

Each partial pressure reflects the concentration of gas involved in the reaction. When you have the partial pressures of all gases involved, you can find the reaction quotient \( Q \), a key step in analyzing the reaction's progress. Using the relation of these pressures in calculations helps in determining important reaction parameters such as the Gibbs free energy changes.

Importance of Standard Conditions

In thermodynamics, standard conditions refer to a set of agreed conditions under which measurements are made. For gases, standard conditions are typically set at a pressure of 1 atmosphere (atm). For reactions, this ensures that calculated values, like Gibbs free energy changes, are comparable across different studies and experiments.Standard Gibbs free energy change, \( \Delta G^\circ \), indicates the spontaneity of a reaction under standard conditions. It is distinct from \( \Delta G \), the actual Gibbs free energy change at particular conditions of temperature and pressure. For example, if \( \Delta G^\circ \) is negative, the reaction under standard conditions tends to proceed spontaneously. Standard conditions provide a baseline, allowing us to predict the direction and extent of chemical processes when conditions vary.

Concept of Chemical Equilibrium

Chemical equilibrium occurs 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, though they are not necessarily equal.The condition of equilibrium is described mathematically by the equilibrium constant \( K \), which is defined similarly to the reaction quotient \( Q \):\[K = \frac{(P_{NF_3})^2}{P_{N_2} \times (P_{F_2})^3}\]At equilibrium, \( Q \) will equal \( K \). Understanding equilibrium is crucial for predicting the outcome of reactions. In a teaching context, showing how \( Q \) relates to \( K \) helps students grasp how systems change to reach equilibrium, and the extent to which a reaction will yield products under given conditions. By mastering this concept, you'll better predict reaction behaviors and conditions necessary to drive reactions toward desired product formation.