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 change in Gibbs free energy (ΔG°) for the reaction at 800K is approximately -39460 J/mol.

Step-by-step solution

Write the expression for the equilibrium constant (K)

The equilibrium constant K can be written in terms of the partial pressures of the components at equilibrium: $$ K = \frac{P_{\mathrm{NF_{3}}}^2}{P_{\mathrm{N_{2}}} \times P_{\mathrm{F_{2}}}^3} $$

Calculate the equilibrium constant (K) using the given partial pressures

Using the provided partial pressures, we can substitute the values into the K expression and calculate its value: $$ K = \frac{(0.48 \mathrm{~atm})^2}{(0.021 \mathrm{~atm}) \times (0.063 \mathrm{~atm})^3} \approx 199.14 $$

Use the relationship ΔG° = -RT ln(K) to calculate ΔG°

Now that we have the value of the equilibrium constant (K), we can use the formula ΔG° = -RT ln(K) to find the standard change in Gibbs free energy for the reaction at 800K. Here, R is the universal gas constant (8.314 J/mol·K), and T is the temperature in Kelvin (800 K). $$ \Delta G^{\circ} = - (8.314 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}) \times (800 \mathrm{~K}) \times \ln(199.14) \approx -39460 \frac{\mathrm{J}}{\mathrm{mol}} $$ Therefore, the standard change in Gibbs free energy (ΔG°) for the reaction at 800K is approximately -39460 J/mol.

Key concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant, denoted as \( K \), is a numerical value that helps us understand the concentration ratio of the products to reactants at equilibrium. It specifically gauges the position of equilibrium, showing which direction is favored when the reaction reaches stability. For a general reaction involving gases, based on their partial pressures, the equilibrium constant expression is derived by taking the partial pressures of the products raised to the power of their coefficients and dividing by the partial pressures of the reactants raised to the power of their coefficients.
This can be seen in the equation:

• For the reaction \( \mathrm{N}_{2}(g) + 3 \mathrm{~F}_{2}(g) \rightarrow 2 \mathrm{NF}_{3}(g) \), \( K = \frac{P_{\mathrm{NF}_3}^2}{P_{\mathrm{N}_2} \times P_{\mathrm{F}_2}^3} \).

Understanding \( K \) is crucial because:

• A large \( K \) value indicates a forward reaction favoring product formation.
• A small \( K \) value suggests that reactants are favored at equilibrium.

This helps chemists predict the extent of reactions and decide on conditions for industrial processes.

Partial Pressure

Partial pressure is a way to measure the pressure contributed by each gas in a mixture of gases. Every gas in a mixture exerts pressure as if it were alone, and the total pressure is the sum of these individual pressures, known as partial pressures.In chemical equilibrium, especially involving gases, knowing the partial pressures is vital for calculating equilibrium constants, such as \( K \). The partial pressure of a gas can be calculated using the ideal gas law and the mole fraction within the mixture.
The significance of partial pressures in the context of equilibrium is:

• They allow us to express the equilibrium constant in terms of pressure, which is especially useful for gases.
• They help in deducing the shifts in equilibria, based on changes in concentration or pressure of the involved gases.

By knowing partial pressures such as \( P_{\mathrm{N}_2} = 0.021 \text{ atm} \), \( P_{\mathrm{F}_2} = 0.063 \text{ atm} \), \( P_{\mathrm{NF}_3} = 0.48 \text{ atm} \) for a specific equilibrium, we can efficiently perform equilibrium calculations.

Standard Change in Gibbs Free Energy

The standard change in Gibbs free energy, represented by \( \Delta G^{ ext{o}} \), is a thermodynamic quantity that reflects the maximum reversible work that is done in a process at constant temperature and pressure.It can be related to the equilibrium constant \( K \) through the equation:\[ \Delta G^{\circ} = -RT \ln(K) \]where

• \( R \) is the universal gas constant \(8.314 \frac{\text{J}}{{\text{mol} \cdot \text{K}}} \)
• \( T \) is the absolute temperature in Kelvin

For the example reaction at \(800 \text{ K}\), \[ \Delta G^{\circ} \] was calculated as \(-39460 \text{ J/mol} \).
This negative value suggests the reaction is spontaneous under standard conditions.
Understanding \( \Delta G^{\circ} \) helps us determine:

• If a reaction will proceed without energy input (negative \( \Delta G^{\circ} \))
• How far a reaction will shift towards products or reactants at equilibrium.

Comparing \( \Delta G^{\circ} \) values is essential in studying reaction feasibilities in various conditions.

Chemical Equilibrium

Chemical equilibrium is the state in which the concentrations of reactants and products remain constant over time, indicating that the forward and reverse reactions occur at equal rates.
In this stable state, no net change is observable, though molecular activity continues at a dynamic pace.The significance of achieving chemical equilibrium is rooted in predicting and controlling the outputs of chemical reactions, especially in industrial and laboratory settings.
The key aspects of chemical equilibrium include:

• The components involved cease to change overall measurements once reached.
• The reaction quotient \( Q \) becomes equal to the equilibrium constant \( K \).

When conditions change (e.g. pressure, temperature, concentration), Le Chatelier’s principle helps predict the direction in which the equilibrium will shift.
These changes, while maintaining new equilibria, help in optimizing processes like Haber-Bosch for ammonia production or even shifting blood oxygen levels with breathing patterns.