Question

What is \(\Delta G\) for the formation of solid uranium hexafluoride from uranium and fluorine at \(25^{\circ} \mathrm{C}\) when the partial pressure of \(\mathrm{F}_{2}\) is \(0.045 \mathrm{~atm} ?\) The standard free energy of formation of \(\mathrm{UF}_{6}(s)\) is \(-2068 \mathrm{~kJ} / \mathrm{mol}\) $$\mathrm{U}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{UF}_{6}(s)$$ Is the reaction spontaneous in the forward or the reverse direction under these conditions?

Short answer

\( \Delta G = -2044.946 \ \text{kJ/mol} \), reaction is spontaneous in the forward direction.

Step-by-step solution

Understand the reaction and terms

First, we need to know that the problem involves calculating the change in Gibbs free energy (\( \Delta G \)) for the given reaction where uranium reacts with fluorine gas to form uranium hexafluoride (\( \mathrm{UF}_6 \)). The provided standard free energy change \( \Delta G^\circ_{f} \) for the formation of \( \mathrm{UF}_6(s) \) is \(-2068 \ \mathrm{kJ/mol}\), which is the value under standard conditions when pressure is 1 atm.

Calculate actual \( \Delta G \) using the reaction quotient

The equation to calculate \( \Delta G \) is given by:\[ \Delta G = \Delta G^\circ + RT \ln Q \]Where \( R \) is the gas constant (\(8.314 \ \mathrm{J/(mol \cdot K)}\) or \(0.008314 \ \mathrm{kJ/(mol \cdot K)}\) for our units), \( T \) is the temperature in Kelvin, and \( Q \) is the reaction quotient. In our case, \( Q = (\mathrm{P}_{F_2})^3 \), where \( \mathrm{P}_{F_2} = 0.045 \ \mathrm{atm} \).

Convert temperature to Kelvin

Convert the temperature from Celsius to Kelvin by adding 273.15: \[ T = 25 + 273.15 = 298.15 \ \mathrm{K} \]

Calculate the reaction quotient \( Q \)

Since \( Q = (\mathrm{P}_{F_2})^3 \), we have: \[ Q = (0.045)^3 = 0.000091125 \]

Calculate \( \Delta G \)

Substitute the values we have into the \( \Delta G \) equation:\[ \Delta G = -2068 + 0.008314 \times 298.15 \times \ln(0.000091125) \]Calculate \( \ln(0.000091125) \approx -9.304 \). Thus,\[ \Delta G = -2068 + 0.008314 \times 298.15 \times (-9.304) \]\[ \Delta G = -2068 + 23.054 \] (\( \Delta G^\circ \) is in kJ/mol, so the resulting energy correction must also be)\[ \Delta G = -2044.946 \ \text{kJ/mol} \]

Determine spontaneity of the reaction

Since \( \Delta G \) is negative, the reaction is spontaneous in the forward direction.

Key concepts

Spontaneity

Spontaneity in chemistry refers to whether a reaction will occur without outside intervention. When we talk about spontaneity, we are looking at the change in Gibbs Free Energy (\( \Delta G \)) during a reaction. \( \Delta G \) is a measure of the total energy change in the system as it transitions to a more stable state.
For a reaction, if \( \Delta G \)

• is negative, the process releases energy and is spontaneous in the forward direction.
• is positive, it requires energy input and is spontaneous in the reverse direction.
• is zero, the system is at equilibrium, and there is no net reaction occurring.

A negative \( \Delta G \) indicates that reactants convert into products, releasing free energy in the process, which can be harnessed to perform work.
In our exercise, the calculated \( \Delta G \) for the formation of uranium hexafluoride from uranium and fluorine is \(-2044.946 \ \text{kJ/mol}.\) This value, being negative, confirms that the formation reaction is spontaneous, meaning it will proceed on its own under the given conditions.

Reaction Quotient

The reaction quotient, denoted as \( Q \), is a crucial concept when assessing reaction spontaneity. It compares the current state of a reaction to its equilibrium state by considering the concentration of products and reactants. The expression for \( Q \) is similar to the equilibrium constant \( K \), but it applies to any point in the reaction, not just equilibrium.
For reactions involving gases, \( Q \) is typically calculated using partial pressures. For the given reaction:\[\mathrm{U}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{UF}_{6}(s) \]\( Q \) is determined solely by the partial pressure of fluorine because it's the only gaseous reactant.
The formula for this reaction is:\[Q = (\text{P}_{F_2})^3\]By substituting the pressure of fluorine (\(0.045\ \text{atm}\)) into the formula, we calculate:\[Q = (0.045)^3 = 0.000091125\]This \( Q \) value helps in determining how the actual reaction conditions affect \( \Delta G \), as it indicates deviations from the standard state used to calculate \( \Delta G^\circ \). If \( Q < K \), the reaction proceeds forward to reach equilibrium.

Standard Free Energy Change

The standard free energy change, symbolized as \( \Delta G^\circ \), represents the change in Gibbs Free Energy when reactants convert into products under standard conditions — usually 1 atmosphere pressure, \( 25^\circ \text{C} \) (298 K), and 1 M concentration for solutions. \( \Delta G^\circ \) provides a reference point for predicting reaction spontaneity.
In the given problem, \( \Delta G^\circ \) is \(-2068 \ \text{kJ/mol}.\) This indicates that under standard conditions, the formation of one mole of uranium hexafluoride from its elements releases 2068 kJ of free energy.
However, real-world conditions often differ from these ideal scenarios, requiring adjustments to the free energy value. The relationship between \( \Delta G^\circ \) and the actual \( \Delta G \) under specific conditions is expressed as:\[\Delta G = \Delta G^\circ + RT \ln Q\]Here, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( Q \) is the reaction quotient. This formula adjusts the standard free energy to reflect actual conditions, showing how deviations from standard pressure or concentration impact the reaction's spontaneity.