Question

For the reaction at 298 K, $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{kJ}\) and \(-176.6 \mathrm{J} / \mathrm{K}\) , respectively. What is the value of \(\Delta G^{\circ}\) at 298 \(\mathrm{K}\) ? Assuming that \(\Delta H^{p}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G^{\circ}\) negative above or below this temperature?

Short answer

The standard Gibbs free energy change (ΔG°) for the given reaction at 298 K is -25.45 kJ. The temperature at which ΔG° becomes zero is 328.3 K. ΔG° is negative below this temperature.

Step-by-step solution

Recall the Gibbs-Helmholtz equation

The Gibbs-Helmholtz equation calculates the relationship between the standard Gibbs free energy change (ΔG°), the standard enthalpy change (ΔH°), and the standard entropy change (ΔS°) for a given reaction: \[ΔG° = ΔH° - TΔS°\] where ΔG° is the standard Gibbs free energy change, ΔH° is the standard enthalpy change, T is the temperature in kelvin (K), and ΔS° is the standard entropy change.

Calculate ΔG° at 298 K

Using the Gibbs-Helmholtz equation, we can find ΔG° at 298 K given the values of ΔH° and ΔS°: ΔH° = -58.03 kJ = -58030 J (convert kJ to J) ΔS° = -176.6 J/K T = 298 K \[ΔG° = ΔH° - TΔS° = -58030 \,\text{J} - (298 \,\text{K} \times -176.6 \, \frac{\text{J}}{\text{K}})\]

Solve for ΔG° at 298 K

Calculate the value of ΔG° at 298 K by solving the equation: ΔG° = -58030 J + (298 K × 176.6 J/K) = -25,445 J = -25.45 kJ (convert J to kJ) So at 298 K, the standard Gibbs free energy change, ΔG°, is -25.45 kJ.

Determine the temperature at which ΔG° = 0

To find the temperature at which the standard Gibbs free energy change is zero, we need to set ΔG° = 0 in the Gibbs-Helmholtz equation and solve for T: \[0 = ΔH° - TΔS°\] Rearrange the equation to solve for T: \[T = \frac{ΔH°}{ΔS°} = \frac{-58030 \, \text{J}}{-176.6 \, \frac{\text{J}}{\text{K}}}\]

Calculate the temperature at which ΔG° = 0

Calculate the temperature at which ΔG° is 0: \(T = \frac{-58030 \, \text{J}}{-176.6 \, \frac{\text{J}}{\text{K}}} = 328.3 \, \text{K}\) So at 328.3 K, ΔG° becomes zero.

Determine whether ΔG° is negative above or below this temperature

Observe the sign of ΔS° to determine where ΔG° is negative: Since ΔS° is negative, ΔG° will be negative when the temperature is below 328.3 K. This is because a lower temperature will make the term TΔS° positive, resulting in a negative ΔG° due to the subtraction in the Gibbs-Helmholtz equation. In summary, at 298 K, the value of ΔG° for the given reaction is -25.45 kJ. The temperature at which ΔG° is zero is 328.3 K, and ΔG° is negative below this temperature.

Key concepts

Gibbs-Helmholtz equation

The Gibbs-Helmholtz equation is a key relationship in thermodynamics that links Gibbs free energy (ΔG°), enthalpy (ΔH°), and entropy (ΔS°) for a reaction. The formula is represented as:\[ΔG° = ΔH° - TΔS°\]where:

• \(ΔG°\) is the Gibbs free energy change.
• \(ΔH°\) is the change in enthalpy.
• \(T\) is the absolute temperature in Kelvin.
• \(ΔS°\) is the entropy change.

This equation provides insight into whether a reaction is spontaneous. If \(ΔG°\) is negative, the reaction tends to be spontaneous under the given conditions. The concept of spontaneity is important as it helps predict reaction behavior without needing to experimentally test it at every condition.

enthalpy change

Enthalpy change, \(ΔH°\), refers to the heat absorbed or released during a chemical reaction at constant pressure. It is measured in kilojoules per mole (kJ/mol) and can indicate whether a reaction is endothermic or exothermic.

• If \(ΔH°\) is positive, the reaction is endothermic, meaning it absorbs heat from its surroundings.
• If \(ΔH°\) is negative, the reaction is exothermic, meaning it releases heat to its surroundings.

In this specific example, the given \(ΔH°\) is \(-58.03 \, ext{kJ/mol}\), which tells us the reaction releases heat, making it exothermic. The enthalpy change is a central component of the Gibbs-Helmholtz equation and helps determine the spontaneity of a reaction.

entropy change

Entropy change, symbolized as \(ΔS°\), is a measure of the disorder or randomness in a system. In thermodynamics, an increase in entropy signals an increase in disorder.

• If \(ΔS°\) is positive, the system's disorder increases during the reaction.
• If \(ΔS°\) is negative, there is a decrease in disorder.

For this reaction, \(ΔS°\) is \(-176.6 \, ext{J/K}\), indicating a decrease in disorder. A negative entropy change suggests that the system becomes more ordered during the reaction. This value influences whether the Gibbs free energy change will be positive or negative when calculated using the Gibbs-Helmholtz equation, especially at different temperatures.

temperature dependence

Gibbs free energy's temperature dependence is crucial for understanding reaction spontaneity at various conditions. Through the Gibbs-Helmholtz equation, we can see how temperature affects \(ΔG°\):\[ΔG° = ΔH° - TΔS°\]

• A higher temperature multiplies the entropy term \(TΔS°\), impacting \(ΔG°\).
• The sign of \(ΔS°\) determines if \(TΔS°\) adds to or subtracts from \(ΔH°\), altering spontaneity conditions.

For instance, in this reaction:

• At 298 K, \(ΔG°\) is \(-25.45\, ext{kJ}\), showing spontaneity.
• At 328.3 K, \(ΔG°\) becomes zero, a crucial point where spontaneity switches.
• Below 328.3 K, the reaction remains spontaneous (negative \(ΔG°\)) due to a negative \(ΔS°\), making \(TΔS°\) positive, supporting negative total \(ΔG°\).

Thus, understanding temperature dependence helps predict reaction direction as conditions change.