Question

For the reaction at \(298 \mathrm{K}\), $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ 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 K? Assuming that \(\Delta H^{\circ}\) 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 value of ΔG° at 298 K is -5373.2 J. The temperature at which ΔG° = 0 is approximately 328.6 K. ΔG° is negative below this temperature (T < 328.6 K).

Step-by-step solution

Calculate ΔG° at 298 K

To calculate the change in Gibbs free energy (ΔG°) at 298 K, we can use the given values of ΔH° and ΔS° with the Gibbs-Helmholtz equation: ΔG° = ΔH° - TΔS° ΔG° = -58.03 kJ - (298 K)(-176.6 J/K) Note that the units for ΔH° are in kJ and for ΔS° are in J/K. We need to convert these units to the same base before calculation. Therefore, we will convert ΔH° into J by multiplying by 1000: ΔH° = -58.03 kJ × 1000 J/kJ = -58030 J Now, we can substitute the converted values into the equation: ΔG° = -58030 J - (298 K)(-176.6 J/K)

Solve for ΔG°

Now, we can solve the equation for ΔG°: ΔG° = -58030 J + (298 K)(176.6 J/K) ΔG° = -58030 J + 52656.8 J ΔG° = -5373.2 J So, the value of ΔG° at 298 K is -5373.2 J.

Find the temperature at which ΔG° = 0

To find the temperature at which ΔG° = 0, we'll set the Gibbs-Helmholtz equation equal to zero: 0 = ΔH° - TΔS° To solve for T, we can rearrange the equation: T = ΔH°/ΔS° Now we can substitute the values of ΔH° and ΔS°: T = (-58030 J)/(-176.6 J/K)

Solve for T

Now, we can solve for T: T = 328.6 K So, the temperature at which ΔG° = 0 is approximately 328.6 K.

Determine if ΔG° is negative above or below this temperature

To determine if ΔG° is negative above or below this temperature, we can look at the signs of ΔH° and ΔS°. If both have the same sign, ΔG° will be negative below the calculated temperature, while if they have opposite signs, ΔG° will be negative above the calculated temperature. In this case, both ΔH° and ΔS° are negative (-58030 J and -176.6 J/K, respectively). Therefore, ΔG° is negative below the temperature at which it is equal to 0: ΔG° < 0 when T < 328.6 K ΔG° becomes negative below 328.6 K.

Key concepts

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat, work, temperature, and energy. It provides a framework for describing how different forms of energy transform and how these changes relate to the physical properties of substances.
In a chemical reaction, thermodynamics helps determine if a process will occur spontaneously, based on the calculation of Gibbs Free Energy ( abla G").
This prediction revolves around three key concepts:

• Enthalpy ( abla H"): Can be described as the total energy content, including internal energy plus the product of pressure and volume.
• Entropy ( abla S"): Measures the disorder or randomness in a system.
• Gibbs Free Energy ( abla G"): Combines both enthalpy and entropy to predict the spontaneity of a process.

When abla G" is negative, the reaction is spontaneous. It means that the process could happen without needing energy from outside.
In other words, thermodynamics doesn’t tell us about the speed of a reaction, only about its potential to happen.

Entropy

Entropy is a measure of the disorder or randomness within a system. It is an essential concept in understanding the second law of thermodynamics, which states that the total entropy of an isolated system will tend to increase over time.
In the context of chemical reactions, entropy can give insight into the molecular arrangement and the feasibility of a reaction. Typically:

• An increase in entropy (abla S") generally favors spontaneity in chemical processes, contributing positively to abla G" becoming negative.
• When entropy decreases, it indicates that the system becomes more ordered, often leading to non-spontaneous reactions unless overcome by enthalpic contributions.

For the given reaction, abla S" is negative, indicating a decrease in disorder. This means molecules become more ordered in forming abla _2O_4(g) from NO2(g).
The balance between entropy and enthalpy changes determines the overall change in Gibbs Free Energy (\( \Delta G \)) for the reaction.

Enthalpy

Enthalpy (abla H) is a measure of the total heat content of a system and reflects the energy needed to form or break bonds during a chemical reaction.
It plays a significant role in determining whether a reaction will release or absorb heat. Enthalpy changes are considered under constant pressure conditions, a typical scenario in thermodynamic evaluations.

• If abla H" is negative, the reaction is exothermic, indicating heat release.
• A positive abla H" indicates an endothermic reaction, where heat is absorbed.

For the reaction under study, abla H" is negative, meaning the formation of abla _2O_4 (g) releases heat, contributing to its spontaneous occurrence. Combined with entropy, this forms a complete picture determining the directionality of chemical processes. Enthalpy changes, alongside entropy, help in calculating the Gibbs Free Energy using the equation:
\[abla G = abla H - Tabla S\]
In this equation, both heat exchange and entropy changes together decide if and how a reaction will proceed.