Problem 67 - Chemistry: The Central Science

Question

Consider the following reaction between oxides of nitrogen: $$ \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g) $$ (a) Use data in Appendix \(\mathrm{C}\) to predict how \(\Delta G^{\circ}\) for the reaction varies with increasing temperature. (b) Calculate \(\Delta G^{\circ}\) at \(800 \mathrm{~K}\), assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with temperature. Under standard conditions is the reaction spontaneous at \(800 \mathrm{~K} ?(\mathrm{c})\) Calculate \(\Delta G^{\circ}\) at \(1000 \mathrm{~K}\). Is the reaction spontaneous under standard conditions at this temperature?

Step-by-Step Solution

Answer

(a) As temperature increases, if ΔS° is positive, ΔG° will decrease, and if ΔS° is negative, ΔG° will increase. (b) Calculate the ΔH° and ΔS° for the reaction using Appendix C data, then find ΔG°(800K) using the Gibbs-Helmholtz Equation. If ΔG°(800K) is negative, the reaction is spontaneous at 800 K; if positive, it's non-spontaneous. (c) Calculate ΔG°(1000K) using the Gibbs-Helmholtz Equation. If ΔG°(1000K) is negative, the reaction is spontaneous at 1000 K; if positive, it's non-spontaneous.

Step 1: (a) Predict how ΔG° varies with increasing temperature

Use the Gibbs-Helmholtz Equation: \(ΔG° = ΔH° - TΔS°\) According to this equation, the change in Gibbs free energy (ΔG°) for the reaction is dependent on the enthalpy change (ΔH°), entropy change (ΔS°), and temperature (T). If the value of ΔG° is negative, the reaction will be spontaneous, and if it's positive, the reaction will be non-spontaneous. With increasing temperature, the term "TΔS°" becomes dominant. Thus, if the entropy change is positive, ΔG° will decrease, and if the entropy change is negative, ΔG° will increase with the increase in temperature.

Step 2: (b) Calculate ΔG° at 800 K and determine spontaneity

First, we need to find the values of ΔH° and ΔS° for the given reaction using the data from Appendix C. ΔH° = [3 × ΔH°(NO)] – [ΔH°(NO₂) + ΔH°(N₂O)] ΔS° = [3 × ΔS°(NO)] – [ΔS°(NO₂) + ΔS°(N₂O)] Now, calculate ΔG° at 800 K using the Gibbs-Helmholtz Equation: ΔG°(800K) = ΔH° - (800 K × ΔS°) If ΔG°(800K) is negative, the reaction is spontaneous at 800 K under standard conditions. If ΔG°(800K) is positive, the reaction is non-spontaneous under these conditions.

Step 3: (c) Calculate ΔG° at 1000 K and determine spontaneity

To find the ΔG° at 1000 K, use the Gibbs-Helmholtz Equation: ΔG°(1000K) = ΔH° - (1000 K × ΔS°) If ΔG°(1000K) is negative, the reaction is spontaneous at 1000 K under standard conditions. If ΔG°(1000K) is positive, the reaction is non-spontaneous under these conditions.

Key Concepts

Gibbs-Helmholtz Equation

The Gibbs-Helmholtz Equation is a fundamental expression in thermodynamics used to estimate the change in Gibbs free energy, noted as \( \Delta G \). It is expressed as:

• \( \Delta G = \Delta H - T\Delta S \)

In this equation:

• \( \Delta G \): Change in Gibbs free energy
• \( \Delta H \): Change in enthalpy
• \( \Delta S \): Change in entropy
• \( T \): Temperature in Kelvin

This relationship shows how Gibbs free energy is influenced by both enthalpy and entropy changes along with temperature. As temperature increases, the \( T\Delta S \) term becomes increasingly significant, potentially leading to either a decrease or increase in \( \Delta G \) depending on the sign of \( \Delta S \). Understanding this balance is crucial as it helps predict how favorable a reaction is under varying conditions.

Enthalpy Change

Enthalpy change, represented by \( \Delta H \), is the heat absorbed or released during a chemical reaction at constant pressure. It is an important parameter for understanding energy changes accompanying reactions. For exothermic reactions, \( \Delta H \) is negative, indicating that the system releases energy to its surroundings. Meanwhile, endothermic reactions have a positive \( \Delta H \) as they absorb energy.
In the context of the given reaction between nitrogen oxides, calculating \( \Delta H \) involves subtracting the enthalpy of the reactants from the enthalpy of the products. This helps determine whether the reaction releases or absorbs energy.
Knowing \( \Delta H \) is also critical when using the Gibbs-Helmholtz Equation, as it influences the overall change in Gibbs free energy and thus the reaction's spontaneity at different temperatures.

Entropy Change

Entropy change, denoted as \( \Delta S \), measures the disorder or randomness within a system. Every reaction results in either an increase or decrease in entropy. A positive \( \Delta S \) implies that the products are more disordered than the reactants, while a negative \( \Delta S \) means the system has become more ordered.

• For example, reactions that produce gases from solids or liquids generally increase entropy due to the increased freedom of movement.

For the reaction of nitrogen oxides given in the exercise, assessing \( \Delta S \) will indicate the direction of entropy change, which plays a crucial role when analyzing Gibbs free energy. As temperature rises, the \( T\Delta S \) component of the Gibbs-Helmholtz Equation gains prominence, highlighting the importance of \( \Delta S \) in determining the spontaneity of the reaction.

Reaction Spontaneity

Reaction spontaneity refers to whether a chemical reaction will proceed on its own without the input of additional energy. It is closely tied to the sign of \( \Delta G \); when \( \Delta G \) is negative, the reaction is considered spontaneous.This spontaneity depends on both enthalpy and entropy changes as indicated by the Gibbs-Helmholtz Equation. Temperature also plays a pivotal role; a reaction may be spontaneous at one temperature but not at another.

• A negative \( \Delta H \) and positive \( \Delta S \) together tend to make a reaction more spontaneous, particularly at higher temperatures due to the \( T\Delta S \) term.
• If both \( \Delta H \) and \( \Delta S \) are negative, or if both are positive, temperature becomes a critical factor in determining spontaneity as it affects the \( T\Delta S \) term's influence on \( \Delta G \).

Thus, understanding the interplay of these factors is essential in predicting reaction behavior under different conditions, as seen in exercises with varying temperature evaluations.