Problem 40

Question

Discuss the effect of temperature change on the spontaneity of the following reactions at 1 atm. (a) $\mathrm{Al}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Fe}(s) \longrightarrow 2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s)$ $$ \Delta H^{\circ}=+851.5 \mathrm{~kJ} ; \Delta S^{\circ}=+38.5 \mathrm{~J} / \mathrm{K} $$ (b) $\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g)$ $$ \Delta H^{\circ}=-50.6 \mathrm{~kJ} ; \Delta S^{\circ}=0.3315 \mathrm{~kJ} / \mathrm{K} $$ (c) $\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$ $$ \Delta H^{\circ}=98.9 \mathrm{~kJ} ; \Delta S^{\circ}=+0.0939 \mathrm{~kJ} / \mathrm{K} $$

Step-by-Step Solution

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Answer

Based on the analysis of Gibbs free energy change for the given reactions, the effect of temperature change on the spontaneity of the reactions can be summarized as follows: - Reaction (a): The temperature needs to be higher than 22117 K for the reaction to be spontaneous. - Reaction (b): This reaction is spontaneous at all temperatures. - Reaction (c): The temperature needs to be higher than 1053 K for the reaction to be spontaneous.

1Step 1: Reaction (a) - Analyzing Gibbs free energy change

For the reaction (a), we have the following values of enthalpy and entropy change: $\Delta H^{\circ} = +851.5 \mathrm{~kJ}$ and $\Delta S^{\circ} = +38.5 \mathrm{~J} / \mathrm{K}$ Both enthalpy and entropy change are positive. Let's plug these values into the Gibbs free energy change formula and calculate the critical temperature at which $\Delta G$ becomes negative: $\Delta G = 851.5 - T(0.0385)$ For the reaction to be spontaneous, $\Delta G < 0$. Therefore: $851.5 - T(0.0385) < 0$ Now, find the critical temperature: $T > \frac{851.5}{0.0385} \approx 22117 \mathrm{K}$ So, for reaction (a) to be spontaneous, the temperature needs to be greater than 22117 K.

2Step 2: Reaction (b) - Analyzing Gibbs free energy change

For the reaction (b), we have the following values of enthalpy and entropy change: $\Delta H^{\circ} = -50.6 \mathrm{~kJ}$ and $\Delta S^{\circ} = +0.3315 \mathrm{~kJ} / \mathrm{K}$ In this case, enthalpy change is negative and entropy change is positive. Let's plug these values into the Gibbs free energy change formula: $\Delta G = -50.6 - T(0.3315)$ For the reaction to be spontaneous, $\Delta G < 0$. Therefore: $-50.6 - T(0.3315) < 0$ In this case, no matter the value of temperature, the $\Delta G$ will always be negative, making the reaction spontaneous at all temperatures.

3Step 3: Reaction (c) - Analyzing Gibbs free energy change

For the reaction (c), we have the following values of enthalpy and entropy change: $\Delta H^{\circ} = +98.9 \mathrm{~kJ}$ and $\Delta S^{\circ} = +0.0939 \mathrm{~kJ} / \mathrm{K}$ Both enthalpy and entropy change are positive. Let's plug these values into the Gibbs free energy change formula and calculate the critical temperature at which $\Delta G$ becomes negative: $\Delta G = 98.9 - T(0.0939)$ For the reaction to be spontaneous, $\Delta G < 0$. Therefore: $98.9 - T(0.0939) < 0$ Now, find the critical temperature: $T > \frac{98.9}{0.0939} \approx 1053 \mathrm{K}$ So, for reaction (c) to be spontaneous, the temperature needs to be greater than 1053 K. In conclusion, the temperature conditions for each reaction to be spontaneous are: - Reaction (a): spontaneous at T > 22117 K - Reaction (b): spontaneous at all temperatures - Reaction (c): spontaneous at T > 1053 K

Key Concepts

Spontaneity of ReactionsEffect of Temperature on ReactionsEnthalpy and Entropy Changes

Spontaneity of Reactions

When we talk about the spontaneity of reactions, we are referring to whether or not a reaction occurs without outside intervention. One of the main tools we use to evaluate this is the Gibbs Free Energy, denoted as $ \Delta G $. Spontaneous reactions tend to occur naturally and result in a decrease in free energy, meaning $ \Delta G $ is less than zero.
To calculate $ \Delta G $, we use the equation: \[ \Delta G = \Delta H - T \Delta S \] where $ \Delta H $ is the change in enthalpy, $ T $ is the temperature in Kelvin, and $ \Delta S $ is the change in entropy. A negative $ \Delta G $ value indicates a spontaneous process. So, the sign and magnitude of $ \Delta G $ help predict whether a process will occur naturally. In general, if the enthalpy change $ \Delta H $ is negative and the entropy change $ \Delta S $ is positive, the reaction tends to be spontaneous at any temperature, as we have seen in reaction (b). However, reactions like (a) and (c), where both $ \Delta H $ and $ \Delta S $ are positive, are dependent on temperature for spontaneity. As temperature increases, the entropy's role becomes more significant because of the $ T \Delta S $ term. Thus, reactions can shift from non-spontaneous to spontaneous depending on the temperature.

Effect of Temperature on Reactions

The temperature of a system plays a crucial role in determining the spontaneity of a reaction. Temperature affects the $ T \Delta S $ component of the Gibbs Free Energy equation. As temperature increases, the entropy change $ \Delta S $ can have a larger effect.
This influence can define whether a reaction becomes spontaneous or not. For instance, in reaction (c), increasing the temperature can make the reaction spontaneous because the positive $ \Delta S $ becomes significant enough with a higher temperature, lowering the $ \Delta G $ to a negative value.

For reactions where $ \Delta H > 0 $ and $ \Delta S > 0 $, like (a) and (c), higher temperatures can favor spontaneity, as shown by the calculated critical temperatures for each reaction.
In contrast, when $ \Delta H < 0 $ and $ \Delta S < 0 $, higher temperatures can make reactions less spontaneous.

This highlights that temperature can be a determining factor for spontaneity and must be considered when analyzing reactions.

Enthalpy and Entropy Changes

Enthalpy ($ \Delta H $) and entropy ($ \Delta S $) are two fundamental thermodynamic quantities that play key roles in the Gibbs Free Energy equation. Each of them provides insight into different aspects of reactions.

Enthalpy ($ \Delta H $)

Enthalpy change refers to the heat absorbed or released in a reaction at constant pressure.

A negative $ \Delta H $ indicates an exothermic reaction where heat is released, often contributing to reaction spontaneity at low temperatures.
A positive $ \Delta H $ signifies an endothermic reaction where heat is absorbed, which may require increased temperature for the reaction to become spontaneous, as seen in reactions (a) and (c).

Entropy ($ \Delta S $)

Entropy change reflects the disorder or randomness of a system.

Positive $ \Delta S $ indicates an increase in disorder, which is often beneficial for spontaneity, especially at high temperatures, as is the case in reactions (a) and (c).
Negative $ \Delta S $ shows a decrease in disorder, which can hinder spontaneity if not compensated by a large $ \Delta H $.

Understanding both $ \Delta H $ and $ \Delta S $ is essential as they collectively dictate the direction and spontaneity of chemical reactions through their impact on $ \Delta G $. This is why they are critical tools for chemists and students when analyzing reactivity and stability of a system.

Problem 39

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