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The reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ is spontaneous at room temperature but becomes nonspontaneous at a much higher temperature. From this fact alone, obtain the signs of \(\Delta H^{\circ},\) and \(\Delta S^{\circ},\) assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change much with temperature. Explain your reasoning.

Short Answer

Expert verified
\(\Delta H^{\circ} > 0\) and \(\Delta S^{\circ} < 0\).

Step by step solution

01

Identify the Equation for Gibbs Free Energy

The spontaneity of a reaction at a given temperature is determined using the Gibbs Free Energy equation: \( \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \). For a reaction to be spontaneous, \( \Delta G^{\circ} < 0 \). Conversely, if the reaction is nonspontaneous, \( \Delta G^{\circ} > 0 \).
02

Analyze Spontaneity at Room Temperature

Given that the reaction is spontaneous at room temperature, \( \Delta G^{\circ} < 0 \). According to the Gibbs Free Energy equation, this implies: \[ \Delta H^{\circ} - T\Delta S^{\circ} < 0 \] Simplifying gives \( \Delta H^{\circ} < T\Delta S^{\circ} \).
03

Analyze Nonspontaneity at High Temperature

At a higher temperature, the reaction becomes nonspontaneous, meaning \( \Delta G^{\circ} > 0 \). Thus, \[ \Delta H^{\circ} - T\Delta S^{\circ} > 0 \] Simplifying gives \( \Delta H^{\circ} > T\Delta S^{\circ} \) as temperature increases.
04

Deduce the Signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\)

For \( \Delta H^{\circ} < T\Delta S^{\circ} \) at low temperatures to become \( \Delta H^{\circ} > T\Delta S^{\circ} \) at higher temperatures, \( \Delta H^{\circ} \) must be positive and \( \Delta S^{\circ} \) negative. This is because increasing temperature \( T\Delta S^{\circ} \) with a negative \( \Delta S^{\circ} \) makes it more positive, hence \( \Delta H^{\circ} \) eventually becomes greater than \( T\Delta S^{\circ} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
In chemical reactions, enthalpy change (\( \Delta H^{\circ} \)) represents the heat absorbed or released. It's like a book of energy where every reaction records how much heat it gains or loses.
- If \( \Delta H^{\circ} \) is **positive**, the reaction absorbs heat from the surroundings, termed endothermic.- If \( \Delta H^{\circ} \) is **negative**, the reaction releases heat, called exothermic.
For the reaction of \( \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g) \), given it is spontaneous at room temperature but becomes nonspontaneous at higher temperatures, indicates a positive \( \Delta H^{\circ} \). This endothermic nature means it requires energy input, but at lower temperatures, other factors make it proceed spontaneously.
Entropy Change
Entropy change (\( \Delta S^{\circ} \)) can be thought of as the universe's way of keeping track of disorder or randomness during a reaction.
- A **positive** \( \Delta S^{\circ} \) suggests an increase in disorder.- A **negative** \( \Delta S^{\circ} \) implies a decrease in disorder.
In our reaction, the conversion of the gases reduces the number of gas molecules, from four moles in the reactants to two moles in the product. This reduction reflects a decrease in disorder, suggesting a negative \( \Delta S^{\circ} \). As a result, even though entropy is lower, spontaneity at room temperature is driven by the absolute values of enthalpy and entropy's contributions to free energy.
Reaction Spontaneity
Spontaneity of a chemical reaction is determined by the Gibbs Free Energy \( \Delta G^{\circ} \) equation. For a reaction to be spontaneous, \( \Delta G^{\circ} \) needs to be less than zero: \( \Delta G^{\circ} < 0 \).
This balance of energy changes (\( \Delta H^{\circ} \)) and disorder (\( \Delta S^{\circ} \)) dictates whether a reaction will simply "happen" or not.
For the ammonia synthesis reaction \( \mathrm{N}_{2} + 3\mathrm{H}_{2} \to 2\mathrm{NH}_{3} \), it is spontaneous at lower temperatures due to this fine balance, even with a positive \( \Delta H^{\circ} \) and negative \( \Delta S^{\circ} \). Reaction conditions play a crucial role in tipping this energy balance.
Temperature Dependence
Temperature acts as an influential player in determining reaction spontaneity through its impact on entropy.
In the Gibbs Free Energy equation, \( \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \), the temperature (\( T \)) multiplies \( \Delta S^{\circ} \).
- Higher temperatures increase the term (\( T\Delta S^{\circ} \)), affecting the spontaneity.- For reactions with negative \( \Delta S^{\circ} \), like the ammonia synthesis, this multiplication diminishes spontaneous conditions as temperature rises.
Thus, initially spontaneous reactions at low heat may turn nonspontaneous as temperature rises, as seen with \( \mathrm{N}_{2}(g) + 3\mathrm{H}_{2}(g) \rightarrow 2\mathrm{NH}_{3}(g) \). Knowing how temperature plays into \( \Delta G^{\circ} \) helps in understanding and predicting reaction behavior.

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Most popular questions from this chapter

Consider a sample of water at \(25^{\circ} \mathrm{C}\) in a beaker in a room at \(50^{\circ} \mathrm{C}\). a.What change do you expect to observe in the water sample? Would this be a spontaneous process or not? b.What are the enthalpy and entropy changes for this change in the water sample? (Just indicate the sign of the changes.) Explain your answers. c.Does the entropy of the water increase or decrease during the change? How do you know? d.Is there a change in free energy for the water sample? If so, indicate the sign of the free-energy change and explain how you arrived at your answer. Consider the same sample of water, but starting at \(75^{\circ} \mathrm{C}\) in a room at \(50^{\circ} \mathrm{C}\) e.What change would you observe in the water sample? Would this change be a spontaneous process or not? f.What are the enthalpy and entropy changes for the water sample? (Just indicate the sign of the changes.) Explain your answers. g.Does the entropy of the water increase or decrease during the change? How do you know? h.Is there a change in free energy for the water sample? If so, indicate the sign of the free-energy change and explain how you arrived at your answer. Finally, consider the same sample of water, starting at \(50^{\circ} \mathrm{C}\) in a room at \(50^{\circ} \mathrm{C}\). i.What would you observe in the water sample? Is this a spontaneous process? j.What are the enthalpy and entropy changes for the water sample? (Just indicate the sign of the changes.) Be sure to justify your answer. k.Did the entropy of the water increase or decrease during the change? How do you know? l.Can you determine the exact free-energy change of the water in this case? If you can make this determination, what is the significance of this value?

18.122 Coal is used as a fuel in some electric-generating plants. Coal is a complex material, but for simplicity we may consider it to be a form of carbon. The energy that can be derived from a fuel is sometimes compared with the enthalpy of the combustion reaction: $$ \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) $$ Calculate the standard enthalpy change for this reaction at \(25^{\circ} \mathrm{C}\). Actually, only a fraction of the heat from this reaction is available to produce electric energy. In electric generating plants, this reaction is used to generate heat for a steam engine, which turns the generator. Basically the steam engine is a type of heat engine in which steam enters the engine at high temperature \(\left(T_{h}\right),\) work is done, and the steam then exits at a lower temperature \(\left(T_{l}\right)\). The maximum fraction, \(f,\) of heat available to produce useful energy depends on the difference between these temperatures (expressed in kelvins), \(f=\left(T_{h}-T_{l}\right) / T_{h} .\) What is the maximum heat energy available for useful work from the combustion of \(1.00 \mathrm{~mol}\) of \(\mathrm{C}(s)\) to \(\mathrm{CO}_{2}(g)\) ? (Assume the value of \(\Delta H^{\circ}\) calculated at \(25^{\circ} \mathrm{C}\) for the heat obtained in the generator.) It is possible to consider more efficient ways to obtain useful energy from a fuel. For example, methane can be burned in a fuel cell to generate electricity directly. The maximum useful energy obtained in these cases is the maximum work, which equals the free-energy change. Calculate the standard free-energy change for the combustion of \(1.00 \mathrm{~mol}\) of \(\mathrm{C}(s)\) to \(\mathrm{CO}_{2}(g)\). Compare this value with the maximum obtained with the heat engine described here.

18.15 Discuss the different sign combinations of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) that are possible for a process carried out at constant temperature and pressure. For each combination, state whether the process must be spontaneous or not, or whether both situations are possible. Explain.

For the reaction $$ \mathrm{HCHO}(g)+\frac{2}{3} \mathrm{O}_{3}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ the value of \(\Delta G^{\circ}\) is \(-618.8 \mathrm{~kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\). Other data are as follows: $$ \begin{array}{lll} & \Delta \mathbf{H}_{\mathrm{f}}^{\circ}(\boldsymbol{k} \text { JImol }) & \mathrm{S}^{\circ}(\text { JImol } \cdot \boldsymbol{K}) \\ & \text { at } 25^{\circ} \mathrm{C} & \text { at } 25^{\circ} \mathrm{C} \\ \mathrm{HCHO} & -117.0 & 219.0 \\ \mathrm{H}_{2} \mathrm{O}(g) & -241.8 & 188.7 \\ \mathrm{CO}_{2}(g) & -393.5 & 213.7 \\ \mathrm{O}_{3}(g) & 142.7 & ? \end{array} $$

Ethanol burns in air or oxygen according to the equation $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) $$ Predict the sign of \(\Delta S^{\circ}\) for this reaction.

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