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Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of \(\Delta S\) for this solution process?

Short Answer

Expert verified
The dissolution of ammonium nitrate in water is an endothermic and spontaneous process at room temperature. Based on the Gibbs free energy equation (\(\Delta G = \Delta H - T\Delta S\)), and given that \(\Delta G < 0\) and \(\Delta H > 0\), we can deduce that the change in entropy (ΔS) for this dissolution process is positive (ΔS > 0).

Step by step solution

01

Understanding the relationship between Gibbs free energy, enthalpy, and entropy

The Gibbs free energy change (ΔG) is related to the changes in enthalpy (ΔH) and entropy (ΔS) by the following equation: \[ \Delta G = \Delta H - T \Delta S\] where T is the temperature in Kelvin. For a spontaneous process, ΔG must be negative.
02

Analyzing given information

We are given that the dissolution of ammonium nitrate in water is endothermic, which means that heat is absorbed in the process, and thus ΔH > 0. The process is also spontaneous at room temperature, so ΔG < 0.
03

Deduce the sign of the change in entropy (ΔS)

As per the given information, ΔH > 0 and ΔG < 0. Now, let's substitute these conditions in the Gibbs free energy equation: \[\Delta G = \Delta H - T \Delta S < 0\] Since ΔH > 0, we need ΔS to be positive in order for ΔG to be negative. Therefore, we can deduce that the change in entropy (ΔS) for this dissolution process is positive (ΔS > 0).

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Key Concepts

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

Gibbs Free Energy
Gibbs free energy, denoted as \( \Delta G \), is a measure used to determine the spontaneity of a process. It combines contributions from both enthalpy and entropy changes. This allows us to predict whether a reaction or physical change will occur on its own. The equation \( \Delta G = \Delta H - T \Delta S \) links these key thermodynamic quantities together:

* \( \Delta G \): Gibbs free energy change
* \( \Delta H \): Enthalpy change, or heat exchanged
* \( \Delta S \): Entropy change, signifying disorder
* \( T \): Absolute temperature in Kelvin

A negative \( \Delta G \) ( \( \Delta G < 0 \) ) indicates a spontaneous process as it can proceed without external input. Conversely, a positive \( \Delta G \) suggests non-spontaneity, requiring energy input to occur. Understanding this relationship helps in predicting the outcome of chemical and physical processes.
Entropy Change
Entropy, denoted as \( \Delta S \), describes the amount of disorder or randomness in a system. When a process results in an increase in randomness, it leads to a positive change in entropy. Conversely, if a process results in decreased disorder, the entropy change is negative. Entropy is a crucial factor in determining whether a process will be spontaneous.

In the dissolution of ammonium nitrate, we observe an endothermic and spontaneous process. This means that despite the energy absorption (\( \Delta H > 0 \)), the disorder contributed by the dissolved ions increases the entropy significantly (\( \Delta S > 0 \)). This increase in entropy helps to drive the process forward, compensating for the absorbed heat and ensuring \( \Delta G < 0 \).

Thus, understanding entropy change helps us recognize why processes occur, even against their energy "cost."
Spontaneous Process
A spontaneous process is one that occurs without continuous external energy input. The key characteristic of such a process is a negative Gibbs free energy change (\( \Delta G < 0 \)). These processes can happen quickly or slowly but tend to favor conditions that promote an increase in entropy and/or a release of energy.

When a substance like ammonium nitrate dissolves in water endothermically, it's still spontaneous due to significant entropy gain. This means even though heat is absorbed, the overall Gibbs free energy calculation indicates spontaneity because of enhanced disorder from the dissolved particles.

In summary, spontaneous processes are driven mainly by the interplay of energy and entropy, highlighting the dynamic nature of chemical reactions and physical changes. Recognizing the conditions for spontaneity allows chemists to control and predict reaction behaviors effectively.

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Most popular questions from this chapter

The standard entropies at \(298 \mathrm{~K}\) for certain group 14 elements are: \(\mathrm{C}(s,\) diamond \()=2.43 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \mathrm{Si}(s)=18.81 \mathrm{~J} /\) \(\mathrm{mol}-\mathrm{K}, \mathrm{Ge}(s)=31.09 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \quad\) a n d \(\quad \mathrm{Sn}(s)=51.818 \mathrm{~J} /\) mol-K. All but \(S\) n have the same (diamond) structure. How do you account for the trend in the \(S^{\circ}\) values?

Using data from Appendix \(\mathrm{C}\), calculate the change in Gibbs free energy for each of the following reactions. In each case, indicate whether the reaction is spontaneous at \(298 \mathrm{~K}\) under standard conditions. (a) \(2 \mathrm{Ag}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AgCl}(s)\) (b) \(\mathrm{P}_{4} \mathrm{O}_{10}(s)+16 \mathrm{H}_{2}(g) \longrightarrow 4 \mathrm{PH}_{3}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{CH}_{4}(g)+4 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g)\) (d) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(I)+\mathrm{O}_{2}(g)\)

Classify each of the following reactions as one of the four possible types summarized in Table 19.3: (i) spontanous at all temperatures; (ii) not spontaneous at any temperature; (iii) spontaneous at low \(T\) but not spontaneous at high \(T ;\) (iv) spontaneous at high T but not spontaneous at low \(T\). $$ \begin{array}{l} \text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g) \\ \Delta H^{\circ}=-249 \mathrm{~kJ} ; \Delta S^{\circ}=-278 \mathrm{~J} / \mathrm{K} \\ \text { (b) } \mathrm{N}_{2}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NCl}_{3}(g) \\ \Delta H^{\circ}=460 \mathrm{~kJ} ; \Delta S^{\circ}=-275 \mathrm{~J} / \mathrm{K} \\ \text { (c) } \mathrm{N}_{2} \mathrm{~F}_{4}(g) \longrightarrow 2 \mathrm{NF}_{2}(g) \\ \Delta H^{\circ}=85 \mathrm{~kJ} ; \Delta S^{\circ}=198 \mathrm{~J} / \mathrm{K} \end{array} $$

Consider the following process: a system changes from state 1 (initial state) to state 2 (final state) in such a way that its temperature changes from \(300 \mathrm{~K}\) to \(400 \mathrm{~K}\). (a) Is this process isothermal? (b) Does the temperature change depend on the particular pathway taken to carry out this change of state? (c) Does the change in the internal energy, \(\Delta E\), depend on whether the process is reversible or irreversible?

(a) Which of the thermodynamic quantities \(p, H, q, w,\) and \(G\) are state functions? (b) Consider a system going from state 1 to state 2 in a reversible and an irreversible way. Compare \(q_{\text {rev }}\) and \(q_{\text {irtev }}\) (c) Consider a system going from state 1 to state 2 in a reversible and an irreversible way. Compare \(w_{\text {rev }}\) and \(w_{\text {trev }}\). (d) For a reversible isothermal process, write an expression for \(\Delta H\) and an expression for \(\Delta G\) in terms of \(q, w\) and \(T, p\) and \(\Delta V\).

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