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The reaction \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) proceeds spontaneously at \(25^{\circ} \mathrm{C}\) even though there is a decrease in entropy in the system (gases are converted to a solid). Explain.

Short Answer

Expert verified
The reaction is exothermic with negative \(\Delta G\), making it spontaneous despite decreased entropy.

Step by step solution

01

Understand the Concept of Entropy

Entropy is a measure of the disorder or randomness in a system. In the given reaction, gases (which have high entropy) are converted to a solid (which has lower entropy), indicating a decrease in entropy for the system.
02

Introduction to Gibbs Free Energy

A reaction can still occur spontaneously if the change in Gibbs free energy (\(\Delta G\)) is negative, even if the entropy of the system decreases. The formula to determine spontaneity is \(\Delta G = \Delta H - T\Delta S\), where \(\Delta H\) is the change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S\) is the change in entropy.
03

Consider the Enthalpy Change

Exothermic reactions usually release heat, which decreases the enthalpy (\(\Delta H < 0\)). The reaction between ammonia and hydrogen chloride is exothermic, suggesting \(\Delta H\) is negative. This contributes to \(\Delta G\) becoming negative and thus the reaction being spontaneous.
04

Analyze the Impact of Temperature

The temperature is given as \(25^{\circ} \mathrm{C}\), which is always positive in Kelvin. This positive temperature means that the effect of \(T\Delta S\) is proportionally small since \(\Delta S\) is negative, thus enhancing the impact of the negative \(\Delta H\) on \(\Delta G\).
05

Conclusion about Reaction Spontaneity

Despite the decrease in entropy, the large negative value for \(\Delta H\) from the exothermic reaction outweighs the \(T\Delta S\) term, leading \(\Delta G < 0\), which means the reaction proceeds spontaneously.

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Key Concepts

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

Entropy
Entropy is a fundamental concept in chemistry and physics, describing the degree of disorder or randomness in a system. In general, gases have higher entropy than solids due to their more random motion and arrangement. This is because the molecules in a gas can move freely and occupy a greater volume, leading to more disorder. On the other hand, solids have tightly packed molecules that are restricted to a fixed structure, resulting in lower entropy.
In the reaction where ammonia gas (\(\mathrm{NH}_3\)) and hydrogen chloride gas (\(\mathrm{HCl}\)) combine to form solid ammonium chloride (\(\mathrm{NH}_4\mathrm{Cl}\)), entropy decreases. This is because the reactants, which are gases and possess high entropy, form a product, a solid, which has lower entropy.
Despite the decrease in entropy, reactions can still be spontaneous. This is because spontaneity is determined by Gibbs free energy, which takes both entropy and enthalpy into account.
Spontaneous Reaction
A spontaneous reaction is one that occurs naturally without any external intervention. Whether a reaction is spontaneous depends on the change in Gibbs free energy (\(\Delta G\)).
If \(\Delta G\) is negative, the reaction proceeds spontaneously. The formula \(\Delta G = \Delta H - T\Delta S\) helps in determining the spontaneity by incorporating both enthalpy and entropy.
When considering spontaneity, it is important to understand that:
  • If \(\Delta H\) is negative (exothermic reaction) and \(\Delta S\) is positive (increase in entropy), \(\Delta G\) is negative, indicating spontaneity.
  • If \(\Delta H\) is negative but \(\Delta S\) is negative, as in the case of our ammonium chloride formation, the spontaneity depends on the magnitude of \(\Delta H\) and \(T\Delta S\).
In the case of the ammonia and hydrogen chloride reaction, despite a decrease in entropy, the reaction is spontaneous because the enthalpy change greatly influences \(\Delta G\).
Exothermic Reaction
Exothermic reactions release heat, indicating that the products have less energy than the reactants. This means that the change in enthalpy (\(\Delta H\)) is negative.
The exothermic nature of a reaction contributes significantly to its spontaneity. When \(\Delta H\) is large and negative, it can outweigh a negative entropy change, leading to a negative Gibbs free energy change (\(\Delta G\)) and thus spontaneous reaction.
For instance, in the reaction between \(\mathrm{NH}_3\) and \(\mathrm{HCl}\), the transformation to \(\mathrm{NH}_4\mathrm{Cl}\) releases heat. This exothermic behavior is a strong factor driving its spontaneity because it creates a large negative \(\Delta H\) that dominates the effect of the entropy decrease.
This demonstrates how, in practice, spontaneous reactions often involve energetic trade-offs between entropy and enthalpy.
Enthalpy Change
Enthalpy change (\(\Delta H\)) is a key factor in understanding chemical reactions and their spontaneity. It represents the total heat content of a system and indicates whether a reaction absorbs or releases heat.
In exothermic reactions, \(\Delta H\) is negative, signifying that heat is released to the surroundings. This release contributes to the decrease in Gibbs free energy, making the reaction more likely to be spontaneous.
Considering our reaction of interest, \(\mathrm{NH}_3(g) + \mathrm{HCl}(g) \rightarrow \mathrm{NH}_4\mathrm{Cl}(s)\), the change in enthalpy is negative. Despite the decrease in entropy, this negative \(\Delta H\) indicates that the reaction releases a considerable amount of heat.
This process exemplifies how the magnitude of enthalpy change can drive the overall spontaneity of a reaction, even when entropy changes are unfavorable.

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Most popular questions from this chapter

Predict whether the entropy change is positive or negative for each of these reactions: (a) \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightleftharpoons \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{O}(g)+\mathrm{O}(g) \rightleftharpoons \mathrm{O}_{2}(g)\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

For each pair of substances listed here, choose the one having the larger standard entropy value at \(25^{\circ} \mathrm{C}\). The same molar amount is used in the comparison. Explain the basis for your choice. (a) \(\operatorname{Li}(s)\) or \(\operatorname{Li}(l),(b) C_{2} H_{5} O H(l)\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}(l)\) (Hint: Which molecule can hydrogen- (d) \(\mathrm{CO}(g)\) or \(\mathrm{CO}_{2}(g),\) (e) \(\mathrm{O}_{2}(g)\) bond?), (c) \(\operatorname{Ar}(g)\) or \(\operatorname{Xe}(g)\), or \(\mathrm{O}_{3}(g),(\mathrm{f}) \mathrm{NO}_{2}(g)\) or \(\mathrm{N}_{2} \mathrm{O}_{4}(g) .\)

A certain reaction is known to have a \(\Delta G^{\circ}\) value of \(-122 \mathrm{~kJ} / \mathrm{mol}\). Will the reaction necessarily occur if the reactants are mixed together?

When a native protein in solution is heated to a high enough temperature, its polypeptide chain will unfold to become the denatured protein. The temperature at which a large portion of the protein unfolds is called the melting temperature. The melting temperature of a certain protein is found to be \(46^{\circ} \mathrm{C},\) and the enthalpy of denaturation is \(382 \mathrm{~kJ} / \mathrm{mol}\). Estimate the entropy of denaturation, assuming that the denaturation is a twostate process; that is, native protein \(\longrightarrow\) denatured protein. The single polypeptide protein chain has 122 amino acids. Calculate the entropy of denaturation per amino acid. Comment on your result.

The standard enthalpy of formation and the standard entropy of gaseous benzene are \(82.93 \mathrm{~kJ} / \mathrm{mol}\) and \(269.2 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\), respectively. Calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) for the given process at \(25^{\circ} \mathrm{C}\). Comment on your answers: $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g) $$

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