Answer :
To determine when the change in entropy, \(\Delta S\), is negative, we need to assess whether the disorder of the system decreases. Entropy, \(\Delta S\), is a measure of disorder or randomness; a negative \(\Delta S\) indicates that the system becomes more ordered.
Let's evaluate each reaction step-by-step:
(a) \(H_2O_{(l)} \longrightarrow H_2O_{(s)}\)
Transitioning from liquid water to solid ice. In the liquid state, water molecules are more disordered compared to the solid state where molecules are arranged in a more orderly structure. Therefore, this process represents a decrease in disorder.
[tex]\[ \Delta S < 0 \quad \text{(negative entropy change)} \][/tex]
(b) \(C_{(s)} + \frac{1}{2} O_{2(g)} \longleftrightarrow CO_{(g)}\)
Combining solid carbon and gaseous oxygen to form gaseous carbon monoxide. Here, we go from 1 solid and a fraction of a gas molecule to 1 gas molecule. The gas phase has higher entropy than solids and the mixed state. Therefore, this process represents an increase in disorder.
[tex]\[ \Delta S > 0 \quad \text{(positive entropy change)} \][/tex]
(c) \(PCl_{5(g)} \longrightarrow PCl_{3(g)}\)
This reaction involves two different gases, but going from one molecule \(PCl_5\) to another \(PCl_3\). The disorder mostly depends on the number of particles, but both a single molecule and resulting molecule are gases. The complexity slightly changes but generally does not lead to decreased disorder.
[tex]\[ \Delta S \approx 0 \quad \text{(approximately no significant change)} \][/tex]
(d) \(NH_4Cl_{(s)} \longleftrightarrow NH_{3(g)} + HCl_{(g)}\)
Decomposing solid ammonium chloride into two gas molecules, ammonia (NH3) and hydrogen chloride (HCl). This transition from a solid to a combination of two gases certainly increases the disorder.
[tex]\[ \Delta S > 0 \quad \text{(positive entropy change)} \][/tex]
Based on this analysis, the only reaction in which \(\Delta S\) is negative (indicating a decrease in disorder) is:
[tex]\[ (a) \quad H_2O_{(l)} \longrightarrow H_2O_{(s)} \][/tex]
Therefore, the correct option is:
[tex]\[ \boxed{1} \][/tex]
Let's evaluate each reaction step-by-step:
(a) \(H_2O_{(l)} \longrightarrow H_2O_{(s)}\)
Transitioning from liquid water to solid ice. In the liquid state, water molecules are more disordered compared to the solid state where molecules are arranged in a more orderly structure. Therefore, this process represents a decrease in disorder.
[tex]\[ \Delta S < 0 \quad \text{(negative entropy change)} \][/tex]
(b) \(C_{(s)} + \frac{1}{2} O_{2(g)} \longleftrightarrow CO_{(g)}\)
Combining solid carbon and gaseous oxygen to form gaseous carbon monoxide. Here, we go from 1 solid and a fraction of a gas molecule to 1 gas molecule. The gas phase has higher entropy than solids and the mixed state. Therefore, this process represents an increase in disorder.
[tex]\[ \Delta S > 0 \quad \text{(positive entropy change)} \][/tex]
(c) \(PCl_{5(g)} \longrightarrow PCl_{3(g)}\)
This reaction involves two different gases, but going from one molecule \(PCl_5\) to another \(PCl_3\). The disorder mostly depends on the number of particles, but both a single molecule and resulting molecule are gases. The complexity slightly changes but generally does not lead to decreased disorder.
[tex]\[ \Delta S \approx 0 \quad \text{(approximately no significant change)} \][/tex]
(d) \(NH_4Cl_{(s)} \longleftrightarrow NH_{3(g)} + HCl_{(g)}\)
Decomposing solid ammonium chloride into two gas molecules, ammonia (NH3) and hydrogen chloride (HCl). This transition from a solid to a combination of two gases certainly increases the disorder.
[tex]\[ \Delta S > 0 \quad \text{(positive entropy change)} \][/tex]
Based on this analysis, the only reaction in which \(\Delta S\) is negative (indicating a decrease in disorder) is:
[tex]\[ (a) \quad H_2O_{(l)} \longrightarrow H_2O_{(s)} \][/tex]
Therefore, the correct option is:
[tex]\[ \boxed{1} \][/tex]
