At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
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]
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.