Answer :
To determine which reactions have an increase in entropy, we consider the change in the number of moles of gas during the reaction. The change in entropy is generally related to the change in the number of gas molecules; if the number of gas molecules increases, the entropy of the system typically increases.
Let's go through each reaction step by step:
### Reaction A:
[tex]\[ 2 \, \text{CO (g)} + 2 \, \text{NO (g)} \rightarrow 2 \, \text{CO}_2 \, \text{(g)} + \text{N}_2 \, \text{(g)} \][/tex]
Calculate the change in the number of gas moles:
- Reactants: [tex]\(2 \, \text{CO (g)} + 2 \, \text{NO (g)} = 4\)[/tex] moles of gas
- Products: [tex]\(2 \, \text{CO}_2 \, \text{(g)} + \text{N}_2 \, \text{(g)} = 3\)[/tex] moles of gas
Change in moles: [tex]\(3 - 4 = -1\)[/tex]
### Reaction B:
[tex]\[ 2 \, \text{H}_2\text{S (g)} + 3 \, \text{O}_2 \, \text{(g)} \rightarrow 2 \, \text{H}_2\text{O (g)} + 2 \, \text{SO}_2 \, \text{(g)} \][/tex]
Calculate the change in the number of gas moles:
- Reactants: [tex]\(2 \, \text{H}_2\text{S (g)} + 3 \, \text{O}_2 \, \text{(g)} = 5\)[/tex] moles of gas
- Products: [tex]\(2 \, \text{H}_2\text{O (g)} + 2 \, \text{SO}_2 \, \text{(g)} = 4\)[/tex] moles of gas
Change in moles: [tex]\(4 - 5 = -1\)[/tex]
### Reaction C:
[tex]\[ \text{CH}_4 \, \text{(g)} + \text{H}_2\text{O (g)} \rightarrow \text{CO (g)} + 3 \, \text{H}_2 \, \text{(g)} \][/tex]
Calculate the change in the number of gas moles:
- Reactants: [tex]\(\text{CH}_4 \, \text{(g)} + \text{H}_2\text{O (g)} = 2\)[/tex] moles of gas
- Products: [tex]\(\text{CO (g)} + 3 \, \text{H}_2 \, \text{(g)} = 4\)[/tex] moles of gas
Change in moles: [tex]\(4 - 2 = 2\)[/tex]
### Reaction D:
[tex]\[ 2 \, \text{NO (g)} + 2 \, \text{H}_2 \, \text{(g)} \rightarrow \text{N}_2 \, \text{(g)} + 2 \, \text{H}_2\text{O (g)} \][/tex]
Calculate the change in the number of gas moles:
- Reactants: [tex]\(2 \, \text{NO (g)} + 2 \, \text{H}_2 \, \text{(g)} = 4\)[/tex] moles of gas
- Products: [tex]\(\text{N}_2 \, \text{(g)} + 2 \, \text{H}_2\text{O (g)} = 3\)[/tex] moles of gas
Change in moles: [tex]\(3 - 4 = -1\)[/tex]
### Summary:
- Reaction A: Change in moles = -1 (Decrease in entropy)
- Reaction B: Change in moles = -1 (Decrease in entropy)
- Reaction C: Change in moles = 2 (Increase in entropy)
- Reaction D: Change in moles = -1 (Decrease in entropy)
Thus, the reaction with an increase in entropy is Reaction C:
[tex]\[ \text{CH}_4 \, \text{(g)} + \text{H}_2\text{O (g)} \rightarrow \text{CO (g)} + 3 \, \text{H}_2 \, \text{(g)} \][/tex]
Let's go through each reaction step by step:
### Reaction A:
[tex]\[ 2 \, \text{CO (g)} + 2 \, \text{NO (g)} \rightarrow 2 \, \text{CO}_2 \, \text{(g)} + \text{N}_2 \, \text{(g)} \][/tex]
Calculate the change in the number of gas moles:
- Reactants: [tex]\(2 \, \text{CO (g)} + 2 \, \text{NO (g)} = 4\)[/tex] moles of gas
- Products: [tex]\(2 \, \text{CO}_2 \, \text{(g)} + \text{N}_2 \, \text{(g)} = 3\)[/tex] moles of gas
Change in moles: [tex]\(3 - 4 = -1\)[/tex]
### Reaction B:
[tex]\[ 2 \, \text{H}_2\text{S (g)} + 3 \, \text{O}_2 \, \text{(g)} \rightarrow 2 \, \text{H}_2\text{O (g)} + 2 \, \text{SO}_2 \, \text{(g)} \][/tex]
Calculate the change in the number of gas moles:
- Reactants: [tex]\(2 \, \text{H}_2\text{S (g)} + 3 \, \text{O}_2 \, \text{(g)} = 5\)[/tex] moles of gas
- Products: [tex]\(2 \, \text{H}_2\text{O (g)} + 2 \, \text{SO}_2 \, \text{(g)} = 4\)[/tex] moles of gas
Change in moles: [tex]\(4 - 5 = -1\)[/tex]
### Reaction C:
[tex]\[ \text{CH}_4 \, \text{(g)} + \text{H}_2\text{O (g)} \rightarrow \text{CO (g)} + 3 \, \text{H}_2 \, \text{(g)} \][/tex]
Calculate the change in the number of gas moles:
- Reactants: [tex]\(\text{CH}_4 \, \text{(g)} + \text{H}_2\text{O (g)} = 2\)[/tex] moles of gas
- Products: [tex]\(\text{CO (g)} + 3 \, \text{H}_2 \, \text{(g)} = 4\)[/tex] moles of gas
Change in moles: [tex]\(4 - 2 = 2\)[/tex]
### Reaction D:
[tex]\[ 2 \, \text{NO (g)} + 2 \, \text{H}_2 \, \text{(g)} \rightarrow \text{N}_2 \, \text{(g)} + 2 \, \text{H}_2\text{O (g)} \][/tex]
Calculate the change in the number of gas moles:
- Reactants: [tex]\(2 \, \text{NO (g)} + 2 \, \text{H}_2 \, \text{(g)} = 4\)[/tex] moles of gas
- Products: [tex]\(\text{N}_2 \, \text{(g)} + 2 \, \text{H}_2\text{O (g)} = 3\)[/tex] moles of gas
Change in moles: [tex]\(3 - 4 = -1\)[/tex]
### Summary:
- Reaction A: Change in moles = -1 (Decrease in entropy)
- Reaction B: Change in moles = -1 (Decrease in entropy)
- Reaction C: Change in moles = 2 (Increase in entropy)
- Reaction D: Change in moles = -1 (Decrease in entropy)
Thus, the reaction with an increase in entropy is Reaction C:
[tex]\[ \text{CH}_4 \, \text{(g)} + \text{H}_2\text{O (g)} \rightarrow \text{CO (g)} + 3 \, \text{H}_2 \, \text{(g)} \][/tex]
