Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
To determine which reaction would cause a decrease in entropy, we need to analyze the number of gas molecules on both the reactant side and the product side of each reaction. Entropy generally decreases when the number of gas molecules decreases.
Let's examine each reaction step-by-step:
### Reaction A
[tex]\[2 CCl_4(g) + O_2(g) \rightarrow 2 COCl_2(g) + 2 Cl_2(g)\][/tex]
- Reactants: [tex]\(2\)[/tex] molecules of [tex]\(CCl_4(g)\)[/tex] and [tex]\(1\)[/tex] molecule of [tex]\(O_2(g)\)[/tex], making a total of [tex]\(2 + 1 = 3\)[/tex] gas molecules.
- Products: [tex]\(2\)[/tex] molecules of [tex]\(COCl_2(g)\)[/tex] and [tex]\(2\)[/tex] molecules of [tex]\(Cl_2(g)\)[/tex], making a total of [tex]\(2 + 2 = 4\)[/tex] gas molecules.
### Reaction B
[tex]\[CO(g) + 3 H_2(g) \rightarrow CH_4(g) + H_2O(g)\][/tex]
- Reactants: [tex]\(1\)[/tex] molecule of [tex]\(CO(g)\)[/tex] and [tex]\(3\)[/tex] molecules of [tex]\(H_2(g)\)[/tex], making a total of [tex]\(1 + 3 = 4\)[/tex] gas molecules.
- Products: [tex]\(1\)[/tex] molecule of [tex]\(CH_4(g)\)[/tex] and [tex]\(1\)[/tex] molecule of [tex]\(H_2O(g)\)[/tex], making a total of [tex]\(1 + 1 = 2\)[/tex] gas molecules.
### Reaction C
[tex]\[2 NOCl(g) \rightarrow 2 NO(g) + Cl_2(g)\][/tex]
- Reactants: [tex]\(2\)[/tex] molecules of [tex]\(NOCl(g)\)[/tex], making a total of [tex]\(2\)[/tex] gas molecules.
- Products: [tex]\(2\)[/tex] molecules of [tex]\(NO(g)\)[/tex] and [tex]\(1\)[/tex] molecule of [tex]\(Cl_2(g)\)[/tex], making a total of [tex]\(2 + 1 = 3\)[/tex] gas molecules.
### Reaction D
[tex]\[2 NH_3(g) \rightarrow N_2(g) + 3 H_2(g)\][/tex]
- Reactants: [tex]\(2\)[/tex] molecules of [tex]\(NH_3(g)\)[/tex], making a total of [tex]\(2\)[/tex] gas molecules.
- Products: [tex]\(1\)[/tex] molecule of [tex]\(N_2(g)\)[/tex] and [tex]\(3\)[/tex] molecules of [tex]\(H_2(g)\)[/tex], making a total of [tex]\(1 + 3 = 4\)[/tex] gas molecules.
Based on these calculations:
- Reaction A: [tex]\(3 \rightarrow 4\)[/tex] gas molecules (increase in entropy)
- Reaction B: [tex]\(4 \rightarrow 2\)[/tex] gas molecules (decrease in entropy)
- Reaction C: [tex]\(2 \rightarrow 3\)[/tex] gas molecules (increase in entropy)
- Reaction D: [tex]\(2 \rightarrow 4\)[/tex] gas molecules (increase in entropy)
Thus, reaction B causes a decrease in entropy because it results in fewer gas molecules on the product side compared to the reactant side.
Therefore, the correct answer is B.
Let's examine each reaction step-by-step:
### Reaction A
[tex]\[2 CCl_4(g) + O_2(g) \rightarrow 2 COCl_2(g) + 2 Cl_2(g)\][/tex]
- Reactants: [tex]\(2\)[/tex] molecules of [tex]\(CCl_4(g)\)[/tex] and [tex]\(1\)[/tex] molecule of [tex]\(O_2(g)\)[/tex], making a total of [tex]\(2 + 1 = 3\)[/tex] gas molecules.
- Products: [tex]\(2\)[/tex] molecules of [tex]\(COCl_2(g)\)[/tex] and [tex]\(2\)[/tex] molecules of [tex]\(Cl_2(g)\)[/tex], making a total of [tex]\(2 + 2 = 4\)[/tex] gas molecules.
### Reaction B
[tex]\[CO(g) + 3 H_2(g) \rightarrow CH_4(g) + H_2O(g)\][/tex]
- Reactants: [tex]\(1\)[/tex] molecule of [tex]\(CO(g)\)[/tex] and [tex]\(3\)[/tex] molecules of [tex]\(H_2(g)\)[/tex], making a total of [tex]\(1 + 3 = 4\)[/tex] gas molecules.
- Products: [tex]\(1\)[/tex] molecule of [tex]\(CH_4(g)\)[/tex] and [tex]\(1\)[/tex] molecule of [tex]\(H_2O(g)\)[/tex], making a total of [tex]\(1 + 1 = 2\)[/tex] gas molecules.
### Reaction C
[tex]\[2 NOCl(g) \rightarrow 2 NO(g) + Cl_2(g)\][/tex]
- Reactants: [tex]\(2\)[/tex] molecules of [tex]\(NOCl(g)\)[/tex], making a total of [tex]\(2\)[/tex] gas molecules.
- Products: [tex]\(2\)[/tex] molecules of [tex]\(NO(g)\)[/tex] and [tex]\(1\)[/tex] molecule of [tex]\(Cl_2(g)\)[/tex], making a total of [tex]\(2 + 1 = 3\)[/tex] gas molecules.
### Reaction D
[tex]\[2 NH_3(g) \rightarrow N_2(g) + 3 H_2(g)\][/tex]
- Reactants: [tex]\(2\)[/tex] molecules of [tex]\(NH_3(g)\)[/tex], making a total of [tex]\(2\)[/tex] gas molecules.
- Products: [tex]\(1\)[/tex] molecule of [tex]\(N_2(g)\)[/tex] and [tex]\(3\)[/tex] molecules of [tex]\(H_2(g)\)[/tex], making a total of [tex]\(1 + 3 = 4\)[/tex] gas molecules.
Based on these calculations:
- Reaction A: [tex]\(3 \rightarrow 4\)[/tex] gas molecules (increase in entropy)
- Reaction B: [tex]\(4 \rightarrow 2\)[/tex] gas molecules (decrease in entropy)
- Reaction C: [tex]\(2 \rightarrow 3\)[/tex] gas molecules (increase in entropy)
- Reaction D: [tex]\(2 \rightarrow 4\)[/tex] gas molecules (increase in entropy)
Thus, reaction B causes a decrease in entropy because it results in fewer gas molecules on the product side compared to the reactant side.
Therefore, the correct answer is B.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.