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Which reaction would cause a decrease in entropy?

A. [tex]\( 2 \text{CCl}_4(g) + \text{O}_2(g) \rightarrow 2 \text{COCl}_2(g) + 2 \text{Cl}_2(g) \)[/tex]

B. [tex]\( \text{CO}(g) + 3 \text{H}_2(g) \rightarrow \text{CH}_4(g) + \text{H}_2\text{O}(g) \)[/tex]

C. [tex]\( 2 \text{NOCl}(g) \rightarrow 2 \text{NO}(g) + \text{Cl}_2(g) \)[/tex]

D. [tex]\( 2 \text{NH}_3(g) \rightarrow \text{N}_2(g) + 3 \text{H}_2(g) \)[/tex]

Sagot :

GinnyAnswer
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.
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