Answer :
To determine which of the given reactions would have a negative entropy change, we need to consider whether the entropy (disorder) of the system increases or decreases during each reaction. Entropy generally increases when a reaction produces gases from solids or liquids and decreases when gases are converted into fewer gas molecules or into solids or liquids.
Let’s analyze each reaction step-by-step:
### A. [tex]\( N_2(g) + 3 H_2(g) \to 2 NH_3(g) \)[/tex]
- Reactants: 1 mole of [tex]\( N_2 \)[/tex] gas and 3 moles of [tex]\( H_2 \)[/tex] gas, for a total of 4 moles of gas.
- Products: 2 moles of [tex]\( NH_3 \)[/tex] gas.
- Comparison: We start with 4 moles of gas and end with 2 moles of gas.
- Entropy Change: The number of gas molecules decreases, which means the disorder of the system decreases. Therefore, the entropy decreases.
### B. [tex]\( Fe_2O_3(s) + 3 C(s) \to 2 Fe(s) + 3 CO(g) \)[/tex]
- Reactants: 1 mole of [tex]\( Fe_2O_3 \)[/tex] solid and 3 moles of carbon solid, for a total of 4 moles of solid.
- Products: 2 moles of iron solid and 3 moles of [tex]\( CO \)[/tex] gas.
- Comparison: We start with solids only and produce some gas.
- Entropy Change: Solid to gas conversion increases disorder. Therefore, the entropy increases.
### C. [tex]\( CaCO_3(s) \to CaO(s) + CO_2(g) \)[/tex]
- Reactants: 1 mole of [tex]\( CaCO_3 \)[/tex] solid.
- Products: 1 mole of [tex]\( CaO \)[/tex] solid and 1 mole of [tex]\( CO_2 \)[/tex] gas.
- Comparison: We start with a solid and produce a gas.
- Entropy Change: The formation of a gas from a solid increases disorder. Therefore, the entropy increases.
### D. [tex]\( H_2O(s) \to H_2O(l) \)[/tex]
- Reactants: 1 mole of [tex]\( H_2O \)[/tex] solid (ice).
- Products: 1 mole of [tex]\( H_2O \)[/tex] liquid (water).
- Comparison: We start with a solid and produce a liquid.
- Entropy Change: Solids are more ordered than liquids, so converting a solid to a liquid increases disorder. Therefore, the entropy increases.
### Conclusion:
By analyzing each reaction, we observe that only reaction A ([tex]\( N_2(g) + 3 H_2(g) \to 2 NH_3(g) \)[/tex]) has a decrease in the number of gas molecules, leading to a decrease in entropy. Therefore, the reaction with a negative entropy change is:
A. [tex]\( N_2(g) + 3 H_2(g) \to 2 NH_3(g) \)[/tex]
Let’s analyze each reaction step-by-step:
### A. [tex]\( N_2(g) + 3 H_2(g) \to 2 NH_3(g) \)[/tex]
- Reactants: 1 mole of [tex]\( N_2 \)[/tex] gas and 3 moles of [tex]\( H_2 \)[/tex] gas, for a total of 4 moles of gas.
- Products: 2 moles of [tex]\( NH_3 \)[/tex] gas.
- Comparison: We start with 4 moles of gas and end with 2 moles of gas.
- Entropy Change: The number of gas molecules decreases, which means the disorder of the system decreases. Therefore, the entropy decreases.
### B. [tex]\( Fe_2O_3(s) + 3 C(s) \to 2 Fe(s) + 3 CO(g) \)[/tex]
- Reactants: 1 mole of [tex]\( Fe_2O_3 \)[/tex] solid and 3 moles of carbon solid, for a total of 4 moles of solid.
- Products: 2 moles of iron solid and 3 moles of [tex]\( CO \)[/tex] gas.
- Comparison: We start with solids only and produce some gas.
- Entropy Change: Solid to gas conversion increases disorder. Therefore, the entropy increases.
### C. [tex]\( CaCO_3(s) \to CaO(s) + CO_2(g) \)[/tex]
- Reactants: 1 mole of [tex]\( CaCO_3 \)[/tex] solid.
- Products: 1 mole of [tex]\( CaO \)[/tex] solid and 1 mole of [tex]\( CO_2 \)[/tex] gas.
- Comparison: We start with a solid and produce a gas.
- Entropy Change: The formation of a gas from a solid increases disorder. Therefore, the entropy increases.
### D. [tex]\( H_2O(s) \to H_2O(l) \)[/tex]
- Reactants: 1 mole of [tex]\( H_2O \)[/tex] solid (ice).
- Products: 1 mole of [tex]\( H_2O \)[/tex] liquid (water).
- Comparison: We start with a solid and produce a liquid.
- Entropy Change: Solids are more ordered than liquids, so converting a solid to a liquid increases disorder. Therefore, the entropy increases.
### Conclusion:
By analyzing each reaction, we observe that only reaction A ([tex]\( N_2(g) + 3 H_2(g) \to 2 NH_3(g) \)[/tex]) has a decrease in the number of gas molecules, leading to a decrease in entropy. Therefore, the reaction with a negative entropy change is:
A. [tex]\( N_2(g) + 3 H_2(g) \to 2 NH_3(g) \)[/tex]