Answer :
To determine which reaction would cause a decrease in entropy, we need to look at the change in the number of gas molecules before and after the reaction. A decrease in the number of gas molecules indicates a decrease in entropy.
Let's analyze each reaction:
Reaction A:
[tex]\[ CO (g) + 3 H_2 (g) \rightarrow CH_4 (g) + H_2O (g) \][/tex]
- Reactants: 1 CO (g) + 3 H_2 (g) = 4 gas molecules
- Products: 1 CH_4 (g) + 1 H_2O (g) = 2 gas molecules
Change in number of gas molecules:
[tex]\[ \Delta n_A = \text{Products} - \text{Reactants} = 2 - 4 = -2 \][/tex]
Reaction B:
[tex]\[ 2 NH_3 (g) \rightarrow N_2 (g) + 3 H_2 (g) \][/tex]
- Reactants: 2 NH_3 (g) = 2 gas molecules
- Products: 1 N_2 (g) + 3 H_2 (g) = 4 gas molecules
Change in number of gas molecules:
[tex]\[ \Delta n_B = \text{Products} - \text{Reactants} = 4 - 2 = 2 \][/tex]
Reaction C:
[tex]\[ 2 NOCl (g) \rightarrow 2 NO (g) + Cl_2 (g) \][/tex]
- Reactants: 2 NOCl (g) = 2 gas molecules
- Products: 2 NO (g) + 1 Cl_2 (g) = 3 gas molecules
Change in number of gas molecules:
[tex]\[ \Delta n_C = \text{Products} - \text{Reactants} = 3 - 2 = 1 \][/tex]
Reaction D:
[tex]\[ 2 CCl_4 (g) + O_2 (g) \rightarrow 2 COCl_2 (g) + 2 Cl_2 (g) \][/tex]
- Reactants: 2 CCl_4 (g) + 1 O_2 (g) = 3 gas molecules
- Products: 2 COCl_2 (g) + 2 Cl_2 (g) = 4 gas molecules
Change in number of gas molecules:
[tex]\[ \Delta n_D = \text{Products} - \text{Reactants} = 4 - 3 = 1 \][/tex]
Now, let's compare the changes in the number of gas molecules:
- [tex]\(\Delta n_A = -2\)[/tex]
- [tex]\(\Delta n_B = 2\)[/tex]
- [tex]\(\Delta n_C = 1\)[/tex]
- [tex]\(\Delta n_D = 1\)[/tex]
Since a decrease in entropy corresponds to a decrease in the number of gas molecules, we can see that Reaction A has the most negative [tex]\(\Delta n\)[/tex] value, which is -2. Therefore, Reaction A would cause a decrease in entropy.
Thus, the correct answer is:
A. [tex]\( CO (g) + 3 H_2 (g) \rightarrow CH_4 (g) + H_2O (g) \)[/tex]
Let's analyze each reaction:
Reaction A:
[tex]\[ CO (g) + 3 H_2 (g) \rightarrow CH_4 (g) + H_2O (g) \][/tex]
- Reactants: 1 CO (g) + 3 H_2 (g) = 4 gas molecules
- Products: 1 CH_4 (g) + 1 H_2O (g) = 2 gas molecules
Change in number of gas molecules:
[tex]\[ \Delta n_A = \text{Products} - \text{Reactants} = 2 - 4 = -2 \][/tex]
Reaction B:
[tex]\[ 2 NH_3 (g) \rightarrow N_2 (g) + 3 H_2 (g) \][/tex]
- Reactants: 2 NH_3 (g) = 2 gas molecules
- Products: 1 N_2 (g) + 3 H_2 (g) = 4 gas molecules
Change in number of gas molecules:
[tex]\[ \Delta n_B = \text{Products} - \text{Reactants} = 4 - 2 = 2 \][/tex]
Reaction C:
[tex]\[ 2 NOCl (g) \rightarrow 2 NO (g) + Cl_2 (g) \][/tex]
- Reactants: 2 NOCl (g) = 2 gas molecules
- Products: 2 NO (g) + 1 Cl_2 (g) = 3 gas molecules
Change in number of gas molecules:
[tex]\[ \Delta n_C = \text{Products} - \text{Reactants} = 3 - 2 = 1 \][/tex]
Reaction D:
[tex]\[ 2 CCl_4 (g) + O_2 (g) \rightarrow 2 COCl_2 (g) + 2 Cl_2 (g) \][/tex]
- Reactants: 2 CCl_4 (g) + 1 O_2 (g) = 3 gas molecules
- Products: 2 COCl_2 (g) + 2 Cl_2 (g) = 4 gas molecules
Change in number of gas molecules:
[tex]\[ \Delta n_D = \text{Products} - \text{Reactants} = 4 - 3 = 1 \][/tex]
Now, let's compare the changes in the number of gas molecules:
- [tex]\(\Delta n_A = -2\)[/tex]
- [tex]\(\Delta n_B = 2\)[/tex]
- [tex]\(\Delta n_C = 1\)[/tex]
- [tex]\(\Delta n_D = 1\)[/tex]
Since a decrease in entropy corresponds to a decrease in the number of gas molecules, we can see that Reaction A has the most negative [tex]\(\Delta n\)[/tex] value, which is -2. Therefore, Reaction A would cause a decrease in entropy.
Thus, the correct answer is:
A. [tex]\( CO (g) + 3 H_2 (g) \rightarrow CH_4 (g) + H_2O (g) \)[/tex]
