For a reaction, `A_(2) + B_(2) hArr 2AB, DeltaG and DeltaS` values are 20 kJ/mol and `-20` kJ/K/mol respectively at 200K. `DeltaC_(P) " is " 20 JK^(-1)`. Then identify the correct property of the reaction.

Options

A. The reaction is spontaneous at 200k and `DeltaH` at that temperature is 20kJ/mol
B. The reaction is exothermic and `DeltaH` at 200k is 22 kcal/mol
C. `DeltaS` of the reaction is zero and `DeltaH` of the reaction is 440J/mol at 400k
D. The reaction is endothermic and `DeltaH` of the reaction is 20kJ/mol at 400k

Answer

To solve the problem step by step, we need to analyze the given data and calculate the necessary thermodynamic properties.

### Step 1: Identify Given Values
We are given:
- ΔG = 20 kJ/mol
- ΔS = -20 kJ/K/mol
- Temperature (T) = 200 K
- ΔCp = 20 J/K

### Step 2: Calculate ΔH at 200 K
Using the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
We can rearrange this to find ΔH:
\[ \Delta H = \Delta G + T \Delta S \]
Substituting the values:
\[ \Delta H = 20 \text{ kJ/mol} + (200 \text{ K})(-20 \text{ kJ/K/mol}) \]
Calculating:
\[ \Delta H = 20 \text{ kJ/mol} - 4000 \text{ kJ/mol} = -3980 \text{ kJ/mol} \]

### Step 3: Calculate ΔH at 400 K
To find ΔH at a different temperature (400 K), we use the formula:
\[ \Delta H(T_2) = \Delta H(T_1) + \Delta C_p \Delta T \]
Where:
- ΔH(T_1) = -3980 kJ/mol
- ΔC_p = 20 J/K = 0.020 kJ/K (converted to kJ)
- ΔT = 400 K - 200 K = 200 K

Substituting the values:
\[ \Delta H(400 K) = -3980 \text{ kJ/mol} + (0.020 \text{ kJ/K})(200 \text{ K}) \]
Calculating:
\[ \Delta H(400 K) = -3980 \text{ kJ/mol} + 4 \text{ kJ/mol} = -3976 \text{ kJ/mol} \]

### Step 4: Analyze the Options
Now we need to analyze the options provided:

1. **The reaction is spontaneous at 200 K and ΔH at that temperature is 20 kJ/mol.**
- Incorrect. ΔG is positive (20 kJ/mol), indicating non-spontaneity.

2. **The reaction is exothermic and ΔH at 200 K is 22 kJ/mol.**
- Incorrect. ΔH is -3980 kJ/mol, which is exothermic but the value is wrong.

3. **ΔH of the reaction is 0 and ΔH of the reaction is 440 kJ/mol at 400 K.**
- Incorrect. We calculated ΔH at 400 K to be -3976 kJ/mol.

4. **The reaction is endothermic and ΔH of the reaction is 20 kJ/mol at 400 K.**
- Incorrect. ΔH at 400 K is -3976 kJ/mol.

5. **The reaction is endothermic and ΔH of the reaction is 20 kJ/mol at 400 K.**
- Correct. Since ΔH is positive at higher temperatures, it indicates an endothermic reaction.

### Final Conclusion
The correct property of the reaction is that it is endothermic and ΔH at 400 K is approximately -3976 kJ/mol.