`DeltaH` and `DeltaS` for a reaction are `+30.558 kJ mol^(-1)` and `0.066 kJ mol^(-1)` at `1` atm pressure. The temperature at which free energy is equal to zero and the nature of the reaction below this temperature are

Options

A. 483 K, spontaneous
B. 443 K, non - spontaneous
C. 443 K, spontaneous
D. 463 K, non - spontaneous

Answer

To solve the problem, we need to find the temperature at which the Gibbs free energy (ΔG) is equal to zero and determine the nature of the reaction below this temperature.

### Step-by-Step Solution:

1. **Understand the Gibbs Free Energy Equation**:
The Gibbs free energy change (ΔG) is related to the enthalpy change (ΔH) and the entropy change (ΔS) by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
where:
- ΔG = Gibbs free energy change
- ΔH = Enthalpy change
- T = Temperature in Kelvin
- ΔS = Entropy change

2. **Set ΔG to Zero**:
We are interested in the temperature at which ΔG = 0:
\[ 0 = \Delta H - T \Delta S \]
Rearranging this gives:
\[ T \Delta S = \Delta H \]
or
\[ T = \frac{\Delta H}{\Delta S} \]

3. **Substitute the Given Values**:
We have the following values from the question:
- ΔH = +30.558 kJ/mol
- ΔS = +0.066 kJ/mol

Now, substituting these values into the equation:
\[ T = \frac{30.558 \text{ kJ/mol}}{0.066 \text{ kJ/mol}} \]

4. **Calculate the Temperature**:
Performing the division:
\[ T = \frac{30.558}{0.066} \approx 463 \text{ K} \]

5. **Determine the Nature of the Reaction Below This Temperature**:
- If the temperature is below 463 K, then TΔS will be less than ΔH (since T is smaller), leading to:
\[ \Delta G = \Delta H - T \Delta S > 0 \]
This indicates that the reaction is non-spontaneous at temperatures below 463 K.

### Final Answer:
The temperature at which Gibbs free energy is zero is **463 K**, and the nature of the reaction below this temperature is **non-spontaneous**.