Enthalpy of a reaction at `27^(@)C` is `"15 kJ mol"^(-1)`. The reaction will be feasible if entropy is

Options

A. `"15 J mol"^(-1)K^(-1)`
B. `-"50 J mol"^(-1)K^(-1)`
C. Greater than `"50 J mol"^(-1)K^(-1)`
D. Less than `"50 J mol"^(-1)K^(-1)`

Answer

To determine the feasibility of a reaction based on its enthalpy and entropy, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

Where:
- \(\Delta G\) = Gibbs free energy change
- \(\Delta H\) = Enthalpy change
- \(T\) = Temperature in Kelvin
- \(\Delta S\) = Entropy change

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin**:
The temperature is given as \(27^\circ C\). To convert this to Kelvin:
\[ T = 27 + 273 = 300 \, K \]

**Hint**: Remember to always convert Celsius to Kelvin by adding 273.

2. **Identify Given Values**:
We are given:
- \(\Delta H = 15 \, kJ/mol\)
- \(T = 300 \, K\)

**Hint**: Ensure that all units are consistent. Here, we will convert \(\Delta H\) to Joules:
\[ \Delta H = 15 \, kJ/mol = 15 \times 10^3 \, J/mol = 15000 \, J/mol \]

3. **Set Up the Gibbs Free Energy Condition**:
For the reaction to be feasible, \(\Delta G\) must be negative:
\[ \Delta G < 0 \implies \Delta H - T \Delta S < 0 \]
Rearranging gives:
\[ \Delta H < T \Delta S \]
or
\[ \frac{\Delta H}{T} < \Delta S \]

**Hint**: This means that the entropy change must be greater than the ratio of enthalpy change to temperature.

4. **Substituting the Values**:
Substitute the values of \(\Delta H\) and \(T\) into the inequality:
\[ \frac{15000 \, J/mol}{300 \, K} < \Delta S \]

5. **Calculate the Left Side**:
\[ \frac{15000}{300} = 50 \, J/mol \cdot K \]
Thus, we have:
\[ 50 < \Delta S \]

**Hint**: Always perform the division carefully to ensure accuracy.

6. **Conclusion**:
The reaction will be feasible if the entropy change (\(\Delta S\)) is greater than \(50 \, J/mol \cdot K\).

### Final Answer:
The reaction will be feasible if the entropy is greater than \(50 \, J/mol \cdot K\).