The enthalpy and entropy of a reaction are `-5.0 kJ//mol and -20 JK^(-1) mol^(-1)` respectively and are independent of temperature . What is the highest temperature unto which the reaction is feasible ?

Options

A. 250 K
B. 240 K
C. 275 K
D. 300 K

Answer

To determine the highest temperature at which the reaction is feasible, we can use the Gibbs free energy equation:

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

For a reaction to be feasible, \(\Delta G\) must be less than or equal to zero. At the highest temperature where the reaction is still feasible, \(\Delta G\) will be equal to zero. Therefore, we can set up the equation:

\[ 0 = \Delta H - T \Delta S \]

Rearranging this gives us:

\[ T = \frac{\Delta H}{\Delta S} \]

### Step 1: Convert the given enthalpy and entropy values to consistent units
- Given \(\Delta H = -5.0 \, \text{kJ/mol}\)
- Convert \(\Delta H\) to joules:
\[ \Delta H = -5.0 \, \text{kJ/mol} = -5.0 \times 10^3 \, \text{J/mol} = -5000 \, \text{J/mol} \]

- Given \(\Delta S = -20 \, \text{J/K} \cdot \text{mol}\)

### Step 2: Substitute the values into the temperature equation
Using the values we have:
\[ T = \frac{\Delta H}{\Delta S} = \frac{-5000 \, \text{J/mol}}{-20 \, \text{J/K} \cdot \text{mol}} \]

### Step 3: Calculate the temperature
\[ T = \frac{5000}{20} = 250 \, \text{K} \]

### Conclusion
The highest temperature up to which the reaction is feasible is **250 K**.