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:

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

Where:
- \(\Delta G\) is the change in Gibbs free energy,
- \(\Delta H\) is the change in enthalpy,
- \(T\) is the temperature in Kelvin,
- \(\Delta S\) is the change in entropy.

The reaction is feasible when \(\Delta G \leq 0\). At the highest temperature where the reaction is still feasible, \(\Delta G\) will be equal to 0. Therefore, we set the equation to zero:

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

Rearranging gives us:

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

From this, we can solve for \(T\):

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

Now, we need to substitute the values given in the problem:

1. Convert \(\Delta H\) from kJ to J:
\[ \Delta H = -5.0 \, \text{kJ/mol} = -5.0 \times 1000 \, \text{J/mol} = -5000 \, \text{J/mol} \]

2. The value of \(\Delta S\) is given as:
\[ \Delta S = -20 \, \text{J/K/mol} \]

Now, substituting these values into the equation for \(T\):

\[ T = \frac{-5000 \, \text{J/mol}}{-20 \, \text{J/K/mol}} = \frac{5000}{20} = 250 \, \text{K} \]

Thus, the highest temperature at which the reaction is feasible is:

\[ \boxed{250 \, \text{K}} \]