A process has `DeltaH=200Jmol^(-1)` and `DeltaS=40JK^(-1)mol^(-1)`. The minimum temperature above which the process will be spontaneous is ___________K.

Options

A. (A)4K
B. (B)5K
C. (C)8K
D. (D)12K

Explanation

To determine the minimum temperature above which the process will be spontaneous, we can use the Gibbs free energy equation:

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

For a process to be spontaneous, \(\Delta G\) must be less than zero:

\[ \Delta G < 0 \]

Substituting the expression for \(\Delta G\):

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

Rearranging this inequality gives:

\[ \Delta H < T \Delta S \]

Now, we can substitute the given values for \(\Delta H\) and \(\Delta S\):

\[ 200 \, \text{J/mol} < T \cdot 40 \, \text{J/K·mol} \]

Dividing both sides by \(40 \, \text{J/K·mol}\):

\[ \frac{200}{40} < T \]

Calculating the left side:

\[ 5 < T \]

This means that the minimum temperature \(T\) at which the process becomes spontaneous is:

\[ T > 5 \, \text{K} \]

Thus, the minimum temperature above which the process will be spontaneous is:

\[ \boxed{5 \, \text{K}} \]