To understand when a reaction is spontaneous, we use the Gibbs free energy equation:
[tex]\[
\Delta G = \Delta H - T \Delta S
\][/tex]
For a reaction to be spontaneous, [tex]\(\Delta G\)[/tex] must be less than 0 ([tex]\(\Delta G < 0\)[/tex]).
Given:
- [tex]\(\Delta H = 620 \, \text{kJ/mol}\)[/tex]
- [tex]\(\Delta S = -0.46 \, \text{kJ/(mol \cdot K)}\)[/tex]
We need to determine the temperature at which the reaction becomes spontaneous. This happens when [tex]\(\Delta G = 0\)[/tex] (the threshold point between spontaneous and non-spontaneous behavior).
Setting [tex]\(\Delta G = 0\)[/tex], we get:
[tex]\[
0 = \Delta H - T \Delta S
\][/tex]
Solving for the temperature [tex]\(T\)[/tex]:
[tex]\[
\Delta H = T \Delta S
\][/tex]
[tex]\[
T = \frac{\Delta H}{\Delta S}
\][/tex]
Substitute the given values into the equation:
[tex]\[
T = \frac{620 \, \text{kJ/mol}}{-0.46 \, \text{kJ/(mol \cdot K)}}
\][/tex]
Perform the division:
[tex]\[
T = \frac{620}{-0.46}
\][/tex]
[tex]\[
T = -1347.83 \, \text{K}
\][/tex]
Since temperature cannot be negative in physical terms, this calculation implies the reaction is non-spontaneous at all positive temperatures. To confirm this, consider the relationship in the Gibbs free energy context:
- The reaction is non-spontaneous if [tex]\(\Delta H > 0\)[/tex] and [tex]\(\Delta S < 0\)[/tex] because for any positive temperature [tex]\(T\)[/tex], the term [tex]\(T \Delta S\)[/tex] will always be negative making [tex]\(\Delta G\)[/tex] always positive.
Thus:
- At any positive temperature, the reaction will not be spontaneous.
So, the correct answer is:
B. It is never spontaneous.
