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What can be said about a reaction with [tex]$\Delta H=620 \text{kJ/mol}$[/tex] and [tex]\Delta S=-0.46 \text{kJ/(mol \cdot K)}[/tex]?

A. It is always spontaneous.
B. It is never spontaneous.
C. It is spontaneous above 1347 K.
D. It is spontaneous at 298 K.


Sagot :

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.