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For a reaction, [tex]\(\Delta H^0 = 176 \text{ kJ/mol}\)[/tex] and [tex]\(\Delta S^0 = 0.285 \text{ kJ/(K·mol)}\)[/tex].

At what temperatures is this reaction spontaneous?

A. [tex]\(T \ \textless \ 50 \text{ K}\)[/tex]
B. [tex]\(T \ \textgreater \ 617 \text{ K}\)[/tex]
C. [tex]\(T \ \textless \ 617 \text{ K}\)[/tex]
D. At no temperature



Answer :

GinnyAnswer
To determine the temperature at which a given reaction becomes spontaneous, we use the Gibbs free energy change equation:

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

For a reaction to be spontaneous, the Gibbs free energy change ([tex]\(\Delta G\)[/tex]) must be negative:

[tex]\[ \Delta G < 0 \][/tex]

Rewriting the above inequality, we get:

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

Rearranging the terms to solve for the temperature ([tex]\(T\)[/tex]):

[tex]\[ T\Delta S > \Delta H \][/tex]

[tex]\[ T > \frac{\Delta H}{\Delta S} \][/tex]

We are given the values:
- [tex]\(\Delta H = 176 \, \text{kJ/mol}\)[/tex]
- [tex]\(\Delta S = 0.285 \, \text{kJ/(K} \cdot \text{mol)}\)[/tex]

Plugging in these values into our inequality, we get:

[tex]\[ T > \frac{176 \, \text{kJ/mol}}{0.285 \, \text{kJ/(K} \cdot \text{mol)}} \][/tex]

[tex]\[ T > 617.5438596491229 \, \text{K} \][/tex]

Therefore, the reaction will be spontaneous at temperatures greater than [tex]\(617 \, \text{K}\)[/tex].

So the correct answer is:
B. [tex]\(T > 617 \, \text{K}\)[/tex]

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