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Question 11 of 30

For a reaction, [tex]\Delta H^0 = -75 \, \text{kJ/mol}[/tex] and [tex]\Delta S^0 = -0.081 \, \text{kJ/(K} \cdot \text{mol)}[/tex]. At what temperatures is this reaction spontaneous?

A. [tex]T \ \textless \ 100 \, \text{K}[/tex]
B. [tex]T \ \textgreater \ 930 \, \text{K}[/tex]
C. At all temperatures
D. [tex]T \ \textless \ 930 \, \text{K}[/tex]



Answer :

GinnyAnswer
To determine at what temperatures the reaction is spontaneous, we use the Gibbs free energy change equation:

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

For a reaction to be spontaneous, [tex]\(\Delta G\)[/tex] must be negative ([tex]\(\Delta G < 0\)[/tex]). Given the enthalpy change ([tex]\(\Delta H\)[/tex]) and the entropy change ([tex]\(\Delta S\)[/tex]), we need to find the temperatures at which:

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

Given values:
[tex]\[ \Delta H = -75 \text{ kJ/mol} \][/tex]
[tex]\[ \Delta S = -0.081 \text{ kJ/(K·mol)} \][/tex]

Substitute the given values into the inequality:

[tex]\[ -75 \text{ kJ/mol} - T (-0.081 \text{ kJ/(K·mol)}) < 0 \][/tex]

Simplify the inequality:

[tex]\[ -75 + 0.081T < 0 \][/tex]

Now, solve for [tex]\( T \)[/tex]:

[tex]\[ 0.081T < 75 \][/tex]

[tex]\[ T < \frac{75}{0.081} \][/tex]

[tex]\[ T < 925.9259259259259 \text{ K} \][/tex]

Therefore, the reaction is spontaneous at temperatures [tex]\( T < 925.93 \text{ K} \)[/tex].

Hence, the correct answer is:

D. [tex]\(T < 930 \text{ K}\)[/tex]