Question

For a particular reaction, ΔH = -32 kJ and ΔS =-98 J/K. Assume that ΔH and ΔS do not vary withtemperature. A) At wha...

For a particular reaction, ΔH = -32 kJ and ΔS =-98 J/K.
Assume that ΔH and ΔS do not vary withtemperature.

A) At what temperature will the reaction have ΔG= 0?
B) If T is increased from that in part A, will the reaction bespontaneous or nonspontaneous?
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Answer #1
a) delG = delH - T*delS = 0

T = delH/delS = (-32 x 10^3 J)/(-98 J/K) = 326.5 K

b) Reaction will be nonspontaneous.

delS is negative, thus T*delS is positive. This positive numbergets larger as T increases, and then it counteracts the negativenumber delH, making delG positive.

When delG is positive, the reaction is nonspontaneous.

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