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PCAT Chemical Processes – Temperature Conditions for Spontaneous Reactions

A certain reaction is associated with ΔH° = −14.0 kJ and ΔS° = −70 J/K. Under what temperature conditions will this reaction be spontaneous, assuming standard pressure?

A. It will be spontaneous at all temperatures.
B. It will be spontaneous at temperatures above 0.2 K.
C. It will be spontaneous at temperatures below 200 K.
D. It will not be spontaneous at any temperature.

Click for Explanation

C is correct. This reaction has a negative ΔH (enthalpy) value, which is favorable, but its ΔS (entropy) value is negative, which is not favorable. Thus, it will be spontaneous at some, but not all, temperatures (eliminate Choices A and D). To solve this problem, we need to use the equation ΔG = ΔH − TΔS. Be careful, however, as the units for ΔH and ΔS are not the same! We can address this by converting −14.0 kJ into −14,000 J.

We can plug in 0 for ΔG, as that is the point at which the reaction switches between being nonspontaneous (positive ΔG) and spontaneous (negative ΔG).


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