For a particular reaction, ΔH° is 34.2 kJ and ΔS is 99.2 J/K. Assuming these values change very little with temperature, over what temperature range is the reaction spontaneous in the forward direction?

The calculation of temperature is done as follows:

T = ΔH°/ΔS°

= 34.2 kJ/mol / 99.2 * 10⁻³ kJ/molK

= 345 K

Since, ΔS° and ΔH° comes out to be positive for the given reaction. Hence, the given reaction is spontaneous at higher temperature and non spontaneous at lower temperature. Thus, it can be said that above 345 K, the given reaction will become spontaneous.

Over 344.76 K the forward reaction will be spontaneous

calculation

ΔG = ΔH° - TΔs

if ΔG = 0

therefore 0 = ΔH°-TΔs

ΔH° = TΔs

divide both side by Δs

T = ΔH°/Δs

convert 34.2 kj into j

that is 1 kj = 1000 j

34.2 kj = ? j

by cross multiplication

=[34.2 kj x 1000j / 1 kj] = 34200 j

T is therefore = 34200 j:99.2 j/k = 344.76 k

if T is greater than 344.76 k the forward reaction will be spontaneous