Question

For some reaction, ΔHº = + 50 kJ and ΔSº = +40 J/K. The reaction will be thermodynamically favorable in the forward directionat all temperatures.at no temperature.when T > 1250 K.when T < 1250 K.

Answer 1

Answer:

T > 1250K

Explanation:

A reaction is thermodynamically favorable when ΔG° is < 0

ΔG° is defined as:

ΔG° = ΔH° - TΔS°

As ΔG° must be < 0:

TΔS° > ΔH°

As ΔHº=+50kJ and ΔSº=+40J/K:

T×40J/K > 50000J

T > 50000J / 40J/K

T > 1250K

I hope it helps!