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.

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!

4,768.57

Explanation:

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