Question

for a certain reaction ∆H=-50 kj and ∆ S =-80 jk/1 at what temperature does the reaction turn from spontaneous to non-spontaneous​

Answer 1

Answer:

625 K

Explanation:

ΔH - TΔS = 0

ΔH = TΔS

T = ΔH/ΔS

= -50,000/(-80)

= 625K