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Asked by Lexi

A certain constant-pressure reaction is nonspontaneous at 49 ∘C. The entropy change for the reaction is 71 J/K .
What can you conclude about the magnitude of ΔH?
answers :ΔH>23kJ
ΔH>220kJ
ΔH<71J/K
ΔH>71J/K
ΔH<23kJ
ΔH<220kJ
5 years ago

Answers

Answered by DrBob222
dG = dH - TdS
To be not spontaneous dG must be -
dH = ?
T is 49+273 = 322
- = dH - 322(71)
- = dH - 22862 J
- = dH - 22,862.
What is the value for dH so that a negative value is obtained when added to 22.3 kJ.
5 years ago

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