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At 1 atm pressure, `DeltaS=75JK^(-1)mol^(-1),DeltaH=30kJ" "mol^(-1)`, the temperature of the reaction at equilibrium isA. 400 KB. 330 KC. 200 KD. 110 K |
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Answer» Correct Answer - A `DeltaG=DeltaH-TDeltaS` At equilibrium, `DeltaG=0` `T=(DeltaH)/(DeltaS)=(30xx10^(3))/(75)=400K` |
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