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At 1 atm pressure, `DeltaS=75JK^(-1)mol^(-1),DeltaH=30kJ" "mol^(-1)`, the temperature of the reaction at equilibrium is

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At 1 atm pressure, `DeltaS=75JK^(-1)mol^(-1),DeltaH=30kJ" "mol^(-1)`, the temperature of the reaction at equilibrium is
A. 400 K
B. 330 K
C. 200 K
D. 110 K
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1 Answer

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Correct Answer - A
`DeltaG=DeltaH-TDeltaS`
At equilibrium, `DeltaG=0`
`T=(DeltaH)/(DeltaS)=(30xx10^(3))/(75)=400K`
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