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DeltaG = DeltaH - TDeltaS

dG = -54.32 kJ/mol - (54'32+273)K(-354.2J/molK)

NB Thevtemperature is quoted in Kelvin(K) and the Entropy must be converted to kJ by dividing by '1000'/

Hence

dG = - 54.32kJ/mol - (327.32K)(-0.3542 kJ/molK)

NB The 'K' cancels out. Then maker the multiplication

dG = -54/32 kJ/mol - - 115.94 kJ/mol Note the double minus; it becomes plus(+).

Hence

dG = -54.32kj/mol + 115.94 kJ/mol

dG = (+)61.61 kJ/mol

Since dG is positive, the reaction is NOT thermodynamically feasible.

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lenpollock

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Q: If a reaction has an enthalpy of -54.32 kJ/mol and an entropy of -354.2 J/(K*mol), what is the Gibbs free Energy at 54.3(degrees c)?
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