Under standard conditions, the Gibbs free energy ofthe reactants G,std(reactants
ID: 682228 • Letter: U
Question
Under standard conditions, the Gibbs free energy ofthe reactants G,std(reactants) in a reaction in the gas phase is231.62 kJ and the Gibbs free energy of the products G,std(products)is 196.20 kJ. Calculate the value of the equilibrium constant forthis reaction under standard conditions at 25degC.Answer: 1.61×106
Calculate the value ofthe equilibrium constant for the reaction described in the previousquestion if the temperature is increased by 375degC to 673.15 K.The pressure remains constant at 1 bar. Under standard conditions,the enthalpy of reaction H,std is 24.8 kJ.
Explanation / Answer
(a) G = 196.20 kJ - 231.62 kJ = -35.42 kJ =-35420 J Since G = -RT ln K, K = exp(-G/RT) = exp(35420/(8.314*298.15)) = 1.61×106 (b) G = -35420 (J) = H - TS =24800 - 298.15*S (at 298.15 K) S = (24800+35420)/298.15 = 201.98(J/K) At 673.15 K, G = H - TS = 24800-673.15*201.9 = -111108.96 (J) K = exp(-G/RT) = exp(111108.96/(8.314*673.15) = 4.19E8
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