Respuesta :
The ratio of k, equilibrium constant at 500°C and at 25°C is 247 : 288.
T1= 500 + 273 = 773K
T2 = 25 + 273 = 298K
What is gibbs free energy?
The gibbs free energy change (∆Gº') of a chemical reaction is defined as the amount of energy which is released during the conversion of reactants into products under standard conditions.
The relationship between gibbs free energy and equilibrium constant k is
∆G° = -RT Ink
For 773K,
∆G° = -773R Ink1 ------- (1)
∆G° = -298R Ink2 -------- (2)
Comparing equation (1) and (2),
-773R Ink1 = -298R Ink2
773 Ink1 = 298 Ink2
773 logk1 = 298 logk2
Multiplying both side by antilog, we get
2.88 k1 = 2.47 k2
k1/ k2 = 247/288
Thus, we calculated that the ratio of k, equilibrium constant at 500°C and at 25°C is 247 : 288.
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