Relation between ΔrxnG0 and equilibrium constant K is
ΔrxnG0 = -RTlnK
ΔrxnG0 =33 x100 J/K.mol , R =8.314J/K.mol
T =1280K ,K =?
lnK = - ΔrxnG0 /RT
= -33 x100 J/K.mol /(8.314J/K.mol *1280K)
= -3.1009
K= 0.0450 at T=1280K
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To find the tempeature at which K>1 ,we have to recall the formula relating the equilibrium constant value at one temperature to the equilibrium constant valu at another temperature ,standard enthalpy change .
i.,e ln(K2/K1) = (-ΔrxnH0/R)[1/T2 -1/T1]
We have , K1 =0.0450 ,T1 =1280K
K2=1 , T2 = ?
ΔrxnH0 = +224 kJmol-1 ,R = 8.314J/K.mol
ln(1/0.0450)=(-224 kJmol-1 / 8.314J/K.mol)[1/T2 - 1/1280K]
3.1010 = -26942.51 *[1/T2 - 1/1280K]
1/T2 = (3.1010 / -26942.51) + (1/1280)
=8.9634x10^-04
T2 = 1115.65K
Therefore ,above 1115.65K ,the equilibrium constant value greater than one .