Homework Answers
According to Clausius-Clayperon equation,
ln (K'/K) = (-L/R) x [(1/T') - (1/T)]
Where
K'= equilibrium constant at T'oC = 3.8*10^3
K = equilibrium constant at ToC = 14.5*10^4
R = gas constant = 8.314*10^-3 kJ/(mol-K)
T' = initial temperature = 20.0 oC = 20.0+273 = 293 K
T = final temperature = 45 oC = 45+273 = 318 K
L = enthalpy change of this substance = ?
Plug the values we get
ln (K'/K) = (-L/R) x [(1/T') - (1/T)]
L = - [ln (K'/K) xR] / [ [(1/T') - (1/T)]] = 58.4 kJ/mole
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