What is the equilibrium constant for a reaction at temperature 78.5 °C if the equilibrium constant at 97.9 °C is 20.64? For this reaction, ΔrH = -7.7 kJ mol-1 .
Given:
T1 = 97.9 oC
=(97.9+273)K
= 370.9 K
T2 = 78.5 oC
=(78.5+273)K
= 351.5 K
K1 = 20.64
Ho = -7.7 KJ/mol
= -7700 J/mol
use:
ln(K2/K1) = (Ho/R)*(1/T1 - 1/T2)
ln(K2/20.64) = (-7700.0/8.314)*(1/370.9 - 1/351.5)
ln(K2/20.64) = -926*(-1.488*10^-4)
K2 = 23.69
Answer: 23.7
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