Part A: The change in enthalpy (ΔHorxn) for a reaction is -25.7 kJ/mol . The equilibrium constant for the reaction is 1.3×103 at 298 K. What is the equilibrium constant for the reaction at 668 K?
Part B: A reaction has an equilibrium constant of 7.4×103 at 298 K. At 753 K , the equilibrium constant is 0.63. Find ΔHorxn for the reaction.
A)
Given:
T1 = 298 K
T2 = 668 K
K1 = 1.3*10^3
Ho = -25.7 KJ/mol
= -25700 J/mol
use:
ln(K2/K1) = (Ho/R)*(1/T1 - 1/T2)
ln(K2/1.3*10^3) = (-25700.0/8.314)*(1/298 - 1/668.0)
ln(K2/1.3*10^3) = -3091*(1.859*10^-3)
ln(K2/1.3*10^3) = -5.746
(K2/1.3*10^3) = e^(-5.746)
(K2/1.3*10^3) = 3.197*10^-3
K2 = 4.156
Answer: 4.16
B)
Given:
T1 = 298 K
T2 = 753 K
K1 = 7.4*10^3
K2 = 0.63
use:
ln(K2/K1) = (Ho/R)*(1/T1 - 1/T2)
ln(0.63/7.4*10^3) = ( Ho/8.314)*(1/298 - 1/753)
-9.3713 = (Ho/8.314)*(2.028*10^-3)
Ho = -38425 J/mol
Ho = -38.4245 KJ/mol
Answer: -38.4 KJ/mol
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