Temperature Effects on Equilibrium: A Study Guide
DH d ln K = - --- d(1/T) R
DG = - R T ln K,
ln K = - DG / R T
Differentiate both sides with respect to (1/T) in the above equation gives,
d(ln K) / (d T) = (- 1 / R (d DG / T) / (dT)
= - DH / R T 2
If K1 and K2 are the equilibrium constant at T1 and T2 respectively, show further that
ln (K1 / K2) = - (DH / R) (1/T1 - 1/T2).
This is achieved by definite integral. This relationship indicates that the plot of ln (K versus 1/(T is a straight line, and the slope is - (DH / R). Thus, DH can be determined by measuring the equilibrium constant at different temperatures.
101,300 N m-2 23.756 ----------- = 3166 Pa 760 mmHg
Go = 8.312 J * 298 ln(3166)
= 20.0 kJ / hr
This example illustrates the evaluation of Gibb's energy when the equilibrium constant is known.
Contributors and Attributions
Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)