Temperature and Equilibrium

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Temperature Effects on Equilibrium: A Study Guide

Show that

           DH
d ln K = - --- d(1/T)
            R

Solution
Since

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

DISCUSSION
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

G^o = 8.312 J * 298 ln(3166)
= 20.0 kJ / hr

DISCUSSION
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)