# 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


Go = 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.