# Temperature and Equilibrium

- Page ID
- 39000

## Temperature Effects on Equilibrium: A Study Guide

Show that

DH d lnK= - --- d(1/T) R

**Solution**

Since

D*G* = - *R T* ln *K*,

ln *K* = - D*G* / *R T*

Differentiate both sides with respect to (1/*T*) in the above equation gives,

d(ln *K*) / (d *T*) = (- 1 / *R* (d D*G* / *T*) / (d*T*)

= - D*H* / *R T* ^{2}

**DISCUSSION**

If *K*_{1} and *K*_{2} are the equilibrium constant at *T*_{1} and *T*_{2} respectively, show further that

ln (*K*_{1} / *K*_{2}) = - (D*H* / *R*) (1/*T*_{1} - 1/*T*_{2}).

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 - (D*H / R*). Thus, D*H* 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)