Question

I want derivation for Van't hoff isotherm equation

Answer 1

I ****LITERALLY***** copy pasted this from wiki. Literally.....

"""
From the definition of Gibbs free energy
$$\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$$
and from the Gibbs free energy isotherm equation
$$\Delta G^\ominus = -RT \ln K_\mathrm{eq} $$
These equations are combined to obtain
$$\ln K_\mathrm{eq} = - \frac{{\Delta H^\ominus}}{RT}+ \frac{{\Delta S^\ominus }}{R}. $$
differentiation of this expression with respect to the variable ''T'' yields the Van 't Hoff equation.
$$\frac{d \ln K_\mathrm{eq}}{dT} = \frac{\Delta H^\ominus}{RT^2}$$
"""(eoq)

Answer 2

is that it? or am I missing something?

Answer 3

Hi.
This seems to be about graduate theoretical biochemical thermodynamics, which is a finger full and a mouthful.
Link below may help, unless of course you are at MIT and have just sat through this lecture.
http://ocw.mit.edu/courses/biology/7-51-graduate-biochemistry-fall-2001/lecture-notes/fa01lec06.pdf
I'll keep an eye out, and if you let me know your progress that would help - TEAMWORK, I think it's called.
Bon voyage
http://perendis.webs.com

Answer 4

There is also an account of the VH "isotherm" equation in a book "Heat and thermodynamics" by Mark W Zemansky, mcgraw-hill. Zemansky refers to it as the van't Hoff (Dutch, european) ISOBAR, not isotherm. To calculate the heat of reaction at any temperature or within any desired temperature range, once the the temperature variation of the equilibrium constant is known.
Since the temperature is varying, it can't be an isotherm - constant temperature - it's at constant pressure. I think that a lot of chemical thermo assumes that reactions take place at constant pressure - in an open ended glass test tube ? ...