Question

Give the thermodynamic derivation of Van’t Hoff reaction isotherms, and explain its significance?

Answer 1

Hint: Gibbs free energy and equilibrium constant is related with the help of an equation which is known as the Van't Hoff equation. This equation is derived at constant temperature therefore, it is known as Van’t Hoff isothermal equation.

Complete answer:
Gibbs free energy of any compound is defined as an energy required during conversion of reactant to product both of them are in their standard states. Mathematically, Gibbs free energy is related to temperature of reaction which is expressed as:
$\mathrm{\Delta }G=\mathrm{\Delta }{G}^{\circ }+nRT\ln P$
Where $\mathrm{\Delta }G$ is change in Gibbs free energy during conversion of reactant to product
$\mathrm{\Delta }{G}^{\circ }$ Standard Gibbs free energy at standard state
$n=$ Mole of compound
$R=$ Gas constant
$T=$ Temperature of reaction
$P=$ Pressure of the reaction
For example: the chemical reaction between $a$ moles of $P,$ and $b$ moles of reactant $Q$ to form $c$ moles of $X$ and $d$ moles of product $Y$ at equilibrium condition is expressed as:
$aP+bQ\rightleftharpoons cX+dY$
For this chemical reaction, equilibrium constant is written as ratio of product concentration to reactant concentration.
$Keq={\displaystyle \frac{{\left[X\right]}^c{\left[Y\right]}^d}{{\left[P\right]}^a{\left[Q\right]}^b}}$

When we express the Gibbs free energy of particular components, it will be written as:
Gibbs free energy of $a$ moles of $P,$ is expressed as
${}_a\mathrm{\Delta }{G}_P{=}_a\mathrm{\Delta }{G_P}^{\circ }+aRT\ln P$
Similarly, standard Gibbs free energy of other components will be-${}_b\mathrm{\Delta }{G_Q}^{\circ }$, $${}_c\mathrm{\Delta }{G}_X$$, $${}_d\mathrm{\Delta }{G}_Y$$.
Total Gibbs free energy of any chemical reaction is the difference of Gibbs free energy of product and reactant.
$\mathrm{\Delta }G=\sum G_P-\sum G_R$
Change in standard Gibbs free energy is expressed as: $\mathrm{\Delta }{G}^{\circ }=$$${}_d\mathrm{\Delta }{G}_Y$$$+$$${}_c\mathrm{\Delta }{G}_X$$ $-$ ${}_b\mathrm{\Delta }{G_Q}^{\circ }$$+$${}_a\mathrm{\Delta }{G}_P$
Now we can write Gibbs free energy in base (e) as:
$\mathrm{\Delta }G=\mathrm{\Delta }{G}^{\circ }+RT\ln {\displaystyle \frac{\left[P{X}^C\times P{Y}^D\right]}{\left[P{P}^A\times P{Q}^B\right]}}$
As the chemical reaction proceeds to equilibrium value of $\mathrm{\Delta }G$ approaches zero.
Concentration of reactant and product with multiplication of pressure is expressed as a constant unit $\left(K_p\right)$.
So now the Gibbs free equation will become-
$0=\mathrm{\Delta }{G}^{\circ }+RT\ln K_P$
On modifying the above equation,
$\mathrm{\Delta }{G}^{\circ }=-RT\ln K_P$
we get we can also convert this equation in terms of log as:
$\mathrm{\Delta }{G}^{\circ }=-RT{\log }_e{K}_P$

Note:
Gibbs free energy plays an important role in correlating standard free Gibbs energy with reaction constant. For exothermic reactions the value of Gibbs free energy is negative and large while it is positive and small in case of endothermic reaction.