Question

The Van’t Hoff reaction isotherm is:
[A]$\mathrm{\Delta }G=RT{\log }_e{K}_p$
[B]$-\mathrm{\Delta }G=RT{\log }_e{K}_p$
[C]$\mathrm{\Delta }G=R{T}^2\ln K_p$
[D] None of these

Answer 1

Hint: The Van’t Hoff isotherm equation gives us a relation between the Gibbs free energy and equilibrium constant. If we take a system of equilibrium and write the change in their free energy and use it in the Gibbs equation, we will get a relation between the standard free energy change and the equilibrium constant, which will be the answer.

Complete step by step answer:
We know that, we can write the Gibbs free energy as a function of temperature and pressure, which gives us the relation-
     $$\mathrm{\Delta }G=\mathrm{\Delta }{G}^{\circ }+nRT\ln P$$
Where, $\mathrm{\Delta }G$ is the change in Gibbs free energy and $\mathrm{\Delta }{G}^{\circ }$ is the change in Gibbs free energy in standard conditions.
R is the universal gas constant, R= 8.314 J/mol.K
T is the temperature and P is the pressure.
As we know that in thermodynamics, the Van't Hoff equation gives us a relation between the equilibrium constant and the change in Gibbs free energy of a chemical reaction where change is in standard conditions.
Let us take a system at equilibrium like-
     $$aA+bB\rightleftharpoons cC+dD$$
The equilibrium constant for the above reaction will be $K_{eq}={\displaystyle \frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}}$
The Gibbs free energy of the component ‘A’ will be $a{G}_A=a{G_A}^{\circ }+aRT\ln P_A$ (by putting ‘a’ in the Gibbs free energy equation)
Similarly, we can write that Gibbs free energy of the component ‘B’, ‘C’ and ‘D’ will be-
$\begin{array}{rl}& b{G}_B=b{G_B}^{\circ }+bRT\ln P_B\\ & c{G}_C=c{G_C}^{\circ }+cRT\ln P_C\\ & d{G}_D=d{G_D}^{\circ }+dRT\ln P_D\end{array}$
As, the total change in Gibbs free energy can be written as the difference of change in free energy of the product and the reactant, which we can write as-
$\mathrm{\Delta }G=\sum G_P-\sum G_R$
We have calculated the free energy of the reactant and the product before, so putting those values in the above equation and rearranging the equation, we will get-
     $$\mathrm{\Delta }G=\{c{G_C}^{\circ }+d{G_D}^{\circ }\}-\{a{G_A}^{\circ }+b{G_B}^{\circ }\}+\{RT\ln \left[{\displaystyle \frac{{P_C}^c\times {P_D}^d}{{P_A}^a\times {P_B}^b}}\right]\}$$
We can write,$$\mathrm{\Delta }{G}^{\circ }=\{c{G_C}^{\circ }+d{G_D}^{\circ }\}-\{a{G_A}^{\circ }+b{G_B}^{\circ }\}$$as it is the change in free energy in standard condition. Therefore, equation becomes
     $$\mathrm{\Delta }G=\mathrm{\Delta }{G}^{\circ }+\{RT\ln \left[{\displaystyle \frac{{P_C}^c\times {P_D}^d}{{P_A}^a\times {P_B}^b}}\right]\}$$
Now at equilibrium,$\mathrm{\Delta }G=0,\left[\left[{\displaystyle \frac{{P_C}^c\times {P_D}^d}{{P_A}^a\times {P_B}^b}}\right]\right]=K_p$
So, we can write the equation as-
     $$\begin{array}{rl}& 0=\mathrm{\Delta }{G}^{\circ }+RT\ln K_p\\ & or,\mathrm{\Delta }{G}^{\circ }=-RT\ln K_p\end{array}$$
This is the Van’t Hoff isotherm equation which gives us a relation between the standard Gibbs free energy and the equilibrium constant.

Therefore, the correct answer is option [B] $-\mathrm{\Delta }G=RT{\log }_e{K}_p$.

Note:
The Van’t Hoff isotherm equation is used for estimating the equilibrium shift during a chemical reaction. We can also derive the Van’t Hoff isotherm equation using the Gibbs-Helmholtz equation which gives a relation between the change in free energy with respect to the changing temperature.