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

Answer Verified Verified
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-
     \[\Delta G=\Delta {{G}^{\circ }}+nRT\ln P\]
Where, $\Delta G$ is the change in Gibbs free energy and $\Delta {{G}^{\circ }}$ 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 ${{K}_{eq}}=\dfrac{{{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}}{{{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}}$
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-
$\begin{align}
  & b{{G}_{B}}=b{{G}_{B}}^{\circ }+bRT\ln {{P}_{B}} \\
 & c{{G}_{C}}=c{{G}_{C}}^{\circ }+cRT\ln {{P}_{C}} \\
 & d{{G}_{D}}=d{{G}_{D}}^{\circ }+dRT\ln {{P}_{D}} \\
\end{align}$
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-
$\Delta G=\sum{{{G}_{P}}-\sum{{{G}_{R}}}}$
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-
     \[\Delta G=\left\{ c{{G}_{C}}^{\circ }+d{{G}_{D}}^{\circ } \right\}-\left\{ a{{G}_{A}}^{\circ }+b{{G}_{B}}^{\circ } \right\}+\left\{ RT\ln \left[ \dfrac{{{P}_{C}}^{c}\times {{P}_{D}}^{d}}{{{P}_{A}}^{a}\times {{P}_{B}}^{b}} \right] \right\}\]
We can write,\[\Delta {{G}^{\circ }}=\left\{ c{{G}_{C}}^{\circ }+d{{G}_{D}}^{\circ } \right\}-\left\{ a{{G}_{A}}^{\circ }+b{{G}_{B}}^{\circ } \right\}\]as it is the change in free energy in standard condition. Therefore, equation becomes
     \[\Delta G=\Delta {{G}^{\circ }}+\left\{ RT\ln \left[ \dfrac{{{P}_{C}}^{c}\times {{P}_{D}}^{d}}{{{P}_{A}}^{a}\times {{P}_{B}}^{b}} \right] \right\}\]
Now at equilibrium,$\Delta G=0,\left[ \left[ \dfrac{{{P}_{C}}^{c}\times {{P}_{D}}^{d}}{{{P}_{A}}^{a}\times {{P}_{B}}^{b}} \right] \right]={{K}_{p}}$
So, we can write the equation as-
     \[\begin{align}
  & 0=\Delta {{G}^{\circ }}+RT\ln {{K}_{p}} \\
 & or,\Delta {{G}^{\circ }}=-RT\ln {{K}_{p}} \\
\end{align}\]
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] $-\Delta G=RT{{\log }_{e}}{{K}_{p}}$.

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.
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