Question

The incorrect expression among the following is:
(A) $\dfrac{\Delta {{G}_{system}}}{\Delta {{S}_{total}}}=-T$
(B) In isothermal process, ${{W}_{reversible}}=-nRT\ln \dfrac{{{V}_{f}}}{{{V}_{i}}}$
(C) $\ln K=\dfrac{\Delta H{}^\circ -T\Delta S{}^\circ }{RT}$
(D) $K={{e}^{{}^{-\Delta G{}^\circ }/{}_{RT}}}$

Answer 1

Hint: Every expression here is related to Gibbs free energy. Above all, three of them are the empirical or the derived expression from the same.

Complete step by step Solution:
For analysing the above expressions, we should know the basic empirical equations of thermodynamics.
-The Gibbs free energy can be defined as,
The energy associated with a chemical reaction that can be used to do work is known as Gibbs free energy. It is the sum of addition of enthalpy and product of temperature and entropy.
It is expressed as,
\[\Delta G=\Delta H-T\Delta S\]
-Also,
$\Delta G=\Delta G{}^\circ +RT\ln Q$
Here, under standard conditions Q=1 and hence, $\Delta G=\Delta G{}^\circ $. Under equilibrium conditions. Q=K and hence, $\Delta G=0$.
So, $\Delta G{}^\circ =-RT\ln K$
Now,
-Enthalpy is defined as H = U+PV. When pressure and volume both are constant,
$\Delta H=\Delta U+\Delta \left( PV \right)$
Now, the criterion of spontaneity is $\Delta {{S}_{total}}>0$. But,
$\Delta {{S}_{total}}=\Delta {{S}_{sys}}-\dfrac{\Delta H}{T}$
From equation ,
$\Delta G=\Delta H-T\Delta {{S}_{sys}}$
Thus, comparing above two equations,
$\Delta G=-T\Delta {{S}_{total}}$ (1.3)
-Now, for reversible isothermal expansion;
$dW=-pdV$
If the volume of the gas changes from initial value to final value, then the total work done will be,
$W=\int{dW=-\int\limits_{{{V}_{i}}}^{{{V}_{f}}}{pdV}}$
For ideal gases,
pV=nRT
$p=\dfrac{nRT}{V}$
Thus,
$W=-nRT\ln \left( \dfrac{{{V}_{f}}}{{{V}_{i}}} \right)$
Illustration-
Comparing all the four equations we tried to derive above with the given options,
For (A)- from equation (1.3),
$\dfrac{\Delta {{G}_{system}}}{\Delta {{S}_{total}}}=-T$ is correct.
For (B)- from equation (1.4),
${{W}_{reversible}}=-nRT\ln \dfrac{{{V}_{f}}}{{{V}_{i}}}$ is correct.
For (C)- from equation (1.1) and (1.2),
equating them (under standard conditions) we get,
$\ln K=\left( \dfrac{\Delta H{}^\circ -T\Delta S{}^\circ }{-\left( RT \right)} \right)$ and hence,
$\ln K=\dfrac{\Delta H{}^\circ -T\Delta S{}^\circ }{RT}$ is incorrect.
For (D)- from equation (1.2),
$K={{e}^{{}^{-\Delta G{}^\circ }/{}_{RT}}}$ is correct.

Therefore, option (C) is the correct option.

Note: Be careful while deriving the empirical equations as one sign or one wrong displacement of any term can change the whole meaning and definition.