Question

In which of the following conditions a chemical reaction can not occur?
A. \[\Delta {\rm{H}}\] and \[\Delta {\rm{S}}\] increase and \[{\rm{T}}\Delta {\rm{S > }}\Delta {\rm{H}}\]
B. \[\Delta {\rm{H}}\] and \[\Delta {\rm{S}}\] decrease and \[\Delta {\rm{H > T}}\Delta {\rm{S}}\]
C. \[\Delta {\rm{H}}\] increases and \[\Delta {\rm{S}}\] decreases
D. \[\Delta {\rm{H}}\]decreases and \[\Delta {\rm{S}}\] increases

Answer 1

Hint: The occurrence of a reaction is decided by its spontaneity. Spontaneity is the tendency of a reaction to promote the formation of products. Gibbs's free energy change is negative for spontaneous reactions.

Formula Used:
The formula for change in Gibbs free energy change is:-
\[\Delta {\rm{G = }}\Delta {\rm{H - T}}\Delta {\rm{S}}\]
where
\[\Delta {\rm{G}}\] = Gibbs free energy change
\[\Delta {\rm{H}}\] = Enthalpy change
\[\Delta {\rm{S}}\] = Entropy change
T = Temperature

Complete Step by Step Solution:
The second law of thermodynamics says that "a reaction which is spontaneous, enhances the entropy of the system." So, a spontaneous reaction will have higher entropy.

Gibbs free energy, G is a thermodynamic function that is an extensive property and a state function. Free energy denotes the amount of energy accessible for work.

It is expressed by the expression:
\[{\rm{G = H - TS}}\]

The formula for change in Gibbs free energy change is:-
\[\Delta {\rm{G = }}\Delta {\rm{H - T}}\Delta {\rm{S}}\]

There is the release of energy in spontaneous reactions. So, the ∆H is negative.
From the second law of thermodynamics, we know that entropy increases for these types of reactions.
So, the product T∆S has a higher value than ∆H.
So, ∆G is negative i.e., Gibbs's free energy change is negative for a spontaneous reaction.

A. \[\Delta {\rm{H}}\] and \[\Delta {\rm{S}}\] increase and \[{\rm{T}}\Delta {\rm{S > }}\Delta {\rm{H}}\]
If \[{\rm{T}}\Delta {\rm{S > }}\Delta {\rm{H}}\], then
\[{\rm{\Delta H - T\Delta S < 0}}\]
\[ \Rightarrow {\rm{\Delta G < 0}}\]
In this case, the reaction is spontaneous as the product T∆S is large enough to surpass ∆H.
So, this reaction is spontaneous.
This option is incorrect.

B. \[\Delta {\rm{H}}\] and \[\Delta {\rm{S}}\] decrease and \[\Delta {\rm{H > T}}\Delta {\rm{S}}\]
It is given that both ∆H and ∆S decrease, so
\[\Delta {\rm{G = }}\left( {{\rm{ - }}\Delta {\rm{H}}} \right){\rm{ - }}\left( {{\rm{ - T}}\Delta {\rm{S}}} \right)\]
\[{\rm{ = - }}\Delta {\rm{H + T}}\Delta {\rm{S}}\]
This reaction can be spontaneous if the value of enthalpy change is greater than the product of temperature and entropy change.
It is also given that
\[\Delta {\rm{H > T}}\Delta {\rm{S}}\]
\[\Delta {\rm{H - T}}\Delta {\rm{S > }}0\]
So, the value of enthalpy change is greater than the product of temperature and entropy change.
Hence, the reaction is spontaneous.
So, B is incorrect.

C. \[\Delta {\rm{H}}\] increases and \[\Delta {\rm{S}}\] decreases
It is given that
\[\Delta {\rm{G = }}\left( {\Delta {\rm{H}}} \right){\rm{ - }}\left( {{\rm{T}}\Delta {\rm{S}}} \right)\]
\[{\rm{ = }}\Delta {\rm{H + T}}\Delta {\rm{S}}\]
Gibbs's free energy change will not be negative making it a nonspontaneous reaction.
So, C is correct.

D. \[\Delta {\rm{H}}\]decreases and \[\Delta {\rm{S}}\] increases
Given that,
\[\Delta {\rm{G = }}\left( {{\rm{ - }}\Delta {\rm{H}}} \right){\rm{ - }}\left( {{\rm{T}}\Delta {\rm{S}}} \right)\]
\[{\rm{ = - }}\Delta {\rm{H - T}}\Delta {\rm{S}}\]
Gibbs's free energy change will be negative. It is a spontaneous reaction.
So, D is incorrect.

So, option C is correct.

Note: If we consider thermodynamics, a spontaneous process is defined as a process that happens without any outward intake taken by the system. The system liberates free energy and shifts to an additionally stable state having less energy. According to the sign convention of thermodynamics, the release of free energy by the system conveys a negative sign to the free energy change. That is why Gibbs's free energy change is negative for spontaneous reactions.