Question

Which of the following conditions may lead to a non-spontaneous change?
A. Positive \[\Delta {\rm{H}}\] and positive \[\Delta {\rm{S}}\]
B. Negative \[\Delta {\rm{H}}\]and negative \[\Delta {\rm{S}}\]
C. Positive \[\Delta {\rm{H}}\]and negative \[\Delta {\rm{S}}\]
D. Negative \[\Delta {\rm{H}}\] and positive \[\Delta {\rm{S}}\]

Answer 1

Hint: A spontaneous process is a process that occurs on its own without any external assistance. These processes favour the formation of products. Most important is the Gibbs's free energy change which 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:
According to the second law, "a spontaneous process leads to the rise of the entropy."
Hence, spontaneous reactions will have increased entropy. Gibbs free energy, G is a thermodynamic function that denotes the amount of energy available for work.

It is formulated 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 energy release in spontaneous reactions as a result, the enthalpy change is negative.
From the second law of thermodynamics, we remember that entropy is greater for these types of reactions.

So, the product \[{\rm{T}}\Delta {\rm{S}}\] has an increased value than \[\Delta {\rm{H}}\].
So, \[\Delta {\rm{G}}\] is negative i.e., Gibbs's free energy change is negative for a spontaneous reaction.

A. Positive \[\Delta {\rm{H}}\] and positive \[\Delta {\rm{S}}\]
In this case, the process will be spontaneous if the product T∆S is large enough to overbalance ∆H.
This option is incorrect.

B. Negative \[\Delta {\rm{H}}\]and negative \[\Delta {\rm{S}}\]
For this,
\[\Delta {\rm{G = }}\left( {{\rm{ - }}\Delta {\rm{H}}} \right){\rm{ - }}\left( {{\rm{ - T}}\Delta {\rm{S}}} \right)\]
\[{\rm{ = - }}\Delta {\rm{H + T}}\Delta {\rm{S}}\]
This process can be spontaneous if the value of enthalpy change is higher than the product of temperature and entropy change.
So, B is incorrect.

C. Positive \[\Delta {\rm{H}}\]and negative \[\Delta {\rm{S}}\]
For this,
\[\Delta {\rm{G = }}\left( {\Delta {\rm{H}}} \right){\rm{ - }}\left( {{\rm{ - T}}\Delta {\rm{S}}} \right)\]
\[{\rm{ = }}\Delta {\rm{H + T}}\Delta {\rm{S}}\]
In this case, Gibbs's free energy change will not be negative. It is a non-spontaneous process.
So, C is correct.

D. Negative \[\Delta {\rm{H}}\] and positive \[\Delta {\rm{S}}\]
For this,
\[\Delta {\rm{G = }}\left( {{\rm{ - }}\Delta {\rm{H}}} \right){\rm{ - }}\left( {{\rm{T}}\Delta {\rm{S}}} \right)\]
\[{\rm{ = - }}\Delta {\rm{H - T}}\Delta {\rm{S}}\]
In this case, Gibbs's free energy change will be negative. It is a spontaneous process.
So, C is correct.

So, option C is correct.

Note: Spontaneous reactions do not need any outward elements or don't require being operated by an external force. All natural processes are spontaneous.
These reactions discharge the free energy from the system resulting in more stability.
For example, melting of ice, bonfire, rusting of iron, etc.
Non-spontaneous reactions need energy infusion or exterior elements to bring about a reaction. For example, the movement of gas from a region of less pressure to a region of high pressure.
- Electrolysis of water
- Photosynthesis