Q20.45 - Chemistry: Molecular Nature Of Matter And Change

Question

What is the advantage of calculating free energy changes rather than entropy changes to determine reaction spontaneity?

Step-by-Step Solution

Answer

The entropy of the surroundings is hardly measurable to identify the spontaneity of the process. But since the entropy of the surroundings can be expressed as entropy using Gibb's free energy, the process or reaction is spontaneous if $\mathrm{\Delta G}<0$.

Step 1: Concept Introduction.

The first law of thermodynamics distinguishes between three types of energy transmission: heat, thermodynamic work, and energy associated with matter transfer, and connects them to an internal energy function of a body. It is a thermodynamic adaption of the idea of conservation of energy.

 

Step 2: The Entropy Change.

 The conservation of energy is based on the $1^{\mathrm{st}}$ law of thermodynamics. As a result, the energy transfer alone cannot determine the spontaneity of the operation. The $2^{\mathrm{nd}}$ Law of Thermodynamics should be used as well.

For a spontaneous process, the entropy should increase $(\mathrm{\Delta }{\mathrm{S}}_{\mathrm{total}}>0)$. However, Gibb's free energy for a spontaneous reaction is $\mathrm{\Delta G}<0$.

The entropy changes for a reversible process then to be spontaneous, where ${\mathrm{q}}_{\mathrm{rev}}$-heat transferred, $T-$absolute temperature –

$\begin{array}{c}\mathrm{\Delta S}= \frac{{\mathrm{q}}_{\mathrm{rev}}}{\mathrm{T}}\\ {\mathrm{\Delta S}}_{\mathrm{total}}= {\mathrm{\Delta S}}_{\mathrm{sys}}+{\mathrm{\Delta S}}_{\mathrm{surr}}>0\\ {\mathrm{\Delta S}}_{\mathrm{total}}= \frac{{\mathrm{q}}_{\mathrm{sys}}}{\mathrm{T}}+\frac{{\mathrm{q}}_{\mathrm{surr}}}{\mathrm{T}}>0\end{array}$

Step 3: Gibbs Free Energy.

However, it is hard to calculate the entropy change of the surroundings. Gibb's Free Energy is used for simplification then –

$\mathrm{\Delta G} = \mathrm{\Delta H}-\mathrm{T\Delta S}$

If the process is isobaric, then –

$\begin{array}{c}{\mathrm{q}}_{\mathrm{surr}}= {\mathrm{\Delta H}}_{\mathrm{surr}}= -{\mathrm{\Delta H}}_{\mathrm{sys}}\\ {\mathrm{\Delta S}}_{\mathrm{total}}= \frac{{\mathrm{q}}_{\mathrm{sys}}}{\mathrm{T}}-\frac{-{\mathrm{\Delta H}}_{\mathrm{sys}}}{\mathrm{T}}\end{array}$

Then, the Gibb's free energy can be rewritten as –

$\begin{array}{l}\mathrm{\Delta G} = \mathrm{\Delta H}-\mathrm{T}·{\mathrm{\Delta S}}_{\mathrm{total}}\\ \mathrm{\Delta G} = \mathrm{\Delta H}-\mathrm{T}·\left({\mathrm{S}}_{\mathrm{sys}}-\frac{-{\mathrm{\Delta H}}_{\mathrm{sys}}}{\mathrm{T}}\right)\end{array}$

 

Therefore, the entropy of the surroundings is not included in the calculation anymore and the reaction spontaneity is determined easier as $\mathrm{\Delta G}<0$.