CHE10002 - Entropy/Gibbs Free Energy¶

Definition

Entropy is s measure of how ordered a system is.
The second law of thermodynamics states that any spontaneous process increases the entropy of the universe.

To decrease the entropy of a system, work needs to be done
The act of decreasing the entropy of one system inevitably means increasing the entropy of another
Entropy, like enthalpy is a state function.

Examples of disorder¶

Systems of a lower energy state of matter have a lower entropy¶

$\ce{H2O_{(l)}}$ has a lower entropy than $\ce{H2O_{(g)}}$ (96.9 vs 188.7)

Atoms of a higher molecular weight have lower entropy¶

$\ce{He_{(g)}}$ has a lower entropy than $\ce{Ne_{(g)}}$ (126.1 vs 146.2 )

Molecules with fewer bonds have less entropy¶

$\ce{C2H2_{(g)}}$ has a lower entropy than $\ce{C2H6_{(g)}}$ (200.9 vs 229.2)

Systems with fewer moles of the same atom have less entropy¶

$\ce{I2_{(g)}}$ has a lower entropy than $\ce{2I_{(g)}}$

This goes for reactions too
$\ce{2ZnO_{(s)}}$ has lower entropy than $\ce{2Zn_{(s)} + O2_{(g)}}$ (2 moles vs 3 moles)

Entropy as a function of time¶

While a spontaneous process occurs, entropy increases until a point at which equilibrium is reached. at this point, there is no net change in the constituents of the reaction, so there can be no further increase in entropy.

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Spontaneity¶

“spontaneous processes increase the entropy”

Chris Cramer

In a chemical sense, is a measure of whether or not a reaction is thermodynamically capable of happening without outside energy input

If an equation has a a positive enthalpy ($\Delta H=+ve$), it is logically unlikely to occur, since heat is required form the environment. We can compare the effects of enthalpy and entropy, to determine which change is driving the reaction:

If $\Delta H>T\Delta S$, then the reaction is enthalpy driven

if $\Delta H<T\Delta S$ then the reaction is entropy driven

E.g.: $$ \ce{2NH4Cl_{(s)} + Ba(OH)2.8H2O_{(s)} -> 2NH3_{(g)} + 10H2O_{(l)} + BaCl2_{(s)}} $$

\[ \begin{gather} \Delta H^\circ=166\:KJ \text{ (endothemric)}\\ \Delta S^\circ=0.594\:KJ\cdot K^{-1}\\ T=298.15\:K\\ \end{gather} \]

So when we calculate $T \Delta S$ we get; $298.15\cdot0.594=177\:KJ$

Since $T \Delta S$ is greater than $\Delta H$, the endothermic nature of the reaction is overcome by the entropy to make it spontaneous

Note

The reason for the much larger contribution of entropy is that the reaction converts 3 moles of solid into one mole of solid, 10 moles of liquid and 2 moles of gas. This is both a massive increase in the number of moles of matter, but also a massive increase in the entropy from species that have more ability to move around.

Gibbs Free Energy¶

Is also a state function
Measures the amount of energy available to do useful chemical work
If $\Delta G$ is $-ve$, then the reaction is spontaneous
If $\Delta G$ is $0$, then the reaction is in equilibrium

\[ \Delta G^\circ=\Delta H^\circ-T\Delta S^\circ \]

E.g.: In the above example, we can calculate the $\Delta G$ as:

\[ \Delta G=166-(298.15\cdot0.594)=-11\:KJ\cdot mol^{-1} \]

Which means that ultimately, despite taking in $166\:KJ\cdot mol^{-1}$, the system has actually lost $11\:KJ\cdot mol^{-1}$ of energy and is more stable as a result.

Non-standard state spontaneity¶

To calculate spontaneity in non standard state conditions we can use the following equation

Where $Q=\frac{\text{products}}{\text{reactants}}$

\[ \Delta G=\Delta G^\circ+RT\ln(Q) \]

Determination of equilibrium constants (rate constants)¶

We can use $\Delta G$ to calculate a $k_x$

Where $R=$ Gas Constant ($8.314\:J\cdot K^{-1}\cdot mol^{-1}$)

\[ \Delta G^\circ=-RT\ln(k) \]