Thermodynamics • Topic 3 of 3

Entropy & Gibbs Energy

The first law tells us energy is conserved, but it does not tell us which way a change will go on its own. A spontaneous process is one that occurs without continuous outside help (heat flowing from hot to cold, a gas expanding into vacuum). Many spontaneous reactions are exothermic, but exothermicity alone is not the criterion — ice melting above 0°C and ammonium nitrate dissolving are both spontaneous yet endothermic. A second factor is at work: disorder.

Entropy $S$ measures the disorder or the number of microscopic ways a system can be arranged. It is a state function. For a reversible process at temperature $T$, the entropy change is $\Delta S=\dfrac{q_{rev}}{T}$, with units $\text{J K}^{-1}\text{mol}^{-1}$. Entropy increases when solids melt, liquids vaporise, gases expand, or the number of gaseous molecules rises in a reaction.

The second law of thermodynamics states that for any spontaneous process the total entropy of the universe increases:

$\Delta S_{total}=\Delta S_{system}+\Delta S_{surroundings}>0$

At equilibrium $\Delta S_{total}=0$. Because tracking the surroundings is awkward, J. W. Gibbs combined enthalpy and entropy into a single system property, the Gibbs energy $G=H-TS$. At constant temperature and pressure its change is

$\Delta G=\Delta H-T\Delta S$

The Gibbs energy is the master criterion for spontaneity at constant $T$ and $P$: if $\Delta G<0$ the process is spontaneous (feasible); if $\Delta G=0$ the system is at equilibrium; if $\Delta G>0$ the process is non-spontaneous (the reverse is spontaneous). The magnitude $-\Delta G$ equals the maximum useful (non-expansion) work obtainable.

The interplay of the two terms decides the outcome. When $\Delta H<0$ and $\Delta S>0$, $\Delta G$ is negative at all temperatures (always spontaneous). When $\Delta H>0$ and $\Delta S<0$, $\Delta G$ is positive at all temperatures (never spontaneous). The remaining two cases are temperature-dependent: an endothermic reaction with $\Delta S>0$ becomes spontaneous only above $T=\Delta H/\Delta S$, while an exothermic reaction with $\Delta S<0$ is spontaneous only below that temperature.

Gibbs energy also connects thermodynamics to chemical equilibrium. The standard Gibbs energy change is related to the equilibrium constant $K$ by

$\Delta G^\circ=-RT\ln K$

A large negative $\Delta G^\circ$ means $K\gg1$ (products favoured); a positive $\Delta G^\circ$ means $K<1$ (reactants favoured). At equilibrium $\Delta G=0$ even though $\Delta G^\circ$ need not be zero.

Finally, the third law of thermodynamics states that the entropy of a perfect crystalline substance is zero at absolute zero ($0$ K). This gives entropy an absolute reference point, so unlike $U$ and $H$ we can tabulate absolute standard entropies $S^\circ$ and use them to compute $\Delta_r S^\circ=\sum\nu_p S^\circ(\text{products})-\sum\nu_r S^\circ(\text{reactants})$.

Sign of $\Delta G=\Delta H-T\Delta S$ and spontaneity
$\Delta H$$\Delta S$$\Delta G$Spontaneity
− (exo)+ (more disorder)always −Spontaneous at all $T$
+ (endo)− (less disorder)always +Never spontaneous
− (exo)− (less disorder)− at low $T$Spontaneous only at low $T$
+ (endo)+ (more disorder)− at high $T$Spontaneous only at high $T$
1
Worked Example
A reaction has $\Delta H=-92.0$ kJ and $\Delta S=-198\ \text{J K}^{-1}$. Calculate $\Delta G$ at 298 K and state whether it is spontaneous.
Solution
  1. $\Delta G=\Delta H-T\Delta S$.
  2. $T\Delta S=298\times(-198\times10^{-3})=-59.0$ kJ.
  3. $\Delta G=-92.0-(-59.0)=-33.0$ kJ.
  4. $\Delta G<0$, so spontaneous at 298 K.

Answer: $\Delta G=-33.0\ \text{kJ}$; spontaneous.

2
Worked Example
For a reaction $\Delta H=+30.0$ kJ and $\Delta S=+100\ \text{J K}^{-1}$. Above what temperature does it become spontaneous?
Solution
  1. Spontaneous when $\Delta G<0$, i.e. $\Delta H
  2. Crossover at $T=\dfrac{\Delta H}{\Delta S}=\dfrac{30000}{100}$.
  3. $T=300$ K.

Answer: spontaneous above $T=300\ \text{K}$.

3
Worked Example
Calculate the entropy change when 36 g of ice melts at 0°C. ($\Delta_{fus} H=6.0$ kJ/mol, molar mass of water = 18 g/mol)
Solution
  1. Moles $=36/18=2$ mol; total $\Delta H=2\times6.0=12.0$ kJ $=12000$ J.
  2. $T=0\degree\text{C}=273$ K (melting is reversible at the melting point).
  3. $\Delta S=\dfrac{q_{rev}}{T}=\dfrac{12000}{273}=43.96\ \text{J K}^{-1}$.

Answer: $\Delta S\approx+44.0\ \text{J K}^{-1}$.

4
Worked Example
The equilibrium constant of a reaction at 298 K is $K=10$. Calculate $\Delta G^\circ$. ($R=8.314\ \text{J K}^{-1}\text{mol}^{-1}$)
Solution
  1. $\Delta G^\circ=-RT\ln K$.
  2. $\ln 10=2.303$.
  3. $\Delta G^\circ=-(8.314)(298)(2.303)=-5705$ J.

Answer: $\Delta G^\circ\approx-5.71\ \text{kJ/mol}$.

5
Worked Example
For a phase change at equilibrium, $\Delta H=40.7$ kJ/mol (vaporisation of water) at 373 K. Find $\Delta S$.
Solution
  1. At equilibrium $\Delta G=0$, so $\Delta H=T\Delta S$.
  2. $\Delta S=\dfrac{\Delta H}{T}=\dfrac{40700}{373}$.
  3. $=109.1\ \text{J K}^{-1}\text{mol}^{-1}$.

Answer: $\Delta S=+109.1\ \text{J K}^{-1}\text{mol}^{-1}$.

6
Worked Example
A reaction has $\Delta H=+50$ kJ and $\Delta S=-80\ \text{J K}^{-1}$. Is it spontaneous at any temperature? Explain.
Solution
  1. $\Delta G=\Delta H-T\Delta S=50-T(-0.080)=50+0.080T$ (kJ).
  2. Since both terms are positive for all $T>0$, $\Delta G>0$ always.
  3. The reaction is never spontaneous.

Answer: Non-spontaneous at all temperatures ($\Delta H>0$, $\Delta S<0$).

Key Points

  • A spontaneous process occurs without external help; exothermicity alone is not the criterion for spontaneity.
  • Entropy $S$ measures disorder; $\Delta S=q_{rev}/T$ and it increases on melting, vaporising, expanding or producing more gas molecules.
  • Second law: for a spontaneous change $\Delta S_{total}=\Delta S_{sys}+\Delta S_{surr}>0$; at equilibrium it is zero.
  • Gibbs energy $\Delta G=\Delta H-T\Delta S$: $\Delta G<0$ spontaneous, $=0$ equilibrium, $>0$ non-spontaneous; also $\Delta G^\circ=-RT\ln K$.
  • Third law: entropy of a perfect crystal is zero at 0 K, giving absolute standard entropies $S^\circ$.
Tap an option to check your answer0 / 4
Q1.The criterion for a spontaneous process at constant $T$ and $P$ is:
Explanation: A process is spontaneous when the Gibbs energy decreases, $\Delta G<0$.
Q2.For a reaction with $\Delta H<0$ and $\Delta S>0$, the reaction is:
Explanation: Both terms make $\Delta G=\Delta H-T\Delta S$ negative at every temperature.
Q3.At equilibrium, the value of $\Delta G$ is:
Explanation: At equilibrium $\Delta G=0$, although $\Delta G^\circ$ need not be zero.
Q4.According to the third law, the entropy of a perfect crystal at 0 K is:
Explanation: The third law sets the entropy of a perfectly ordered crystal to zero at absolute zero.