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$\Delta G = \Delta H - T\Delta S$: spontaneity as a condition set by enthalpy, entropy and temperature, and its link to $K$ and to cell potentials.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to calculate $\Delta G$ from $\Delta H$, $\Delta S$ and the temperature, decide from the signs of $\Delta H$ and $\Delta S$ when a change is spontaneous, find the crossover temperature and give the spontaneous temperatures as an interval, read $\Delta G$ against temperature as a straight line, and connect the sign of $\Delta G^\circ$ to $K$ and to a cell potential.
Last lesson a change was spontaneous when the total entropy of system and surroundings increased, and the surroundings' share was $-\Delta H \div T$. From Chemistry 1 you have enthalpy changes; from the equilibrium unit, $K$; from the redox unit, cell potentials. This lesson puts all of them on one scale.
The Gibbs energy change, $\Delta G$, of a reaction is $\Delta H - T\Delta S$, with $T$ in kelvin and $\Delta S$ in kJ/(K mol) so that everything is in kJ/mol. A change is spontaneous (thermodynamically feasible) when $\Delta G$ is negative. The crossover temperature is where $\Delta G = 0$, found from $T = \Delta H \div \Delta S$. $\Delta G^\circ$ is the value under standard conditions.
From the last lesson, a change is spontaneous when
$$\Delta S_{total} = \Delta S_{sys} - \frac{\Delta H}{T} > 0$$
Multiply through by $-T$ (which turns the inequality round) and it becomes
$$\Delta G = \Delta H - T\Delta S < 0$$
That is all Gibbs energy is: the second law, rewritten so that only the system's own quantities appear. Two drives are inside it. A negative $\Delta H$ helps a change happen, because the heat released raises the entropy of the surroundings. A positive $\Delta S$ helps, because the system's own arrangements increase. Temperature decides how much the entropy term counts.
| $\Delta H$ | $\Delta S$ | $\Delta G = \Delta H - T\Delta S$ | Spontaneous |
|---|---|---|---|
| $-$ | $+$ | always negative | at every temperature |
| $+$ | $-$ | always positive | at no temperature |
| $-$ | $-$ | negative while $T$ is small | below the crossover |
| $+$ | $+$ | negative once $T$ is large | above the crossover |
The last two rows are where chemistry gets interesting. Ice melting has $\Delta H = +6.01$ kJ/mol and $\Delta S = +22.0$ J/(K mol). The crossover is $6.01 \div 0.0220 = 273$ K, 0 °C: above it melting is spontaneous, below it freezing is. An endothermic change goes by itself — whenever the entropy gain, multiplied by the temperature, pays for the heat taken in.
The units need care. Enthalpies are tabulated in kJ/mol and entropies in J/(K mol). Divide the entropy by 1000 before putting it into $\Delta G = \Delta H - T\Delta S$, or every answer is out by a factor of a thousand in the entropy term.
Another way: picture
Think of a decision with two pressures: money ($\Delta H$) and freedom ($\Delta S$). Temperature is how much you value freedom. A job that pays well and gives freedom is taken at once; one that costs you and confines you never is. When they pull opposite ways, what you decide depends on how much freedom counts, and there is a point where you are exactly torn — the crossover.
Another way: steps
To decide spontaneity:
Limestone, calcium carbonate, decomposes to quicklime and carbon dioxide with $\Delta H = +178$ kJ/mol and $\Delta S = +159$ J/(K mol). The crossover is $178 \div 0.159 = 1119$ K, about $846$ °C, which is why lime kilns run at around $900$ °C: below the crossover no amount of waiting turns limestone into lime.
The Haber process has $\Delta H = -92$ kJ/mol and $\Delta S = -198$ J/(K mol). It is spontaneous only below $92 \div 0.198 = 465$ K. Yet industry runs it at about $700$ K. The reason is the rest of this course: at $465$ K the reaction is far too slow, even with an iron catalyst, so a higher temperature is used for rate, accepting a smaller equilibrium yield, and the ammonia is removed as it forms to pull the equilibrium on. Thermodynamics says which way; kinetics says how fast; Le Chatelier says how far, and a chemical engineer has to satisfy all three.
Plotted against temperature, $\Delta G$ is a straight line with intercept $\Delta H$ and gradient $-\Delta S$. Where it crosses zero is the crossover temperature.
The standard Gibbs energy change connects to the equilibrium constant and to the cell potential:
$$\Delta G^\circ = -RT\ln K \qquad \Delta G^\circ = -nFE^\circ_{cell}$$
where $R = 8.31$ J/(K mol), $n$ is the moles of electrons transferred and $F = 96\,500$ C/mol. The minus signs mean all three tell the same story:
| $\Delta G^\circ$ | $K$ | $E^\circ_{cell}$ | Products favoured? |
|---|---|---|---|
| negative | greater than 1 | positive | yes |
| zero | 1 | zero | neither |
| positive | less than 1 | negative | no |
For the zinc–copper cell, $\Delta G^\circ = -2 \times 96\,500 \times 1.10 = -212\,300$ J/mol, about $-212$ kJ/mol: the most electrical work one mole of that reaction can do.
What $\Delta G$ never says is how fast. Diamond turning into graphite has a negative $\Delta G$ at room temperature and an activation energy so high that diamonds last for ever. Spontaneous means the change is allowed, not that it is quick.
An exothermic change is always spontaneous. Only if its entropy does not fall too far. The Haber reaction is exothermic and not spontaneous above 465 K.
An endothermic change can never be spontaneous. Ice melts at room temperature and salt dissolves in cold water; the entropy term pays for the heat.
Mixing kJ and J. $\Delta S$ must be divided by 1000 before it goes into $\Delta G = \Delta H - T\Delta S$ with $\Delta H$ in kJ.
Temperature in °C. $T$ must be in kelvin; at 25 °C it is 298 K.
Spontaneous means fast. $\Delta G$ gives the direction a change is allowed to go; the activation energy sets its rate.
A reaction has $\Delta H = -50$ kJ/mol and $\Delta S = -120$ J/(K mol). Is it spontaneous at $298$ K?
Opposite drives: enthalpy helps, entropy hinders.
$T\Delta S = 298 \times (-0.120) = -35.8$ kJ/mol.
Entropy in kJ first.
$\Delta G = -50 - (-35.8) = -14.2$ kJ/mol: negative, so yes. It stops being spontaneous above $50 \div 0.120 = 417$ K.
Below the crossover, the enthalpy term wins.
Dinitrogen tetroxide splits into nitrogen dioxide: $\Delta H = +58$ kJ/mol, $\Delta S = +177$ J/(K mol).
Both positive: spontaneous only at high temperature.
$T = 58 \div 0.177 = 328$ K.
Where $\Delta H = T\Delta S$.
Above about $328$ K, $55$ °C, the brown nitrogen dioxide is favoured — the colour change the equilibrium unit showed in hot and cold flasks.
The same fact seen from thermodynamics.
$T\Delta S = 250 \times 0.100 = 25$ kJ/mol.
Entropy in kJ/(K mol).
$\Delta G = 30 - 25 = \ldots$
…$+5$ kJ/mol, positive, so it is not spontaneous; it would be above $300$ K.
A reaction has $\Delta H = 113$ kJ/mol and $\Delta S = -160$ J/(K mol). What is $\Delta G$ at $300$ K?
Answer: unit: J / kJ
For each combination of signs, when is the change spontaneous?
| spontaneous | |
|---|---|
| $\Delta H$ negative, $\Delta S$ positive | |
| $\Delta H$ positive, $\Delta S$ negative | |
| $\Delta H$ negative, $\Delta S$ negative | |
| $\Delta H$ positive, $\Delta S$ positive |
A decomposition has $\Delta H = +42$ kJ/mol and $\Delta S = +40$ J/(K mol), and neither changes much with temperature. At which temperatures, in kelvin, is it spontaneous?
This task has no paper form; do it on a device.
A reaction has $\Delta H = -100$ kJ/mol and $\Delta S = -0.4$ kJ/(K mol). Plotted against temperature, $\Delta G$ is a straight line. Give its gradient and its intercept at $T = 0$.
Gradient, in kJ/(K mol):
Intercept, in kJ/mol:
Making ammonia, $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$, has $\Delta H = -92$ kJ/mol and $\Delta S = -198$ J/(K mol). Is it spontaneous at $800$ K?
Match each value of the standard Gibbs energy change to what it means for the equilibrium constant and, for a redox reaction, the standard cell potential.
| K greater than 1, and a positive cell potential | K equal to 1, and a cell potential of zero | K less than 1, and a negative cell potential | |
|---|---|---|---|
| $\Delta G^\circ$ negative | |||
| $\Delta G^\circ$ zero | |||
| $\Delta G^\circ$ positive |
A battery designer wants the maximum electrical work available from the magnesium–copper cell, $\mathrm{Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu}$, which has $E^\circ_{cell} = 2.71$ V and moves $2$ mol of electrons per mole of reaction. Using $\Delta G^\circ = -nFE^\circ_{cell}$ with $F = 96\,500$ C/mol, what is $\Delta G^\circ$ per mole of reaction?
Answer: unit: J / kJ
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A gas-phase reaction that joins two molecules into one has $\Delta H = -38.5$ kJ/mol and $\Delta S = -70$ J/(K mol). At which temperatures, in kelvin, is it spontaneous?
This task has no paper form; do it on a device.
You can decide spontaneity from enthalpy, entropy and temperature together. Say out loud why ice melts although melting is endothermic, and why the Haber process is run above the temperature where it stops being spontaneous. That closes the course: how fast, how far, and whether at all.
10. Your turn: a reaction has $\Delta H = +30$ kJ/mol and $\Delta S = +100$ J/(K mol). Is it spontaneous at $250$ K?, step 3