Chemistry of Solutions 202-SN2-RE • CEGEP, Montreal

Revision sheet: chemical thermodynamics (202-SN2-RE)

Thermodynamics says what is POSSIBLE, never what is fast. Two quantities are enough, enthalpy and entropy, and their combination $\Delta G=\Delta H-T\Delta S$ settles the question of spontaneity. Temperature acts as the referee, which explains why the same reaction can be impossible when cold and easy when hot.

This sheet gives the table of the four cases, the way to predict the sign of $\Delta S$ without calculation, the relation between free energy and the equilibrium constant, and the list of unit mistakes that cost the most. It is written for CEGEP students in Chemistry of Solutions, 202-SN2-RE (formerly 202-NYB).

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The thread of the chapter

$\Delta H$ is in kilojoules per mole and $\Delta S$ in joules per mole per kelvin. Subtracting the two without converting gives a $\Delta G$ a thousand times too large, and this is by far the most frequent mistake of the chapter.

This chapter is part of 202-SN2-RE, Chemistry of Solutions

The essentials

The three quantities and their units

• $\Delta H^{\circ}_{r}=\sum n\Delta H^{\circ}_{f}(\text{products})-\sum n\Delta H^{\circ}_{f}(\text{reactants})$ , in kJ/mol. Zero for an element in its standard state.
• $\Delta S^{\circ}_{r}=\sum nS^{\circ}(\text{products})-\sum nS^{\circ}(\text{reactants})$ , in J/(mol·K). $S^{\circ}$ is TABULATED and always POSITIVE, even for an element.
• $\Delta G=\Delta H-T\Delta S$ , with $T$ in KELVINS and both terms in the SAME unit.
• Spontaneous if $\Delta G<0$ ; at equilibrium if $\Delta G=0$ ; nonspontaneous if $\Delta G>0$ .
• Link with equilibrium: $\Delta G^{\circ}=-RT\ln K$ , and away from standard conditions $\Delta G=\Delta G^{\circ}+RT\ln Q$ .

The two tables differ in nature: the enthalpy of FORMATION is zero for an element and can be negative; ABSOLUTE entropy never is, by virtue of the third law.

Predicting the sign of $\Delta S$ without calculation

• What decides first is the change in the NUMBER OF MOLES OF GAS. More gas on the right: $\Delta S>0$ .
• Increasing order of disorder: solid $<$ liquid $<$ gas. A change of state toward the right of this list increases entropy.
• Dissolving an ionic solid generally increases entropy; dissolving a gas DECREASES it, since free molecules are being trapped.
• With an equal number of moles of gas, compare the complexity of the molecules: the more atoms a molecule has, the higher its entropy.
• Heating always increases the entropy of a substance, without changing the sign of $\Delta S$ for the reaction.

Solid, liquid, gas: disorder, and so entropy, grows in this order, and the change to gas is by far the biggest jump.

A reaction that consumes gases to give a solid necessarily has $\Delta S<0$ : this is the case for solid ammonium chloride formed from gaseous $NH_{3}$ and $HCl$ . The Haber process, for its part, produces a gas: its $\Delta S$ is negative because four moles of gas give two.

The four cases, and the crossover temperature

• $\Delta H<0$ and $\Delta S>0$ : spontaneous at EVERY temperature. The most favourable case.
• $\Delta H>0$ and $\Delta S<0$ : NEVER spontaneous, at any temperature.
• $\Delta H<0$ and $\Delta S<0$ : spontaneous at LOW temperature, the term $-T\Delta S$ eventually winning as the temperature rises.
• $\Delta H>0$ and $\Delta S>0$ : spontaneous at HIGH temperature, which is the case for most decompositions.
• Crossover temperature: $T=\dfrac{\Delta H}{\Delta S}$ , the one where $\Delta G=0$ . It exists ONLY in the last two cases, those where the two signs are identical.

Only the two cells of the DIAGONAL, where the two signs coincide, have a crossover temperature. The other two are settled in advance, whatever the temperature.

In the first two cases, the calculation $\dfrac{\Delta H}{\Delta S}$ gives a negative number, hence an impossible absolute temperature: this is how arithmetic signals that there is no crossover.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Subtracting joules from kilojoules in $\Delta G=\Delta H-T\Delta S$

the whole question, 6 points, and a reversed conclusion

What not to write

“ $\Delta G=178.3-298\times 160.6=-47{,}680$ kJ/mol.”

What to write

“ $\Delta S$ is in J/(mol·K): it must be converted to kJ. $\Delta G=178.3-298.15\times 0.1606=+130.4$ kJ/mol, and the reaction is NOT spontaneous.”

Why: The factor of a thousand changes not only the value but the SIGN of $\Delta G$ , and so the conclusion about spontaneity. This is the costliest mistake of the chapter.

2. Setting the standard entropy of an element to zero

the whole question, 5 points

What not to write

“ $S^{\circ}(O_{2})=0$ , like its enthalpy of formation.”

What to write

“ $S^{\circ}(O_{2})=205.0$ J/(mol·K). Standard entropy is ABSOLUTE and always positive: only $\Delta H^{\circ}_{f}$ is zero for an element.”

The enthalpy of formation of

O_{2}

is zero, its standard entropy is

205.0

J/(mol·K): in the table of

S^{\circ}

, no bar is at zero.

Why: The two tables do not measure the same thing: one is a difference relative to the elements, the other an absolute value referred to absolute zero, where every entropy vanishes.

3. Leaving the temperature in degrees Celsius

3 points, and a

\Delta G

too large by

44

kJ/mol here

What not to write

“At $25$ °C: $\Delta G=178.3-25\times 0.1606=174.3$ kJ/mol.”

What to write

“ $T$ in KELVINS: $\Delta G=178.3-298.15\times 0.1606=130.4$ kJ/mol.”

On the same line,

T=25

reads the red point, almost at

\Delta H

; the true temperature,

298.15

K, gives the green point,

44

kJ/mol lower.

Why: Entropy is defined on an absolute scale. Using degrees Celsius would give a $\Delta G$ equal to $\Delta H$ at $0$ °C, which makes no physical sense.

4. Confusing spontaneous and fast

4 points on the reasoning question

What not to write

“ $\Delta G^{\circ}$ is very negative, so the reaction is fast.”

What to write

“ $\Delta G^{\circ}<0$ only says that the reaction is POSSIBLE. The rate depends on the activation energy, which appears nowhere in thermodynamics.”

The products are much lower,

\Delta G^{\circ}

very negative, but the barrier is high: thermodynamics says where the reaction would go, kinetics says whether it goes there.

Why: The combustion of diamond has a very negative $\Delta G^{\circ}$ at room temperature and yet never happens spontaneously: its kinetic barrier cannot be crossed without heating.

5. Looking for a crossover temperature where there is none

3 points, and the direction of the conclusion reversed

What not to write

“ $\Delta H=-92$ kJ/mol and $\Delta S=-198$ J/(mol·K): the crossover temperature is $\dfrac{-92{,}000}{-198}=465$ K, and beyond it the reaction becomes spontaneous.”

What to write

“The two signs are identical, so the crossover does exist, but in the OTHER direction: the reaction is spontaneous BELOW $465$ K and stops being so above.”

Same formula

T=\dfrac{\Delta H}{\Delta S}

, opposite directions: the Haber line RISES and is negative only below

465

K; the limestone line goes down and is negative only above

1110

Why: The formula gives the temperature of the switch, not the direction of the switch. That is read off the signs: $\Delta H<0$ means that the reaction is favoured when cold.

6. Using $\Delta G^{\circ}$ away from standard conditions

the whole question, 5 points

What not to write

“ $\Delta G^{\circ}=+130$ kJ/mol at $25$ °C, so the decomposition cannot happen at any pressure.”

What to write

“ $\Delta G=\Delta G^{\circ}+RT\ln Q$ : by sweeping away the carbon dioxide, we lower $Q$ , hence $\Delta G$ , and the decomposition becomes possible sooner.”

The same reaction, the same enthalpy and the same entropy: only the pressure of the product gas changes, and the temperature needed drops by more than two hundred degrees.

Why: The small circle means “all species at $1$ mol/L or $1$ bar”. As soon as a partial pressure departs from this value, the $\ln Q$ term must be added, and the industrial effect is considerable.

7. Counting total moles to predict the sign of $\Delta S$

2 points, but the question on the sign comes up every time

What not to write

“ $CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)$ : one mole becomes two, so $\Delta S$ increases a little.”

What to write

“What counts is the number of moles of GAS: we go from ZERO to ONE. That is a considerable jump, and $\Delta S=+160.6$ J/(mol·K).”

Two solids count for almost nothing: it is the appearance of the gas, from zero to one mole, that makes the entropy jump.

Why: The entropy of a gas is about twice that of the corresponding solid. Two solids count for almost nothing in the balance; it is the appearance of a gas that wins.

8. Believing that a negative system entropy forbids the reaction

4 points on the reasoning question

What not to write

“ $\Delta S_{system}<0$ , so the reaction violates the second law and is impossible.”

What to write

“It is the entropy of the UNIVERSE that must increase: $\Delta S_{universe}=\Delta S_{system}+\dfrac{-\Delta H}{T}$ . An exothermic reaction increases the entropy of the surroundings, which can compensate.”

Synthesis of ammonia at

298

K: the system loses

198

J/K, but the heat given up makes the surroundings gain

309

, and the entropy of the universe increases.

Why: The criterion $\Delta G<0$ is nothing other than $\Delta S_{universe}>0$ multiplied by $-T$ . This is why an ordering reaction, such as the formation of a crystal, remains possible if it releases enough heat.

9. Confusing a spontaneous reaction with a complete one

3 points

What not to write

“ $\Delta G^{\circ}<0$ , so the reaction goes to completion and no reactant is left.”

What to write

“ $\Delta G^{\circ}<0$ only means $K>1$ . A $\Delta G^{\circ}$ of $-2$ kJ/mol gives $K=2.2$ , so an equilibrium where a lot of reactant remains.”

Why: The link is logarithmic: about $-6$ kJ/mol is needed for a $K$ of $10$ , and $-34$ kJ/mol for a $K$ of $10^{6}$ . A weakly spontaneous reaction is not a complete reaction.

10. Forgetting to multiply by the coefficients in an entropy balance

3 points

What not to write

“ $\Delta S^{\circ}=S^{\circ}(NH_{3})-S^{\circ}(N_{2})-S^{\circ}(H_{2})$ for $N_{2}+3H_{2}\rightarrow 2NH_{3}$ .”

What to write

“Each term carries its coefficient: $\Delta S^{\circ}=2S^{\circ}(NH_{3})-S^{\circ}(N_{2})-3S^{\circ}(H_{2})$ .”

Why: The tabulated quantities are molar: they refer to one mole of the species. The reaction balance consumes or produces several, so each term is weighted.

11. Getting the sign wrong in the relation between $\Delta G^{\circ}$ and $K$

3 points, and a reversed conclusion

What not to write

“ $\Delta G^{\circ}=RT\ln K$ , so a negative $\Delta G^{\circ}$ gives a $K$ below $1$ .”

What to write

“ $\Delta G^{\circ}=-RT\ln K$ : a NEGATIVE $\Delta G^{\circ}$ gives a $K$ ABOVE $1$ , which is consistent with a favoured reaction.”

With the minus sign, a negative

\Delta G^{\circ}

gives

\ln K>0

, so

K>1

: the green line. The red line would pair a favoured reaction with

K<1

Why: The minus sign is what makes the two criteria agree: spontaneous means both $\Delta G^{\circ}<0$ and $K>1$ . Without it, the two would systematically contradict each other.

Which method to choose

Is the reaction spontaneous?

The signs of $\Delta H$ and $\Delta S$ , before any calculation

Each case is a line with intercept

\Delta H

and slope

-\Delta S

: only the two that change sign have a crossover temperature.

If $\Delta H<0$ and $\Delta S>0$ → spontaneous at EVERY temperature, no calculation needed

Example: combustion of a hydrocarbon producing more gas
If $\Delta H>0$ and $\Delta S<0$ → NEVER spontaneous, whatever the temperature

Example: the formation of ozone from oxygen
If $\Delta H<0$ and $\Delta S<0$ → spontaneous BELOW $T=\dfrac{\Delta H}{\Delta S}$

Example: the Haber process, favoured when cold but too slow
If $\Delta H>0$ and $\Delta S>0$ → spontaneous ABOVE $T=\dfrac{\Delta H}{\Delta S}$

Example: the decomposition of limestone, above $837$ °C
If the conditions are NOT standard → $\Delta G=\Delta G^{\circ}+RT\ln Q$ , which can change the conclusion

sweeping away a gaseous product lowers $Q$ and makes the reaction possible much sooner

The table of the four cases is remembered by its diagonal: where the two signs coincide, temperature is the referee; where they are opposite, the answer is the same at every temperature.

Which quantity to calculate, depending on the question

What the problem asks

If “is the reaction spontaneous at such a temperature” → $\Delta G=\Delta H-T\Delta S$ , after converting the units

Example: $+130.4$ kJ/mol at $298$ K: nonspontaneous
If “from what temperature” → $T=\dfrac{\Delta H}{\Delta S}$ , in consistent units, then interpret the direction from the signs

Example: $\dfrac{178{,}300}{160.6}=1110$ K
If “what is the equilibrium constant” → $K=e^{-\Delta G^{\circ}/RT}$ , with $\Delta G^{\circ}$ IN JOULES

Example: $\Delta G^{\circ}=+130$ kJ/mol gives $K\approx 10^{-23}$
If the problem gives pressures or concentrations that are not unity → $\Delta G=\Delta G^{\circ}+RT\ln Q$

Example: sweeping away the product gas lowers the required temperature by two hundred degrees
If the problem gives several equations → Hess's law on $\Delta H$ , on $\Delta S$ and on $\Delta G$ : all three are state functions

reversing changes the sign, multiplying multiplies, adding adds

The three quantities combine through Hess's law in exactly the same way. This is what makes it possible to make an impossible reaction spontaneous by COUPLING it to another, very favourable one, in an industrial process.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Calculating spontaneity and the crossover temperature

When to use it: The problem gives an equation, a table of enthalpies of formation and standard entropies, and asks whether the reaction is spontaneous.

1 Calculate $\Delta H^{\circ}_{r}$ in kJ/mol, products minus reactants, each term multiplied by its coefficient.
2 Calculate $\Delta S^{\circ}_{r}$ in J/(mol·K), in the same way, remembering that elements have a NONZERO entropy.
3 CONVERT $\Delta S$ to kJ/(mol·K) by dividing by one thousand, on a visible line, then write $\Delta G=\Delta H-T\Delta S$ with $T$ in kelvins.
4 Conclude on spontaneity at the requested temperature, and check consistency with the signs in the table of the four cases.
5 If the two signs coincide, calculate $T=\dfrac{\Delta H}{\Delta S}$ and state whether spontaneity is obtained above or below it.

Concluding sentence

“With $\Delta H^{\circ}=+178.3$ kJ/mol and $\Delta S^{\circ}=+160.6$ J/(mol·K), we obtain $\Delta G^{\circ}=178.3-298.15\times 0.1606=+130.4$ kJ/mol: the reaction is not spontaneous at $25$ °C, but it becomes so above $T=\dfrac{178{,}300}{160.6}=1110$ K.”

The trap: Forgetting to convert $\Delta S$ . Writing the division by one thousand explicitly on its own line is the only reliable way not to skip it under exam pressure.

Marking: Typically 2 points for $\Delta H$ , 2 for $\Delta S$ , 1 for the conversion, 2 for $\Delta G$ and the conclusion, 2 for the crossover temperature and its direction.

Handling nonstandard conditions

When to use it: The problem imposes a partial pressure or a concentration different from the standard value, or asks for the effect of sweeping away a gas.

1 First calculate $\Delta G^{\circ}$ under standard conditions, as usual.
2 Write $Q$ as for an equilibrium, including only the gases and the dissolved species.
3 Apply $\Delta G=\Delta G^{\circ}+RT\ln Q$ , with $R$ in J/(mol·K) and $\Delta G^{\circ}$ converted to joules.
4 If the temperature that makes the reaction possible is wanted, set $\Delta G=0$ and solve for $T$ , keeping the $\ln Q$ term.
5 Comment on the gap from the standard temperature: this is usually the real point of the question.

Concluding sentence

“By keeping the pressure of carbon dioxide at $0.010$ bar, the condition $\Delta G=0$ is written $178{,}300=T(160.6-8.314\ln 0.010)$ , so $T=897$ K, that is $623$ °C instead of $837$ °C.”

The trap: Mixing $\Delta G^{\circ}$ in kilojoules and $RT\ln Q$ in joules. The whole calculation must be done in a single unit, and the safest is to convert everything to joules before adding.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Consistent units before the subtraction

Go back over the line for $\Delta G$ and check that $\Delta H$ and $T\Delta S$ are in the same unit. A gap of three orders of magnitude between the two terms signals the oversight.

A $\Delta G=-47{,}680$ kJ/mol: $\Delta S$ was left in joules.
Consistency with the table of the four cases

Compare the sign of the $\Delta G$ obtained with what the signs of $\Delta H$ and $\Delta S$ predict. A contradiction signals a unit or temperature error.

A $\Delta G<0$ when cold for an endothermic, ordering reaction: impossible.
The sign of the crossover temperature

An absolute temperature is always positive. A negative result simply says there is no crossover, which is what you should write rather than correcting it.

A $T=-570$ K: the two signs are opposite, the answer is the same at every temperature.
The order of magnitude of the equilibrium constant

A $\Delta G^{\circ}$ of $\pm 6$ kJ/mol corresponds to a factor of $10$ on $K$ . A $\Delta G^{\circ}$ of $130$ kJ/mol should give a $K$ of the order of $10^{-23}$ .

A $K=0.5$ for a $\Delta G^{\circ}$ of $+130$ kJ/mol: the sign of the relation has been reversed.
Coefficients present in both balances

Check that each entropy AND enthalpy term carries the coefficient of its species in the balanced equation.

An entropy balance without the factor $3$ in front of the hydrogen of the Haber process.

The typical problem, taken apart

The lime kiln, and why it works well below the predicted temperature

The decomposition of limestone is written $CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)$ .

Calculate $\Delta H^{\circ}$ , $\Delta S^{\circ}$ and $\Delta G^{\circ}$ at $25$ °C, then the crossover temperature. Then determine the equilibrium constant at $25$ °C, and finally the temperature required if the kiln is swept so as to keep the pressure of carbon dioxide at $0.010$ bar.

Data in kJ/mol for $\Delta H^{\circ}_{f}$ : $CaCO_{3}=-1206.9$ , $CaO=-635.1$ , $CO_{2}=-393.5$ . In J/(mol·K) for $S^{\circ}$ : $CaCO_{3}=92.9$ , $CaO=39.8$ , $CO_{2}=213.7$ . $R=8.314$ J/(mol·K).

The kiln is swept by a stream of air: the carbon dioxide is carried away as it forms and its pressure stays at

0.010

bar.

Step 1

$\Delta H^{\circ}=[(-635.1)+(-393.5)]-(-1206.9)=-1028.6+1206.9=+178.3$ kJ/mol.

Why

The result is POSITIVE: the decomposition absorbs heat, which is consistent with the fact that a lime kiln has to be heated continuously.

Step 2

$\Delta S^{\circ}=(39.8+213.7)-92.9=+160.6$ J/(mol·K).

Why

The entropy of limestone is not zero even though it appears alone on the left: it is an ABSOLUTE entropy, not an entropy of formation. The gain comes from the appearance of a gas.

Step 3

$\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}=178.3-298.15\times\dfrac{160.6}{1000}=178.3-47.9=+130.4$ kJ/mol.

Why

The division by one thousand is written on the line itself: it is the only gesture that prevents the costliest mistake of the chapter, the one that changes the sign of the conclusion.

Step 4

$\Delta G^{\circ}>0$ : at $25$ °C, the decomposition is NOT spontaneous. The signs $\Delta H>0$ and $\Delta S>0$ announce a reaction favoured at HIGH temperature.

Why

The table of the four cases gives the answer even before the calculation: the only remaining question is from what temperature, which is the object of the next step.

Step 5

Crossover temperature: $T=\dfrac{\Delta H^{\circ}}{\Delta S^{\circ}}=\dfrac{178{,}300}{160.6}=1110$ K, that is $837$ °C.

Why

Both quantities are brought to joules so that the ratio gives kelvins. The direction of the switch is read off the signs: ABOVE this temperature, the reaction goes.

Step 6

Constant at $25$ °C: $\ln K=-\dfrac{\Delta G^{\circ}}{RT}=-\dfrac{130{,}400}{8.314\times 298.15}=-52.6$ , so $K\approx 1.4\times 10^{-23}$ .

Why

This value means that at room temperature the equilibrium pressure of carbon dioxide is of the order of $10^{-23}$ bar: in other words, nothing decomposes.

Step 7

With sweeping: $\Delta G=\Delta G^{\circ}+RT\ln Q$ with $Q=P_{CO_{2}}=0.010$ . The condition $\Delta G=0$ gives $178{,}300=T\left(160.6-8.314\ln 0.010\right)=T(160.6+38.3)=198.9\,T$ .

Why

The logarithm of a number below one is negative, so the term $-R\ln Q$ ADDS to the entropy: it is as if the reaction gained $38$ units of effective entropy.

Step 8

$T=\dfrac{178{,}300}{198.9}=897$ K, that is $623$ °C, against $837$ °C under standard conditions.

Why

Two hundred and fourteen degrees saved by a single change to the process: this is exactly the reason why industrial kilns are ventilated continuously.

Step 9

Check: at $1110$ K and under standard conditions, $\Delta G^{\circ}=178.3-1110\times 0.1606=0.0$ kJ/mol, so $K=1$ and the equilibrium pressure of the gas is indeed $1$ bar.

Why

This check ties the three results together: the crossover temperature is exactly the one where the constant equals one, which validates both the calculation of $\Delta G^{\circ}$ and that of $T$ at the same time.

The conclusion, written out

“The decomposition of limestone absorbs $178.3$ kJ/mol and gains $160.6$ J/(mol·K) of entropy: it is not spontaneous at $25$ °C, where $\Delta G^{\circ}$ is $+130.4$ kJ/mol and $K$ about $10^{-23}$ , but it becomes so above $837$ °C. By sweeping the kiln to keep the carbon dioxide at $0.010$ bar, this temperature falls to $623$ °C.”

The classic mistake on this problem: Writing $\Delta G^{\circ}=178.3-298.15\times 160.6$ and finding a very negative number, hence concluding that limestone decomposes spontaneously at room temperature. The world would then have no limestone cliffs, which is enough to reject the result.

Learn by heart

• $\Delta H$ in kJ/mol, $\Delta S$ in J/(mol·K): CONVERT before subtracting.
• $\Delta G=\Delta H-T\Delta S$ , with $T$ in KELVINS. Spontaneous if $\Delta G<0$ .
• $S^{\circ}$ is absolute and always POSITIVE, even for an element; only $\Delta H^{\circ}_{f}$ is zero there.
• The sign of $\Delta S$ is read first off the change in the number of moles of GAS.
• Four cases: two opposite signs give an answer valid at every temperature; two identical signs give a crossover temperature $T=\dfrac{\Delta H}{\Delta S}$ .
• $\Delta G^{\circ}=-RT\ln K$ : the MINUS sign makes $\Delta G^{\circ}<0$ equivalent to $K>1$ .
• Away from standard conditions: $\Delta G=\Delta G^{\circ}+RT\ln Q$ .
• Spontaneous does not mean fast, nor complete.

Frequently asked questions

Why must entropy be converted before calculating free energy?

Because enthalpies are tabulated in kilojoules per mole and entropies in joules per mole per kelvin. Subtracting directly amounts to mixing two units that differ by a factor of one thousand, which changes not only the value but the sign of the result, and so the conclusion about spontaneity.

Is the standard entropy of an element zero?

No, never. It is the enthalpy of formation of an element that is zero, by convention. Standard entropy is an absolute value measured from absolute zero, where every entropy vanishes. Gaseous oxygen, for example, has a standard entropy above two hundred joules per mole per kelvin.

How can the sign of the entropy change be predicted without calculation?

First look at the change in the number of moles of gas. If there is more gas on the product side, entropy increases. Solids and liquids weigh much less in the balance. With an equal number of gas moles, compare the complexity of the molecules: the more atoms a molecule has, the greater its entropy.

Is a spontaneous reaction fast?

Not at all. Thermodynamics only says that a reaction is possible, never how fast it happens. The combustion of diamond is very favourable at room temperature and yet never happens spontaneously, for lack of enough energy to cross the activation barrier. The rate belongs to kinetics.

When does a crossover temperature exist?

Only when the enthalpy and the entropy of reaction have the same sign. If the enthalpy is negative and the entropy positive, the reaction is spontaneous at every temperature; in the opposite case, it never is. The ratio then gives a negative number, which is how arithmetic signals that there is no crossover.

Why does a lime kiln work below the calculated temperature?

Because the calculated temperature assumes a carbon dioxide pressure of one bar, which corresponds to standard conditions. By ventilating the kiln, the gas is removed as it forms, which lowers the reaction quotient and the free energy with it. The decomposition becomes possible a couple of hundred degrees sooner.

Practise it

Corrected exercises: Chemical thermodynamics, 202-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-chem-sn2-thermodynamics. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: chemical thermodynamics (202-SN2-RE)

The essentials

The three quantities and their units

Predicting the sign of ΔS\Delta SΔS without calculation

The four cases, and the crossover temperature

The mistakes that cost marks

1. Subtracting joules from kilojoules in ΔG=ΔH−TΔS\Delta G=\Delta H-T\Delta SΔG=ΔH−TΔS

2. Setting the standard entropy of an element to zero

3. Leaving the temperature in degrees Celsius

4. Confusing spontaneous and fast

5. Looking for a crossover temperature where there is none

6. Using ΔG∘\Delta G^{\circ}ΔG∘ away from standard conditions

7. Counting total moles to predict the sign of ΔS\Delta SΔS

8. Believing that a negative system entropy forbids the reaction

9. Confusing a spontaneous reaction with a complete one

10. Forgetting to multiply by the coefficients in an entropy balance

11. Getting the sign wrong in the relation between ΔG∘\Delta G^{\circ}ΔG∘ and KKK

Which method to choose

Is the reaction spontaneous?

Which quantity to calculate, depending on the question

How the answer is expected to be written

Calculating spontaneity and the crossover temperature

Handling nonstandard conditions

Check before you hand in

The typical problem, taken apart

The lime kiln, and why it works well below the predicted temperature

Learn by heart

Frequently asked questions

Why must entropy be converted before calculating free energy?

Is the standard entropy of an element zero?

How can the sign of the entropy change be predicted without calculation?

Is a spontaneous reaction fast?

When does a crossover temperature exist?

Why does a lime kiln work below the calculated temperature?

Corrected exercises: Chemical thermodynamics, 202-SN2-RE

See also

Looking for a 202-SN2-RE tutor in Montreal?

Predicting the sign of $\Delta S$ without calculation

1. Subtracting joules from kilojoules in $\Delta G=\Delta H-T\Delta S$

6. Using $\Delta G^{\circ}$ away from standard conditions

7. Counting total moles to predict the sign of $\Delta S$

11. Getting the sign wrong in the relation between $\Delta G^{\circ}$ and $K$