General Chemistry 202-SN1-RE • CEGEP, Montreal

Revision sheet: thermochemistry and the first law (202-SN1-RE)

Q: Why is the heat of an exothermic reaction negative if the temperature goes up?

Because the sign is counted from the point of view of the system, that is, the reaction itself. The reaction gives off energy, so its heat is negative. It is the calorimeter that receives it, and it is the calorimeter whose temperature rises. The two quantities are opposites, which is written as heat of the reaction equals minus heat of the calorimeter.

Q: What is the difference between internal energy and enthalpy of a reaction?

Internal energy is what you measure at constant volume, in a bomb calorimeter, where no work is exchanged. Enthalpy is what you measure at constant pressure, in an open calorimeter, where the system can expand. The two differ by the product of the change in the number of moles of gas, the gas constant and the temperature.

Q: In what order should bond energies be subtracted?

Bonds broken minus bonds formed. Breaking a bond costs energy, forming one pays energy back. This is exactly the reverse of the order for enthalpies of formation, where you write products minus reactants, and it is the most frequent source of sign reversal in this chapter.

Q: Why is the enthalpy of formation of an element zero?

Because it measures the energy needed to form a compound from its elements in their standard state. For an element already in that state, there is nothing to form. Beware, however: only the most stable form counts, so graphite carbon and not diamond, dioxygen and not ozone.

Q: Do I need to account for the heat capacity of the calorimeter?

Yes, as soon as the problem supplies it. The container, the stirrer and the thermometer also absorb heat, and their contribution is added to that of the contents. Leaving it out systematically underestimates the energy released. As always, a piece of data supplied and not used signals a missing term.

Q: What can be changed in an equation when applying Hess's law?

You can reverse it, which changes the sign of its enthalpy, and multiply it by any factor, which multiplies the enthalpy by the same factor. You must then check that the sum of the equations reproduces the target equation exactly, with all the intermediate species cancelling.

Thermochemistry measures energy transfers, and it always does so from the point of view of the system. This single convention, applied without exception, settles half of the difficulties of the chapter: an exothermic reaction gives off heat, so its $q$ is negative, even though the thermometer goes up.

This sheet fixes the conventions, separates internal energy from enthalpy, and brings together the three methods for calculating an enthalpy of reaction, with the traps specific to each: Hess's law, enthalpies of formation, bond energies.

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The thread of the chapter

The whole chapter rests on a single sign convention: $q$ and $w$ are counted from the point of view of the SYSTEM. The heat the calorimeter receives is the heat the reaction gave off, and going from one to the other is a simple change of sign that nonetheless costs half of the marks lost.

This chapter is part of 202-SN1-RE, General Chemistry

The essentials

The conventions, from the system's point of view

• $q>0$ if the system RECEIVES heat, $q<0$ if it gives some off. An exothermic reaction therefore has $q<0$ , even though the thermometer goes up.
• $w>0$ if the system RECEIVES work, that is, if it is compressed. Pressure-volume work: $w=-P\,\Delta V$ , and $1$ kPa·L $=1$ J.
• First law: $\Delta U=q+w$ .
• At constant VOLUME: $\Delta V=0$ , so $w=0$ and $q_{V}=\Delta U$ . This is what a bomb calorimeter measures.
• At constant PRESSURE: $q_{P}=\Delta H$ . Link between the two: $\Delta H=\Delta U+\Delta n_{\text{gas}}RT$ , where $\Delta n_{\text{gas}}$ counts ONLY gaseous species.

The arrow ALWAYS goes from the reactants to the products. When it points down, the system has given off energy and

\Delta H

is negative; when it points up, it has absorbed energy.

The heat measured by a calorimeter is the heat it RECEIVES. That of the reaction is the opposite: $q_{\text{reaction}}=-q_{\text{calorimeter}}$ . Writing this line in black and white saves many lost signs.

The three ways to calculate an enthalpy of reaction

• Enthalpies of FORMATION: $\Delta H^{\circ}=\sum n\Delta H_{f}^{\circ}(\text{products})-\sum n\Delta H_{f}^{\circ}(\text{reactants})$ . It is ZERO for an element in its standard state.
• HESS's law: add known equations, reversing those that need it (the sign changes) and multiplying those that need it (the enthalpy is multiplied too).
• Bond ENERGIES: $\Delta H\approx\sum E(\text{broken})-\sum E(\text{formed})$ . Breaking costs, forming pays back. Valid only for GASEOUS species, and only as an estimate.
• The physical state is part of the data: $H_{2}O(l)$ and $H_{2}O(g)$ differ by the enthalpy of vaporization, $44$ kJ/mol.
• Calorimetry: $q=mc\Delta T$ for the contents, PLUS $C_{cal}\Delta T$ for the calorimeter itself when its heat capacity is given.

Both paths start from the elements, whose enthalpy is zero:

\Delta H^{\circ}

is the gap between the products and the reactants, hence “products minus reactants”. Drawn to scale.

The three methods should give the same result, up to the approximation of bond energies. When a problem provides two sets of data, it often expects both calculations and their comparison.

Bomb calorimeter or constant pressure

• Bomb calorimeter: RIGID container, so $\Delta V=0$ , $w=0$ and the measurement gives $\Delta U$ .
• Open calorimeter, an insulated cup in open air: the pressure is constant, so the measurement directly gives $\Delta H$ .
• $\Delta H=\Delta U+\Delta n_{\text{gas}}RT$ , with $R=8.314\times 10^{-3}$ kJ/(mol·K) if the enthalpies are in kilojoules.
• $\Delta n_{\text{gas}}$ is calculated from the coefficients of the GASES only: a solid or a liquid never enters it.
• The gap between $\Delta U$ and $\Delta H$ is a few kilojoules per mole, often less than $1\%$ of an enthalpy of combustion, but it is required when the problem asks for it.

The bomb is RIGID: no work, and the measurement gives

\Delta U

. The cup is open to the air, at constant pressure: the measurement directly gives

\Delta H

The word “bomb” in a problem always signals constant volume, hence $\Delta U$ . The words “cup”, “coffee cup” or “constant-pressure calorimeter” signal $\Delta H$ .

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. Confusing the heat received by the calorimeter with that of the reaction

the whole question, 5 marks, and a sign that reverses the conclusion

What not to write

“The calorimeter absorbed $22.7$ kJ, so $\Delta H$ for the reaction is $+22.7$ kJ.”

What to write

“ $q_{\text{reaction}}=-q_{\text{calorimeter}}=-22.7$ kJ. The reaction GAVE OFF this energy, it is exothermic.”

The arrow goes from the reaction to the water: what the calorimeter gains, the reaction loses. Same number, opposite sign.

Why: The calorimeter and the reaction are two systems exchanging the same energy in opposite directions. A combustion that absorbed heat would be physically absurd anyway.

2. Reversing the subtraction of bond energies

the whole question, 5 marks, with the sign reversed

What not to write

“ $\Delta H=\sum E(\text{formed})-\sum E(\text{broken})$ .”

What to write

“ $\Delta H=\sum E(\text{BROKEN})-\sum E(\text{FORMED})$ : breaking COSTS energy, forming PAYS energy back.”

Breaking the bonds goes UP to the isolated atoms, forming bonds comes back down:

\Delta H=2500-3200=-700

kJ, an exothermic reaction. Drawn to scale.

Why: The order is the reverse of that for enthalpies of formation, which explains the confusion. The foolproof check: a combustion must come out exothermic, hence negative.

3. Forgetting the physical state in an enthalpy of formation

3 marks per mole of water, often 6 marks on the question

What not to write

“The combustion of methane produces water: $\Delta H_{f}^{\circ}(H_{2}O)=-241.8$ kJ/mol.”

What to write

“ $-241.8$ kJ/mol is the value for $H_{2}O(g)$ . If the problem writes $H_{2}O(l)$ , it is $-285.8$ kJ/mol, a difference of $44.0$ kJ/mol per mole of water.”

Liquid water sits

44.0

kJ/mol below gaseous water: taking one for the other throws the enthalpy off by

44

kJ per mole of water formed.

Why: The difference is exactly the enthalpy of vaporization. On a combustion producing two moles of water, the error reaches $88$ kJ/mol, which is far from negligible.

4. Multiplying an equation without multiplying its enthalpy

the whole question, 6 marks

What not to write

“I double the equation to balance the Hess cycle, but $\Delta H$ stays the same.”

What to write

“ $\Delta H$ is an EXTENSIVE quantity: doubling the equation doubles its enthalpy, and reversing it changes its sign.”

Why: The enthalpy of reaction refers to the equation as it is written. Twice as much matter transformed releases twice as much energy, which is also why the unit is the kilojoule per mole of reaction.

5. Forgetting the heat capacity of the calorimeter

3 marks, and an enthalpy that is systematically underestimated

What not to write

“ $q=mc\Delta T$ for the $200$ g of water, and that is all.”

What to write

“You must add $C_{cal}\Delta T$ : the container, the thermometer and the stirrer warm up too, and their contribution is given in the problem.”

Water is not alone in warming up: the container, the thermometer and the stirrer also receive heat, and the term

C_{cal}\Delta T

counts it.

Why: Any heat capacity supplied in the problem must be used. An unused piece of data is almost always the sign of a term forgotten in the balance.

6. Counting solids and liquids in $\Delta n_{\text{gas}}$

3 marks, and a wrong sign on the correction

What not to write

“ $CH_{3}OH(l)+\tfrac{3}{2}O_{2}(g)\rightarrow CO_{2}(g)+2H_{2}O(l)$ : $\Delta n=3-2.5=0.5$ .”

What to write

“Only GASES count: $\Delta n_{\text{gas}}=1-1.5=-0.5$ . The liquid methanol and the liquid water do not enter this count.”

Only the boxes of gas count: three molecules of dioxygen give two of carbon dioxide. The methanol and the water, being liquids, stay out of the count.

Why: The term $\Delta n_{\text{gas}}RT$ comes from the pressure-volume work, and only gases have a volume that changes appreciably. The volume of condensed phases is negligible next to that of gases.

7. Giving a nonzero enthalpy of formation to an element in its standard state

2 marks, and a table looked up for nothing

What not to write

“ $\Delta H_{f}^{\circ}(O_{2})$ is small but nonzero, you have to look it up in the table.”

What to write

“It is EXACTLY zero by definition, for any element in its standard state: $O_{2}(g)$ , $N_{2}(g)$ , $C(\text{graphite})$ , $Br_{2}(l)$ , $Hg(l)$ .”

Why: The enthalpy of formation measures the formation of a compound FROM its elements: for an element, there is nothing to form. Beware, however: $C(\text{diamond})$ and $O_{3}(g)$ are NOT standard states, and their enthalpy is not zero.

8. Getting the sign of pressure-volume work wrong

2 marks, and a wrong

\Delta U

What not to write

“The gas expands, it does work, so $w=+P\Delta V$ .”

What to write

“ $w=-P\Delta V$ : an expansion gives $\Delta V>0$ , so $w<0$ , which correctly shows that the system GIVES UP work.”

The gas expands and pushes the piston:

\Delta V>0

, it GIVES UP work, so

w=-P\Delta V

is negative.

Why: Every quantity is counted from the system's point of view. The minus sign is not an arbitrary convention: it ensures that $w$ is negative when the system works against its surroundings.

9. Mixing joules and kilojoules in the same balance

3 marks, and a factor of

1000

What not to write

“ $q=mc\Delta T=4.18\times 200\times 3.5=2926$ kJ.”

What to write

“ $c=4.18$ J/(g·°C), so $q=2926$ J $=2.93$ kJ. Enthalpies of reaction, on the other hand, are tabulated in kJ/mol.”

Why: Specific heat is in joules, enthalpies of formation in kilojoules per mole. Writing the unit on every line is the only protection against this mix-up, which is the most frequent one in the chapter.

10. Adding Hess equations without checking what cancels

the whole question, 6 marks

What not to write

“I add the three given equations and add the three enthalpies, without checking whether the intermediate species disappear.”

What to write

“First adjust the equations so that ALL the intermediate species cancel, and only then add the enthalpies so adjusted.”

Both paths start from the same state and arrive at the same state: their enthalpies are therefore rigorously equal. This is the whole content of Hess's law, and what allows you to reverse and multiply equations.

Why: The sum of the equations must reproduce the target equation EXACTLY, no more and no less. A species that remains on both sides with different coefficients signals a wrong factor.

Which method to choose

Which method for calculating an enthalpy of reaction

The data supplied by the problem

If a table of standard enthalpies of FORMATION → $\Delta H^{\circ}=\sum n\Delta H_{f}^{\circ}(\text{products})-\sum n\Delta H_{f}^{\circ}(\text{reactants})$

Example: combustion of methanol, $-726$ kJ/mol
If several EQUATIONS with their enthalpy → Hess's law: reverse, multiply, add until you reproduce the target equation

Example: three combustions to obtain an enthalpy of formation
If a table of BOND ENERGIES → $\sum E(\text{broken})-\sum E(\text{formed})$ , as an estimate only, and on gaseous species

Example: estimate for a gas-phase reaction
If a temperature measurement in a calorimeter → $q=mc\Delta T+C_{cal}\Delta T$ , then $q_{\text{reaction}}=-q_{\text{calorimeter}}$ , then division by the amount of substance

Example: a bomb gives $\Delta U$ , a cup gives $\Delta H$
If a lattice enthalpy and an enthalpy of hydration → $\Delta H_{\text{solution}}=E_{\text{lattice}}+\Delta H_{\text{hydration}}$ , the difference of two large opposite numbers

this is why a dissolution can be exothermic or endothermic depending on the salt

If two sets of data are supplied, the problem almost always expects both calculations AND their comparison. A gap between a calorimetric value and a tabulated value is commented on, not hidden.

$\Delta U$ or $\Delta H$

The apparatus described in the problem

If "bomb calorimeter", "rigid container", "at constant volume" → the measurement gives $\Delta U$ ; to get $\Delta H$ use $\Delta H=\Delta U+\Delta n_{\text{gas}}RT$

Example: combustion of a food or a fuel
If "insulated cup", "open to the air", "at constant pressure" → the measurement gives $\Delta H$ directly, no correction

Example: acid-base neutralization, dissolution of a salt
If the reaction involves no gas → $\Delta n_{\text{gas}}=0$ , so $\Delta H=\Delta U$ exactly

Example: precipitation in aqueous solution
If the problem supplies a table of enthalpies of formation → the values are $\Delta H^{\circ}$ values, hence at constant pressure, at $25$ °C

comparing them with a measurement made in a bomb requires the correction $\Delta n_{\text{gas}}RT$

The gap between $\Delta U$ and $\Delta H$ rarely exceeds a few kilojoules per mole. It is nonetheless required as soon as the problem explicitly asks for one after measuring the other.

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.

Working up a calorimetric measurement

When to use it: The problem gives a change in temperature and asks for an enthalpy or an internal energy of reaction.

1 Write the balance of the CALORIMETER: $q_{cal}=mc\Delta T+C_{cal}\Delta T$ , including every term supplied.
2 Go to the reaction by the change of sign, explicitly writing $q_{\text{reaction}}=-q_{cal}$ .
3 Divide by the amount of substance actually transformed to get a MOLAR value, with its unit in kJ/mol.
4 Identify the apparatus: a rigid bomb gives $\Delta U$ , an open calorimeter gives $\Delta H$ .
5 If needed, correct with $\Delta H=\Delta U+\Delta n_{\text{gas}}RT$ , counting only the gases.
6 Conclude with the sign interpreted: exothermic if negative, endothermic if positive.

Concluding sentence

“The calorimeter received $q=C_{cal}\Delta T=10.4\times 2.18=22.7$ kJ, so the reaction gave off $-22.7$ kJ for $1.00$ g, that is $\Delta U=-726$ kJ/mol; since $\Delta n_{\text{gas}}=-0.5$ , $\Delta H=-728$ kJ/mol.”

The trap: Forgetting to express the value per mole. The raw result is in kilojoules for the sample burned, never in kilojoules per mole, and it is the molar value that can be compared with a table.

Marking: Typically 2 marks for the calorimeter balance, 1 for the change of sign, 2 for the molar value, 2 for the correction, 1 for the interpretation.

Writing up a Hess's law calculation

When to use it: The problem supplies several equations with their enthalpy and asks for the enthalpy of a target equation.

1 Write the TARGET equation at the top of the page and note where each species must end up, as a reactant or as a product.
2 For each equation supplied, decide whether it must be reversed and by what factor it must be multiplied, noting this beside it.
3 Rewrite the transformed equations one under the other, changing the sign of the enthalpy of those that were reversed and multiplying those that were multiplied.
4 Add them up and CHECK that all the intermediate species cancel, crossing them out one by one.
5 Add the transformed enthalpies and conclude.

Concluding sentence

“By reversing the second equation, which changes the sign of its enthalpy, and multiplying the third by two, which doubles its own, the sum reproduces the target equation exactly and gives $\Delta H=-726$ kJ/mol.”

The trap: Adding the enthalpies before checking the cancellation. An intermediate species that remains signals a wrong factor, and the calculated enthalpy then has no meaning at all.

Check before you hand in

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

The sign against the physical meaning

A combustion, a neutralization and most precipitations are exothermic: their enthalpy must be negative. A positive result calls for an explanation.

A combustion found at $+726$ kJ/mol: the sign of the calorimetric balance was lost.
The molar value against a known order of magnitude

The enthalpy of combustion of a simple organic compound is counted in hundreds of kilojoules per mole; a neutralization is worth about $-57$ kJ/mol.

A $\Delta H=-22.7$ kJ/mol for a combustion: the division by the amount of substance was not done.
The intermediate species crossed out one by one

In a Hess calculation, go back over the sum of the equations and check that each intermediate appears as many times on the left as on the right.

A $CO(g)$ that remains in the final sum: one of the multiplying factors is wrong.
The agreement between two methods

When the problem allows two calculations, compare them. A gap of a few percent is normal with bond energies; a gap of more than $10\%$ signals an error.

A calorimetric value of $-728$ kJ/mol against $-726$ kJ/mol from the enthalpies of formation: the agreement is excellent.
The units written on every line

Go back over the calculation and check that the joules of the specific heat were converted to kilojoules before any comparison with a table.

A $q=2926$ reported in kilojoules when the specific heat was in joules per gram.

The typical problem, taken apart

The combustion of methanol, by two independent routes

We burn $1.00$ g of methanol $CH_{3}OH(l)$ in a bomb calorimeter with a heat capacity of $10.4$ kJ/°C. The temperature rises by $2.18$ °C.

Determine the molar internal energy of combustion, then the corresponding molar enthalpy. Then compare with the value obtained from the standard enthalpies of formation.

Data: $M(CH_{3}OH)=32.04$ g/mol; $\Delta H_{f}^{\circ}$ in kJ/mol: $CH_{3}OH(l)=-238.7$ , $CO_{2}(g)=-393.5$ , $H_{2}O(l)=-285.8$ , $O_{2}(g)=0$ ; $R=8.314\times 10^{-3}$ kJ/(mol·K); $T=298.15$ K.

One gram of methanol burns in the rigid bomb; the thermometer in the water bath rises by

2.18

°C.

Step 1

Equation: $CH_{3}OH(l)+\tfrac{3}{2}O_{2}(g)\rightarrow CO_{2}(g)+2\,H_{2}O(l)$ .

Why

The physical states are written from the first line: both the count of gases and the choice of enthalpies of formation in the table will depend on them.

Step 2

Heat received by the calorimeter: $q_{cal}=C_{cal}\Delta T=10.4\times 2.18=22.7$ kJ. So $q_{\text{reaction}}=-22.7$ kJ for $1.00$ g.

Why

The change of sign is written out explicitly, on its own line. It is the only way not to carry a positive value through a combustion, which is exothermic by nature.

Step 3

$n=\dfrac{1.00}{32.04}=0.03121$ mol, so $\Delta U=\dfrac{-22.67}{0.03121}=-726.4$ kJ/mol.

Why

Dividing by the amount of substance turns a measurement on a sample into a molar quantity, the only kind that can be compared with a table.

Step 4

The bomb is RIGID, so $\Delta V=0$ , $w=0$ and the measurement does give $\Delta U$ , not $\Delta H$ .

Why

This sentence justifies the next step. Without it, the correction $\Delta n_{\text{gas}}RT$ would look like a formula applied at random.

Step 5

$\Delta n_{\text{gas}}=1-1.5=-0.5$ : only carbon dioxide and dioxygen count, the methanol and the water being liquids. So $\Delta H=\Delta U+\Delta n_{\text{gas}}RT=-726.4-0.5(8.314\times 10^{-3})(298.15)=-727.6$ kJ/mol.

Why

The count of gases is exactly where students add the condensed phases. Writing it out species by species, with its state, makes the error impossible.

Step 6

From the enthalpies of formation: $\Delta H^{\circ}=[(-393.5)+2(-285.8)]-[(-238.7)+\tfrac{3}{2}(0)]=-965.1+238.7=-726.4$ kJ/mol.

Why

The order products minus reactants is the reverse of that for bond energies: writing the two brackets separately, with their signs, avoids swapping them.

Step 7

Comparison: $-727.6$ kJ/mol by calorimetry against $-726.4$ kJ/mol from the tables, a difference of $1.2$ kJ/mol, that is $0.2\%$ .

Why

Two completely independent methods converge: this is the best possible validation of the calculation, and commenting on the gap is almost always the last mark of the scheme.

Step 8

Sign check: the combustion is exothermic, $\Delta H<0$ , and the gap between $\Delta U$ and $\Delta H$ is negative since $\Delta n_{\text{gas}}$ is.

Why

The direction of the correction can be predicted before calculating it: the reaction consumes more gas than it produces, so the surroundings do work on the system, and the enthalpy is more negative than the internal energy.

The conclusion, written out

“The combustion of methanol releases $726$ kJ per mole as internal energy and $728$ kJ per mole as enthalpy; the value calculated from the standard enthalpies of formation, $-726.4$ kJ/mol, confirms the calorimetric measurement to within $0.2\%$ .”

The classic mistake on this problem: Reporting $\Delta H=+22.7$ kJ. The sign comes from the confusion between the calorimeter, which receives, and the reaction, which gives off; and the missing division by the amount of substance also makes the result thirty-two times too small.

Learn by heart

• $q>0$ if the system RECEIVES; an exothermic reaction has $q<0$ .
• $w=-P\Delta V$ , and $1$ kPa·L $=1$ J. $\Delta U=q+w$ .
• Constant volume: $q_{V}=\Delta U$ . Constant pressure: $q_{P}=\Delta H$ .
• $\Delta H=\Delta U+\Delta n_{\text{gas}}RT$ , with $\Delta n_{\text{gas}}$ counted on GASES only.
• $q_{\text{reaction}}=-q_{\text{calorimeter}}$ , and $q_{cal}=mc\Delta T+C_{cal}\Delta T$ .
• Formation: products MINUS reactants. Bonds: broken MINUS formed. The two orders are the reverse of each other.
• $\Delta H_{f}^{\circ}=0$ for an element in its STANDARD STATE, and the physical state is always to be specified.
• Hess: reversing changes the sign, multiplying multiplies the enthalpy.

Frequently asked questions

Why is the heat of an exothermic reaction negative if the temperature goes up?

Because the sign is counted from the point of view of the system, that is, the reaction itself. The reaction gives off energy, so its heat is negative. It is the calorimeter that receives it, and it is the calorimeter whose temperature rises. The two quantities are opposites, which is written as heat of the reaction equals minus heat of the calorimeter.

What is the difference between internal energy and enthalpy of a reaction?

Internal energy is what you measure at constant volume, in a bomb calorimeter, where no work is exchanged. Enthalpy is what you measure at constant pressure, in an open calorimeter, where the system can expand. The two differ by the product of the change in the number of moles of gas, the gas constant and the temperature.

In what order should bond energies be subtracted?

Bonds broken minus bonds formed. Breaking a bond costs energy, forming one pays energy back. This is exactly the reverse of the order for enthalpies of formation, where you write products minus reactants, and it is the most frequent source of sign reversal in this chapter.

Why is the enthalpy of formation of an element zero?

Because it measures the energy needed to form a compound from its elements in their standard state. For an element already in that state, there is nothing to form. Beware, however: only the most stable form counts, so graphite carbon and not diamond, dioxygen and not ozone.

Do I need to account for the heat capacity of the calorimeter?

Yes, as soon as the problem supplies it. The container, the stirrer and the thermometer also absorb heat, and their contribution is added to that of the contents. Leaving it out systematically underestimates the energy released. As always, a piece of data supplied and not used signals a missing term.

What can be changed in an equation when applying Hess's law?

You can reverse it, which changes the sign of its enthalpy, and multiply it by any factor, which multiplies the enthalpy by the same factor. You must then check that the sum of the equations reproduces the target equation exactly, with all the intermediate species cancelling.

Practise it

Corrected exercises: Thermochemistry and the first law, 202-SN1-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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Revision sheet: thermochemistry and the first law (202-SN1-RE)

The essentials

The conventions, from the system's point of view

The three ways to calculate an enthalpy of reaction

Bomb calorimeter or constant pressure

The mistakes that cost marks

1. Confusing the heat received by the calorimeter with that of the reaction

2. Reversing the subtraction of bond energies

3. Forgetting the physical state in an enthalpy of formation

4. Multiplying an equation without multiplying its enthalpy

5. Forgetting the heat capacity of the calorimeter

6. Counting solids and liquids in Δngas\Delta n_{\text{gas}}Δngas​

7. Giving a nonzero enthalpy of formation to an element in its standard state

8. Getting the sign of pressure-volume work wrong

9. Mixing joules and kilojoules in the same balance

10. Adding Hess equations without checking what cancels

Which method to choose

Which method for calculating an enthalpy of reaction

ΔU\Delta UΔU or ΔH\Delta HΔH

How the answer is expected to be written

Working up a calorimetric measurement

Writing up a Hess's law calculation

Check before you hand in

The typical problem, taken apart

The combustion of methanol, by two independent routes

Learn by heart

Frequently asked questions

Why is the heat of an exothermic reaction negative if the temperature goes up?

What is the difference between internal energy and enthalpy of a reaction?

In what order should bond energies be subtracted?

Why is the enthalpy of formation of an element zero?

Do I need to account for the heat capacity of the calorimeter?

What can be changed in an equation when applying Hess's law?

Corrected exercises: Thermochemistry and the first law, 202-SN1-RE

See also

Looking for a 202-SN1-RE chemistry tutor in Montreal?

6. Counting solids and liquids in $\Delta n_{\text{gas}}$

$\Delta U$ or $\Delta H$