Thermochemistry | IB - Wyatt's Notes

Intuition

Thermochemistry is like tracking energy bookkeeping — every bond broken costs energy, every bond formed releases energy: Enthalpy changes determine whether reactions release heat (exothermic) or absorb it (endothermic)

Why it matters: Thermochemistry guides fuel selection, industrial process design, and understanding metabolic energy flow

The key insight: Enthalpy changes determine whether reactions release heat (exothermic) or absorb it (endothermic)

Enthalpy Changes

Definitions

Enthalpy ( $H$ ) is the heat content of a system at constant pressure.

An enthalpy change ( $\Delta H$ ) is the heat energy exchanged with the surroundings during a Reaction at constant pressure.

Sign Convention

Type	Sign	Energy Flow
Exothermic	$\Delta H \lt 0$	System releases heat to surroundings
Endothermic	$\Delta H \gt 0$	System absorbs heat from surroundings

Standard Conditions

Standard enthalpy changes are measured under standard conditions:

Pressure: $100\mathrm{ kPa}$ (IB standard)
Concentration: $1\mathrm{ mol/L}$ for solutions
Temperature: $298\mathrm{ K}$ ( $25\degree\mathrm{C}$ )
All substances in their standard states

Types of Enthalpy Change

Type	Symbol	Definition
Standard enthalpy of formation	$\Delta H_f^\circ$	Enthalpy change when 1 mol of compound forms from its elements in standard states
Standard enthalpy of combustion	$\Delta H_c^\circ$	Enthalpy change when 1 mol of substance burns completely in oxygen
Standard enthalpy of neutralisation	$\Delta H_{\mathrm{neut}}^\circ$	Enthalpy change when 1 mol of water forms from acid-base reaction
Standard enthalpy of atomisation	$\Delta H_{\mathrm{at}}^\circ$	Enthalpy change to form 1 mol of gaseous atoms from element in standard state

Exam Tip $\Delta H_f^\circ$ for an element in its standard state is always zero (by definition). For example, $\Delta H_f^\circ$ of O $_2$ (g) = 0, $\Delta H_f^\circ$ of C(graphite) = 0.

Calorimetry

Principle

Calorimetry measures the heat exchanged during a chemical or physical process.

Specific Heat Capacity

The amount of energy required to raise the temperature of $1\mathrm{ g}$ of a substance by $1\degree\mathrm{C}$ :

Q = mc\Delta T

Where:

$q$ = heat energy (J)
$m$ = mass (g)
$c$ = specific heat capacity (J/g/ $\degree$ C)
$\Delta T$ = temperature change ( $\degree$ C)

Substance	$c$ (J/g/ $\degree$ C)
Water	4.18
Ice	2.09
Aluminium	0.90
Copper	0.39
Iron	0.45

Measuring Enthalpy of Reaction

For a reaction in solution:

\Delta H = -\frac{mc\Delta T}{n}

The negative sign accounts for the convention: heat lost by the reaction is gained by the solution.

Assumptions in Calorimetry

No heat loss to the surroundings (use a calorimeter).
The calorimeter itself has negligible heat capacity.
The solution has the same density and specific heat capacity as water.

Example $50.0\mathrm{ mL}$ of $1.0\mathrm{ M}$ HCl is mixed with $50.0\mathrm{ mL}$ of $1.0\mathrm{ M}$ NaOH In a calorimeter. The temperature rises from $21.0\degree\mathrm{C}$ to $27.5\degree\mathrm{C}$ . Calculate the enthalpy of neutralisation.

M = 100.0\mathrm{ g} \mathrm{ (assuming density of water)}

Q = mc\Delta T = 100.0 \times 4.18 \times 6.5 = 2717\mathrm{ J} = 2.717\mathrm{ kJ}

N(\mathrm{H}_2\mathrm{O}) = 0.050\mathrm{ mol} \mathrm{ (limited by the reagent volumes)}

\Delta H = -\frac{2.717}{0.050} = -54.3\mathrm{ kJ/mol}

Bomb Calorimetry

Used for combustion reactions. The calorimeter constant $C$ accounts for the heat absorbed by the Calorimeter:

Q_{\mathrm{reaction}} = -(mc\Delta T + C\Delta T)

Hess”s Law

Statement

The total enthalpy change for a reaction is independent of the route taken. Only on the Initial and final states.

\Delta H_{\mathrm{total}} = \Delta H_1 + \Delta H_2 + \cdots

Using Enthalpy Cycles

To find an unknown enthalpy change, construct a cycle with known enthalpy changes and apply Hess’s Law.

Using Standard Enthalpies of Formation

\Delta H_r^\circ = \sum \Delta H_f^\circ(\mathrm{products}) - \sum \Delta H_f^\circ(\mathrm{reactants})

Example Calculate $\Delta H_r^\circ$ for: CH $_4$ (g) + 2O $_2$ (g) $\to$ CO $_2$ (g) + 2H $_2$ O(l)

Given:

$\Delta H_f^\circ$ (CH $_4$ ) = $-74.8\mathrm{ kJ/mol}$
$\Delta H_f^\circ$ (CO $_2$ ) = $-393.5\mathrm{ kJ/mol}$
$\Delta H_f^\circ$ (H $_2$ O) = $-285.8\mathrm{ kJ/mol}$

\Delta H_r^\circ = [(-393.5) + 2(-285.8)] - [(-74.8) + 2(0)]

= (-393.5 - 571.6) - (-74.8) = -965.1 + 74.8 = -890.3\mathrm{ kJ/mol}

Using Standard Enthalpies of Combustion

\Delta H_r^\circ = \sum \Delta H_c^\circ(\mathrm{reactants}) - \sum \Delta H_c^\circ(\mathrm{products})

Note the reversed order compared to formation enthalpies.

Bond Enthalpies

Definition

The bond enthalpy (bond dissociation energy) is the enthalpy change when one mole of covalent Bonds in the gas phase is broken.

Average Bond Enthalpies

Since bond enthalpies vary slightly depending on the molecular environment, tables give average Values.

Bond	Average Enthalpy (kJ/mol)
C—C	347
C=C	612
C $\equiv$ C	838
C—H	413
C—O	358
C=O	743
O—H	463
O=O	495
H—H	436
N $\equiv$ N	945
N—H	391

Calculating Enthalpy Change from Bond Enthalpies

\Delta H = \sum (\mathrm{bonds broken}) - \sum (\mathrm{bonds formed})

Bonds broken (positive — energy absorbed) and bonds formed (negative — energy released).

Exam Tip Bond enthalpy calculations give approximate values because average bond enthalpies are used. Values From Hess’s law with formation data are more accurate. Bond enthalpy calculations only apply to Gases.

Born-Haber Cycles

Purpose

Born-Haber cycles determine the lattice energy of an ionic compound using a thermodynamic cycle.

Lattice Energy

The lattice energy $\Delta H_{\mathrm{latt}}$ is the enthalpy change when one mole of an ionic Compound is formed from its gaseous ions.

Steps in a Born-Haber Cycle

For an ionic compound MX:

$\Delta H_f^\circ$ : Standard enthalpy of formation of MX(s)
$\Delta H_{\mathrm{at}}(M)$ : Enthalpy of atomisation of M(s) $\to$ M(g)
$\Delta H_{\mathrm{at}}(X_2)$ : Enthalpy of atomisation of $\frac{1}{2}$ X $_2$ (g) $\to$ X(g)
IE $_1$ IE $_2$ …: Ionisation energies of M
EA $_1$ : Electron affinity of X (energy released when X gains an electron)
$\Delta H_{\mathrm{latt}}$ : Lattice energy (exothermic)

Applying Hess’s Law

\Delta H_f^\circ = \Delta H_{\mathrm{at}}(M) + \frac{1}{2}\Delta H_{\mathrm{at}}(X_2) + \mathrm{IE} + \mathrm{EA} + \Delta H_{\mathrm{latt}}

Example Calculate the lattice energy of NaCl.

Given:

$\Delta H_f^\circ$ (NaCl) = $-411\mathrm{ kJ/mol}$
$\Delta H_{\mathrm{at}}$ (Na) = $+108\mathrm{ kJ/mol}$
$\frac{1}{2}\Delta H_{\mathrm{at}}$ (Cl $_2$ ) = $+122\mathrm{ kJ/mol}$
IE $_1$ (Na) = $+496\mathrm{ kJ/mol}$
EA $_1$ (Cl) = $-349\mathrm{ kJ/mol}$

\Delta H_{\mathrm{latt}} = \Delta H_f^\circ - \Delta H_{\mathrm{at}}(\mathrm{Na}) - \frac{1}{2}\Delta H_{\mathrm{at}}(\mathrm{Cl}_2) - \mathrm{IE}_1 - \mathrm{EA}_1

= -411 - 108 - 122 - 496 - (-349) = -411 - 108 - 122 - 496 + 349 = -788\mathrm{ kJ/mol}

Factors Affecting Lattice Energy

Factor	Effect
Larger ionic charges	Higher lattice energy
Smaller ionic radii	Higher lattice energy
Higher charge density	Higher lattice energy

Entropy

Definition

Entropy ( $S$ ) is a measure of the disorder or randomness of a system.

Factors Affecting Entropy

Factor	Effect on Entropy
More particles	Higher entropy
Higher temperature	Higher entropy
Gas $\to$ liquid $\to$ solid	Decreasing entropy
Dissolving solid in solvent	Increasing entropy

Standard Entropy Change

\Delta S^\circ = \sum S^\circ(\mathrm{products}) - \sum S^\circ(\mathrm{reactants})

Example Calculate $\Delta S^\circ$ for: CaCO $_3$ (s) $\to$ CaO(s) + CO $_2$ (g)

Given: $S^\circ$ (CaCO $_3$ ) = $92.9\mathrm{ J/(mol}\cdot\mathrm{K)}$ , $S^\circ$ (CaO) = $39.7\mathrm{ J/(mol}\cdot\mathrm{K)}$ , $S^\circ$ (CO $_2$ ) = $213.7\mathrm{ J/(mol}\cdot\mathrm{K)}$ .

\Delta S^\circ = (39.7 + 213.7) - 92.9 = 253.4 - 92.9 = 160.5\mathrm{ J/(mol}\cdot\mathrm{K)}

The positive $\Delta S$ is expected because a gas is produced from a solid.

Gibbs Free Energy

Definition

Gibbs free energy combines enthalpy and entropy to predict spontaneity:

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

Where $T$ is the temperature in Kelvin.

Spontaneity

$\Delta G$	Spontaneity
$\Delta G \lt 0$	Spontaneous (thermodynamically favourable)
$\Delta G = 0$	At equilibrium
$\Delta G \gt 0$	Non-spontaneous

Standard Gibbs Free Energy Change

\Delta G^\circ = \sum \Delta G_f^\circ(\mathrm{products}) - \sum \Delta G_f^\circ(\mathrm{reactants})

Relationship to Equilibrium Constant

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

Where $R = 8.314\mathrm{ J/(mol}\cdot\mathrm{K)}$ and $K$ is the equilibrium constant.

Temperature Dependence

A reaction that is non-spontaneous at low temperature may become spontaneous at high temperature if $\Delta S \gt 0$ (and vice versa).

$\Delta H$	$\Delta S$	Spontaneous when
Negative	Positive	Always spontaneous
Negative	Negative	Low temperature
Positive	Positive	High temperature
Positive	Negative	Never spontaneous

Example For the reaction: CaCO $_3$ (s) $\to$ CaO(s) + CO $_2$ (g)

$\Delta H = +178\mathrm{ kJ/mol}$ , $\Delta S = +160.5\mathrm{ J/(mol}\cdot\mathrm{K)}$

Find the temperature at which the reaction becomes spontaneous.

\Delta G = 0 \mathrm{ when } T = \frac{\Delta H}{\Delta S} = \frac{178000}{160.5} = 1109\mathrm{ K}

The reaction is spontaneous above $1109\mathrm{ K}$ .

IB Exam-Style Questions

Question 1 (Paper 1 style)

Using bond enthalpies, calculate $\Delta H$ for: H $_2$ (g) + Cl $_2$ (g) $\to$ 2HCl(g)

Bonds broken: H—H ( $436$ ) + Cl—Cl ( $242$ ) $= 678\mathrm{ kJ/mol}$

Bonds formed: $2 \times$ H—Cl ( $431$ ) $= 862\mathrm{ kJ/mol}$

\Delta H = 678 - 862 = -184\mathrm{ kJ/mol}

Question 2 (Paper 2 style)

$25.0\mathrm{ cm}^3$ of $1.0\mathrm{ M}$ HCl is added to $25.0\mathrm{ cm}^3$ of $1.0\mathrm{ M}$ NaOH in a polystyrene cup. The temperature increases from $20.0\degree\mathrm{C}$ to $26.5\degree\mathrm{C}$ .

(a) Calculate the enthalpy change of neutralisation per mole of water formed.

Q = 50.0 \times 4.18 \times 6.5 = 1359\mathrm{ J}

N = 0.025 \times 1.0 = 0.025\mathrm{ mol}

\Delta H = -\frac{1.359}{0.025} = -54.4\mathrm{ kJ/mol}

(b) Explain why the experimental value differs from the theoretical value of $-57.1\mathrm{ kJ/mol}$ .

Heat loss to the surroundings, calorimeter absorbs some heat, incomplete reaction, or the assumption That the solution has the same properties as pure water.

Question 3 (Paper 2 style)

Given the following data, calculate the lattice energy of MgO:

$\Delta H_f^\circ$ (MgO) = $-602\mathrm{ kJ/mol}$
$\Delta H_{\mathrm{at}}$ (Mg) = $+148\mathrm{ kJ/mol}$
$\frac{1}{2}\Delta H_{\mathrm{at}}$ (O $_2$ ) = $+249\mathrm{ kJ/mol}$
IE $_1$ (Mg) + IE $_2$ (Mg) = $+2188\mathrm{ kJ/mol}$
EA $_1$ (O) + EA $_2$ (O) = $+603\mathrm{ kJ/mol}$

\Delta H_{\mathrm{latt}} = -602 - 148 - 249 - 2188 - 603 = -3790\mathrm{ kJ/mol}

Question 4 (Paper 1 style)

For which reaction is $\Delta S$ positive?

A. 2H $_2$ (g) + O $_2$ (g) $\to$ 2H $_2$ O(g) B. NH $_4$ Cl(s) $\to$ NH $_3$ (g) + HCl(g) C. CaO(s) + H $_2$ O(l) $\to$ Ca(OH) $_2$ (s) D. N $_2$ (g) + 3H $_2$ (g) $\to$ 2NH $_3$ (g)

Answer: B — a solid produces two gases, increasing disorder.

flowchart TD
    A[1_Thermochemistry] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

Formula	Expression
Heat energy	$q = mc\Delta T$
Enthalpy from formation	$\Delta H_r = \sum \Delta H_f(\mathrm{products}) - \sum \Delta H_f(\mathrm{reactants})$
From bond enthalpies	$\Delta H = \sum(\mathrm{bonds broken}) - \sum(\mathrm{bonds formed})$
Entropy change	$\Delta S = \sum S(\mathrm{products}) - \sum S(\mathrm{reactants})$
Gibbs free energy	$\Delta G = \Delta H - T\Delta S$
Equilibrium relation	$\Delta G^\circ = -RT\ln K$

Exam Strategy For Hess’s law questions, draw the energy cycle . For calorimetry, always account for the Total mass of the solution. For Gibbs free energy, pay attention to units — $\Delta H$ is In kJ/mol while $\Delta S$ is in J/(mol $\cdot$ K), so convert one before combining.

Thermochemistry Extended

Enthalpy of Solution

The enthalpy change when one mole of solute dissolves in a solvent to form an infinitely dilute Solution:

\Delta H_{\mathrm{sol}} = \Delta H_{\mathrm{lattice}} + \Delta H_{\mathrm{hydration}}

If $\Delta H_{\mathrm{sol}} \gt 0$ : endothermic (solution cools, e.g., NH $_4$ NO $_3$ ).
If $\Delta H_{\mathrm{sol}} \lt 0$ : exothermic (solution warms, e.g., NaOH).

Enthalpy of Hydration

The enthalpy change when gaseous ions are surrounded by water molecules:

\Delta H_{\mathrm{hyd}} = \Delta H_{\mathrm{at}} + \Delta H_{\mathrm{EA}} + \mathrm{other terms}

Trends in Lattice Energy

Trend	Effect on Lattice Energy
Higher ionic charge	Increases
Smaller ionic radius	Increases
Higher charge density	Increases

Comparison	Higher Lattice Energy
NaCl vs NaBr	NaCl (Br $^-$ is larger)
NaCl vs MgCl $_2$	MgCl $_2$ (Mg $^{2+}$ has higher charge)
NaCl vs Na $_2$ O	Na $_2$ O (O $^{2-}$ has higher charge)

Using Born-Haber Cycles to Compare Compounds

Example Explain why the lattice energy of MgO ( $-3791\mathrm{ kJ/mol}$ ) is much more negative than that of NaCl ( $-788\mathrm{ kJ/mol}$ ).

MgO has Mg $^{2+}$ and O $^{2-}$ ions (both doubly charged), while NaCl has Na $^+$ and Cl $^-$ ions (singly charged). The electrostatic attraction between ions is proportional to the product of their Charges, so the doubly charged ions in MgO have much stronger attraction (four times stronger by Coulomb’s law). Additionally, Mg $^{2+}$ is smaller than Na $^+$ Further increasing lattice energy.

Entropy: Extended Analysis

Predicting Entropy Changes

Process	$\Delta S$	Reason
Solid $\to$ liquid	Positive	More disorder
Liquid $\to$ gas	Positive	Much more disorder
Dissolving ionic solid in water	positive	Ions become dispersed
Gas $\to$ solid	Negative	Much less disorder
Decreasing volume of gas	Negative	Fewer microstates
Increasing temperature	Positive	More molecular motion

Calculating $\Delta S$ for Reactions

Example Calculate $\Delta S^\circ$ for: 2H $_2$ (g) + O $_2$ (g) $\to$ 2H $_2$ O(l)

$S^\circ$ (H $_2$ ) $= 131\mathrm{ J/(mol}\cdot\mathrm{K)}$ , $S^\circ$ (O $_2$ ) $= 205\mathrm{ J/(mol}\cdot\mathrm{K)}$ , $S^\circ$ (H $_2$ O) $= 70\mathrm{ J/(mol}\cdot\mathrm{K)}$ .

\Delta S^\circ = 2(70) - [2(131) + 205] = 140 - 467 = -327\mathrm{ J/(mol}\cdot\mathrm{K)}

The large negative $\Delta S$ is expected: 3 moles of gas produce 2 moles of liquid.

Gibbs Free Energy: Extended

Calculating $\Delta G$ from $\Delta G_f^\circ$ Values

\Delta G_r^\circ = \sum \Delta G_f^\circ(\mathrm{products}) - \sum \Delta G_f^\circ(\mathrm{reactants})

Example Calculate $\Delta G^\circ$ for: C(s) + CO $_2$ (g) $\to$ 2CO(g) at $298\mathrm{ K}$ .

Given: $\Delta G_f^\circ$ (CO $_2$ ) $= -394\mathrm{ kJ/mol}$ , $\Delta G_f^\circ$ (CO) $= -137\mathrm{ kJ/mol}$ .

\Delta G^\circ = 2(-137) - (-394) = -274 + 394 = +120\mathrm{ kJ/mol}

The reaction is not spontaneous at $298\mathrm{ K}$ (but is at higher temperatures since $\Delta S \gt 0$ for this reaction).

Using $\Delta G$ to Predict the Equilibrium Constant

K = e^{-\Delta G^\circ/RT}

Example For a reaction with $\Delta G^\circ = -5.4\mathrm{ kJ/mol}$ at $298\mathrm{ K}$ :

K = e^{-(-5400)/(8.314 \times 298)} = e^{2.18} = 8.85

Since $K \gt 1$ Products are favoured at equilibrium.

Additional IB Exam-Style Questions

Question 5 (Paper 2 style)

The enthalpy of combustion of methanol is $-726\mathrm{ kJ/mol}$ . A spirit burner containing Methanol is used to heat $200\mathrm{ g}$ of water from $20.0\degree\mathrm{C}$ to $65.0\degree\mathrm{C}$ . The mass of methanol burned is $1.50\mathrm{ g}$ .

(a) Calculate the experimental enthalpy of combustion.

Q = 200 \times 4.18 \times 45 = 37620\mathrm{ J} = 37.62\mathrm{ kJ}

N(\mathrm{CH}_3\mathrm{OH}) = \frac{1.50}{32.04} = 0.0468\mathrm{ mol}

\Delta H_c = -\frac{37.62}{0.0468} = -804\mathrm{ kJ/mol}

(b) Calculate the percentage error compared to the literature value.

\mathrm{Percentage error} = \frac{|-804 - (-726)|}{726} \times 100\% = \frac{78}{726} \times 100\% = 10.7\%

(c) Explain two sources of error.

Heat loss to the surroundings; incomplete combustion of methanol; not all heat transferred to the Water; the calorimeter absorbs some heat.

Question 6 (Paper 1 style)

For the reaction: N $_2$ O $_4$ (g) $\rightleftharpoons$ 2NO $_2$ (g), $\Delta H = +57\mathrm{ kJ/mol}$ And $\Delta S = +176\mathrm{ J/(mol}\cdot\mathrm{K)}$ .

At what temperature does the reaction become spontaneous?

\Delta G = 0 \mathrm{ when } T = \frac{\Delta H}{\Delta S} = \frac{57000}{176} = 324\mathrm{ K}

The reaction is spontaneous above $324\mathrm{ K}$ ( $51\degree\mathrm{C}$ ).

Question 7 (Paper 2 style)

Using the data below, calculate the lattice energy of CaF $_2$ :

$\Delta H_f^\circ$ (CaF $_2$ ) $= -1220\mathrm{ kJ/mol}$
$\Delta H_{\mathrm{at}}$ (Ca) $= +178\mathrm{ kJ/mol}$
$\frac{1}{2}\Delta H_{\mathrm{at}}$ (F $_2$ ) $= +79\mathrm{ kJ/mol}$ (per mole of F atoms)
IE $_1$ (Ca) + IE $_2$ (Ca) $= +1735\mathrm{ kJ/mol}$
EA $_1$ (F) $= -328\mathrm{ kJ/mol}$ (per mole of F atoms)

\Delta H_f^\circ = \Delta H_{\mathrm{at}}(\mathrm{Ca}) + 2\left[\frac{1}{2}\Delta H_{\mathrm{at}}(\mathrm{F}_2) + \mathrm{EA}(\mathrm{F})\right] + \mathrm{IE}_1 + \mathrm{IE}_2 + \Delta H_{\mathrm{latt}}

-1220 = 178 + 2(79 - 328) + 1735 + \Delta H_{\mathrm{latt}}

-1220 = 178 + 2(-249) + 1735 + \Delta H_{\mathrm{latt}}

-1220 = 178 - 498 + 1735 + \Delta H_{\mathrm{latt}} = 1415 + \Delta H_{\mathrm{latt}}

\Delta H_{\mathrm{latt}} = -1220 - 1415 = -2635\mathrm{ kJ/mol}

Thermochemistry: Extended Topics

Enthalpy of Atomisation of Elements

The enthalpy of atomisation is the enthalpy change to form one mole of gaseous atoms from the Element in its standard state under standard conditions.

Element	Standard State	$\Delta H_{\mathrm{at}}$ (kJ/mol)
Na	Solid	$+108$
Mg	Solid	$+148$
Al	Solid	$+330$
Cl $_2$	Gas	$+122$
H $_2$	Gas	$+218$
C (graphite)	Solid	$+717$
O $_2$	Gas	$+249$

Electron Affinity Trends

Electron affinity is the enthalpy change when one mole of gaseous atoms gains one electron.

Trend	Effect
Across a period (left to right)	Generally becomes more negative
Down a group	Generally becomes less negative
Noble gases	Positive (unfavourable)
Group 17	Most negative (most favourable)

Ionisation Energy Trends

Trend	Effect
Across a period	Increases (nuclear charge increases, same shielding)
Down a group	Decreases (atomic radius increases)
Large jumps	Occur when removing electron from new shell

Successive Ionisation Energies

Each successive ionisation energy is larger than the previous one. Large jumps indicate removal from A new shell.

Example $50\mathrm{ cm}^3$ of $1.0\mathrm{ M}$ HCl and $50\mathrm{ cm}^3$ of $1.0\mathrm{ M}$ NaOH are mixed In a calorimeter with heat capacity $15\mathrm{ J/K}$ . The temperature rises from $20.0\degree\mathrm{C}$ to $26.8\degree\mathrm{C}$ .

Q_{\mathrm{total}} = (100)(4.18)(6.8) + 15(6.8) = 2842.4 + 102 = 2944.4\mathrm{ J}

N(\mathrm{H}_2\mathrm{O}) = 0.050\mathrm{ mol}

\Delta H = -\frac{2944.4}{0.050} = -58888\mathrm{ J/mol} = -58.9\mathrm{ kJ/mol}

Additional IB Exam-Style Questions

Question 8 (Paper 1 style)

Which process has a positive entropy change?

A. $\mathrm{Ca}^{2+}(aq) + \mathrm{CO}_3^{2-}(aq) \to \mathrm{CaCO}_3(s)$ B. $\mathrm{NH}_4\mathrm{Cl}(s) \to \mathrm{NH}_3(g) + \mathrm{HCl}(g)$ C. $2\mathrm{H}_2(g) + \mathrm{O}_2(g) \to 2\mathrm{H}_2\mathrm{O}(l)$ D. $\mathrm{NaOH}(aq) + \mathrm{HCl}(aq) \to \mathrm{NaCl}(aq) + \mathrm{H}_2\mathrm{O}(l)$

Answer: B. A solid produces two gases, increasing the number of particles and the disorder.

Question 9 (Paper 2 style)

Using the following data, calculate the enthalpy of reaction for:

$\mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \to \mathrm{CO}_2(g) + 2\mathrm{H}_2\mathrm{O}(l)$

Given bond enthalpies (kJ/mol): C—H $= 413$ O=O $= 495$ C=O $= 743$ O—H $= 463$ .

Bonds broken: $4(\mathrm{C--H}) + 2(\mathrm{O=O}) = 4(413) + 2(495) = 1652 + 990 = 2642\mathrm{ kJ/mol}$

Bonds formed: $2(\mathrm{C=O}) + 4(\mathrm{O--H}) = 2(743) + 4(463) = 1486 + 1852 = 3338\mathrm{ kJ/mol}$

\Delta H = 2642 - 3338 = -696\mathrm{ kJ/mol}

Question 10 (Paper 2 style)

For the reaction: $\mathrm{N}_2\mathrm{O}(g) \to \mathrm{N}_2(g) + \mathrm{O}_2(g)$ $\Delta H = -163\mathrm{ kJ/mol}$ and $\Delta S = +149\mathrm{ J/(mol}\cdot\mathrm{K)}$ .

(a) Calculate $\Delta G^\circ$ at $298\mathrm{ K}$ and state whether the reaction is Spontaneous.

\Delta G^\circ = -163000 - 298 \times 149 = -163000 - 44402 = -207402\mathrm{ J/mol} = -207.4\mathrm{ kJ/mol}

Since $\Delta G^\circ \lt 0$ The reaction is spontaneous at $298\mathrm{ K}$ .

(b) At what temperature does $\Delta G^\circ$ become positive?

\Delta G^\circ = 0 \mathrm{ when } T = \frac{\Delta H}{\Delta S} = \frac{-163000}{149} = -1094\mathrm{ K}

Since both $\Delta H$ and $\Delta S$ are negative, the reaction is spontaneous at low temperatures. It becomes non-spontaneous above $1094\mathrm{ K}$ . Since the calculated “temperature” is negative, $\Delta G^\circ$ is negative at all positive temperatures — the reaction is always spontaneous.

Practice Problems

Question 1: Calorimetry Calculation

$50.0\mathrm{ cm}^3$ of $1.0\mathrm{ M}$ $\mathrm{HCl}$ is mixed with $50.0\mathrm{ cm}^3$ of $1.0\mathrm{ M}$ $\mathrm{NaOH}$ in a calorimeter. The temperature increases from $22.0\degree\mathrm{C}$ to $28.8\degree\mathrm{C}$ . Calculate the enthalpy of neutralisation per Mole of water formed.

Answer

$q = mc\Delta T = 100.0 \times 4.18 \times 6.8 = 2842\mathrm{ J} = 2.842\mathrm{ kJ}$

$n(\mathrm{H}_2\mathrm{O}) = 0.0500 \times 1.0 = 0.0500\mathrm{ mol}$

$\Delta H = -\frac{2.842}{0.0500} = -56.8\mathrm{ kJ/mol}$

Question 2: Hess's Law with Formation Enthalpies

Using standard enthalpies of formation, calculate $\Delta H_r^\circ$ for the combustion of propane:

$\mathrm{C}_3\mathrm{H}_8(g) + 5\mathrm{O}_2(g) \to 3\mathrm{CO}_2(g) + 4\mathrm{H}_2\mathrm{O}(l)$

Given: $\Delta H_f^\circ(\mathrm{C}_3\mathrm{H}_8) = -104\mathrm{ kJ/mol}$ $\Delta H_f^\circ(\mathrm{CO}_2) = -394\mathrm{ kJ/mol}$ $\Delta H_f^\circ(\mathrm{H}_2\mathrm{O}) = -286\mathrm{ kJ/mol}$ .

Answer

$\Delta H_r^\circ = \sum \Delta H_f^\circ(\mathrm{products}) - \sum \Delta H_f^\circ(\mathrm{reactants})$

$= [3(-394) + 4(-286)] - [(-104) + 5(0)]$

$= (-1182 - 1144) - (-104) = -2326 + 104 = -2222\mathrm{ kJ/mol}$

Question 3: Bond Enthalpy Calculation

Using average bond enthalpies, calculate $\Delta H$ for the reaction:

$\mathrm{N}_2(g) + 3\mathrm{H}_2(g) \to 2\mathrm{NH}_3(g)$

Given: $\mathrm{N} \equiv \mathrm{N} = 945\mathrm{ kJ/mol}$ $\mathrm{H}-\mathrm{H} = 436\mathrm{ kJ/mol}$ , $\mathrm{N}-\mathrm{H} = 391\mathrm{ kJ/mol}$ .

Answer

Bonds broken: $1(\mathrm{N} \equiv \mathrm{N}) + 3(\mathrm{H}-\mathrm{H}) = 945 + 3(436) = 945 + 1308 = 2253\mathrm{ kJ/mol}$

Bonds formed: $6(\mathrm{N}-\mathrm{H}) = 6 \times 391 = 2346\mathrm{ kJ/mol}$

$\Delta H = 2253 - 2346 = -93\mathrm{ kJ/mol}$

The actual value is $-92\mathrm{ kJ/mol}$ So the bond enthalpy approximation is close.

Question 4: Gibbs Free Energy and Spontaneity

For the decomposition of calcium carbonate:

$\mathrm{CaCO}_3(s) \to \mathrm{CaO}(s) + \mathrm{CO}_2(g)$

$\Delta H = +178\mathrm{ kJ/mol}$ , $\Delta S = +161\mathrm{ J/(mol \cdot K)}$ .

(a) Calculate $\Delta G$ at $298\mathrm{ K}$ and state whether the reaction is spontaneous.

(b) Calculate the minimum temperature at which the reaction becomes spontaneous.

Answer

(a) $\Delta G = \Delta H - T\Delta S = 178000 - 298 \times 161 = 178000 - 47978 = +130\,022\mathrm{ J/mol} = +130\mathrm{ kJ/mol}$

Since $\Delta G \gt 0$ The reaction is not spontaneous at $298\mathrm{ K}$ .

(b) At $\Delta G = 0$ :

$T = \frac{\Delta H}{\Delta S} = \frac{178000}{161} = 1106\mathrm{ K}$

The reaction becomes spontaneous above $1106\mathrm{ K}$ (approximately $833\degree\mathrm{C}$ ).

Question 5: Entropy Change Prediction

Predict the sign of $\Delta S$ for each of the following processes and explain:

(a) $\mathrm{NH}_4\mathrm{Cl}(s) \to \mathrm{NH}_3(g) + \mathrm{HCl}(g)$

(b) $2\mathrm{NO}(g) + \mathrm{O}_2(g) \to 2\mathrm{NO}_2(g)$

Answer

(a) Positive $\Delta S$ : One mole of solid produces two moles of gas, significantly increasing Disorder.

(b) Negative $\Delta S$ : Three moles of gas produce two moles of gas, decreasing the number of Gaseous particles and thus disorder.

(c) Positive $\Delta S$ : An ordered solid lattice breaks apart into freely moving hydrated ions In solution, increasing disorder.

For the A-Level treatment of this topic, see Thermodynamics & Energetics.

Common Pitfalls

Assuming that a strong acid always has a lower pH than a weak acid without considering concentration.
Drawing structural formulae incorrectly. Check the number of bonds each atom can form and the overall charge.
Writing half-equations without balancing charges or atoms. Always check electrons, hydrogen ions, and water molecules.
Confusing enthalpy of formation with enthalpy of combustion, or using the wrong sign convention.

Cross-References

Topic	Site	Link
[Thermodynamics (Chemistry)]	A-Level	View
[Thermodynamics (Chemistry)]	IB	View
[Thermodynamics (Chemistry)]	DSE	View

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Thermochemistry | IB - Wyatt's Notes

Intuition

Enthalpy Changes

Definitions

Sign Convention

Standard Conditions

Types of Enthalpy Change

See Also