IB Chemistry SL — Thermodynamics & Gibbs Free Energy

Complete Study Guide

Topics Covered

1. Energy Changes in Chemical Reactions (R1.1)
2. Calorimetry — Measuring Enthalpy Changes
3. Enthalpy of Combustion, Formation, and Neutralisation
4. Hess’s Law & Energy Cycles (R1.2)
5. Bond Enthalpies
6. Entropy & the Second Law (R1.3)
7. Gibbs Free Energy & Spontaneity (R1.3)
8. Quick Reference — Key Formulas and Sign Conventions

Videos on this page: Enthalpy Changes & Calorimetry

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1. Energy Changes in Chemical Reactions (R1.1)

Every chemical reaction involves energy. When bonds break, energy is absorbed; when bonds form, energy is released. The net balance of these two processes determines whether a reaction releases energy to its surroundings or absorbs energy from them. This is one of the most fundamental ideas in chemistry, and it underpins everything from why fuels burn to why ice melts.

Exothermic and Endothermic Reactions

	Exothermic	Endothermic
Energy change	Energy released to surroundings	Energy absorbed from surroundings
Temperature of surroundings	Increases	Decreases
Sign of $\Delta H$	Negative ($\Delta H < 0$)	Positive ($\Delta H > 0$)
Bond energy balance	Energy released forming bonds > Energy absorbed breaking bonds	Energy absorbed breaking bonds > Energy released forming bonds
Examples	Combustion, neutralisation, most oxidation reactions	Photosynthesis, thermal decomposition, dissolving ammonium nitrate

Enthalpy Diagrams

Enthalpy diagrams (also called energy profile diagrams) show the enthalpy of the reactants and products on a vertical energy axis, with the reaction pathway on the horizontal axis.

Exothermic reaction:

• Reactants are at a higher enthalpy level than products.
• The arrow for $\Delta H$ points downwards.
• There is an activation energy ($E_a$) hump that must be overcome for the reaction to start.

Endothermic reaction:

• Reactants are at a lower enthalpy level than products.
• The arrow for $\Delta H$ points upwards.
• The activation energy hump is measured from the reactant level to the top of the curve.

Activation Energy

Activation energy ($E_a$) is the minimum energy that colliding particles must possess for a reaction to occur. Even exothermic reactions need activation energy to get started — think of a match: once lit (activation energy supplied), the combustion reaction releases far more energy than was initially needed.

A catalyst lowers the activation energy by providing an alternative reaction pathway. It does not change $\Delta H$ — the enthalpy difference between reactants and products stays the same.

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2. Calorimetry — Measuring Enthalpy Changes

Calorimetry is the experimental technique used to measure enthalpy changes. The basic idea is simple: carry out a reaction in or near water, measure the temperature change of the water, and use this to calculate the energy transferred.

The Calorimetry Equation

$q = mc\Delta T$

where:

• $q$ = energy transferred (in joules, J)
• $m$ = mass of the solution being heated (in grams, g) — usually assumed to be the mass of water
• $c$ = specific heat capacity (for water, $c = 4.18 \text{ J g}^{-1}\text{K}^{-1}$)
• $\Delta T$ = temperature change (in K or $^\circ\text{C}$ — the magnitude is the same)

Sources of Error in Calorimetry

In practice, experimental $\Delta H$ values are always less accurate than literature values because:

• Heat loss to surroundings — the calorimeter is not perfectly insulated.
• Incomplete combustion (for combustion experiments) — soot formation means not all fuel reacted completely.
• Assumption that the solution has the density and specific heat capacity of water — dilute solutions are close, but not exact.
• Evaporation of volatile substances — some fuel or solvent may evaporate.

Watch: Enthalpy Changes & Calorimetry

MSJChem · 5 min · Exothermic and endothermic reactions — energy profile diagrams and enthalpy sign conventions

MSJChem · 7 min · Calorimetry calculations — applying q = mcΔT and ΔH = −Q/n with worked examples

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3. Enthalpy of Combustion, Formation, and Neutralisation

The IB syllabus requires you to know several standard enthalpy definitions. All are measured under standard conditions: 298 K (25$^\circ$C), 100 kPa, solutions at 1 mol dm$^{-3}$. The symbol is $\Delta H^\circ$ (the superscript $^\circ$ means standard).

Key fact: The standard enthalpy of formation of any element in its standard state is zero by definition. For example, $\Delta H_f^\circ[\text{O}_2\text{(g)}] = 0$, $\Delta H_f^\circ[\text{C(graphite)}] = 0$.

Quick Recall — Sections 1-3

Try to answer without scrolling up:

1. What is the sign of delta H for an exothermic reaction?
2. Write the formula for calculating enthalpy change from calorimetry data.
3. What is the standard enthalpy of formation of an element in its standard state?

Reveal answers

1. Negative (the system releases energy to the surroundings).
2. $q = mc\Delta T$, where $m$ is mass of water, $c$ is specific heat capacity (4.18 J/g/K), and $\Delta T$ is the temperature change.
3. Zero, by definition.

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4. Hess’s Law & Energy Cycles (R1.2)

What is Hess’s Law?

Hess’s Law states that the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same. In plain language: it does not matter whether a reaction happens in one step or in five steps — the total energy change is the same.

This is a consequence of the law of conservation of energy and is enormously useful because it lets us calculate enthalpy changes for reactions that are difficult or impossible to measure directly.

Using Enthalpy of Formation to Calculate Reaction Enthalpy

$\Delta H_{\text{rxn}}^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})$

In words: add up the enthalpies of formation of all the products, subtract the sum of the enthalpies of formation of all the reactants.

Using Enthalpy of Combustion to Calculate Reaction Enthalpy

$\Delta H_{\text{rxn}}^\circ = \sum \Delta H_c^\circ (\text{reactants}) - \sum \Delta H_c^\circ (\text{products})$

Practice: Hess’s Law — Fading Sequence

These problems build your confidence step by step. The first shows the complete solution, the second hides the final calculation, and the third gives only the setup.

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5. Bond Enthalpies

What are Bond Enthalpies?

A bond enthalpy (also called bond energy) is the energy required to break one mole of a specific covalent bond in the gaseous state, averaged over many different molecules. Because they are averages, calculations using bond enthalpies give approximate values for $\Delta H$.

Breaking bonds is endothermic (requires energy input). Forming bonds is exothermic (releases energy).

Calculating $\Delta H$ from Bond Enthalpies

$\Delta H \approx \sum (\text{bond enthalpies of bonds broken}) - \sum (\text{bond enthalpies of bonds formed})$

In words: add up the energy needed to break all bonds in the reactants, then subtract the energy released when all bonds in the products form.

Why Bond Enthalpy Calculations are Only Estimates

1. Bond enthalpies are average values — the actual energy of a C-H bond in methane is slightly different from a C-H bond in ethane.
2. They apply to the gaseous state only — if reactants or products are liquids or solids, additional energy changes (vaporisation, condensation) are not included.
3. In molecules with resonance or delocalisation (like benzene), the actual bond strengths differ from the tabulated average.

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6. Entropy & the Second Law (R1.3)

What is Entropy?

Entropy ($S$) is a measure of the disorder or dispersal of energy in a system. A system with high entropy has many possible arrangements of its particles and energy — it is more “spread out” or disordered. A system with low entropy is highly ordered with fewer possible arrangements.

Think of it this way: a tidy room has low entropy (everything in its place, only one arrangement). A messy room has high entropy (items scattered in many possible positions). Nature tends towards the messy room — systems naturally move towards greater disorder.

Predicting Entropy Changes

You can predict whether entropy increases or decreases by looking at:

Change	Effect on entropy
Solid $\rightarrow$ Liquid $\rightarrow$ Gas	Entropy increases (particles become more disordered)
Fewer moles of gas $\rightarrow$ More moles of gas	Entropy increases
More moles of gas $\rightarrow$ Fewer moles of gas	Entropy decreases
Dissolving a solid in water	Entropy usually increases (particles become more dispersed)
Temperature increases	Entropy increases (particles have more energy, more possible arrangements)

Calculating Entropy Changes

$\Delta S_{\text{rxn}}^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})$

Note: unlike enthalpy of formation, the standard entropy of elements is not zero. Every substance has a positive standard entropy value (the third law of thermodynamics states that entropy is zero only at absolute zero, 0 K).

Quick Recall — Sections 4-6

Try to answer without scrolling up:

1. State Hess’s Law in one sentence.
2. How do you calculate reaction enthalpy from bond enthalpies?
3. What does a positive change in entropy mean?

Reveal answers

1. The total enthalpy change for a reaction is independent of the route taken, provided initial and final conditions are the same.
2. $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds formed})$ — energy in minus energy out.
3. The system has become more disordered (e.g. solid to liquid, or fewer moles of gas to more moles of gas).

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7. Gibbs Free Energy & Spontaneity (R1.3)

What is Gibbs Free Energy?

Gibbs free energy ($G$) combines enthalpy and entropy into a single quantity that tells us whether a reaction is spontaneous (thermodynamically favourable) at a given temperature. A spontaneous reaction is one that can occur without continuous external input of energy — but “spontaneous” does not mean “fast.” Diamond converting to graphite is spontaneous but takes millions of years.

The Gibbs Equation

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

where:

• $\Delta G$ = Gibbs free energy change (kJ mol$^{-1}$)
• $\Delta H$ = enthalpy change (kJ mol$^{-1}$)
• $T$ = temperature in kelvin (K = $^\circ$C + 273)
• $\Delta S$ = entropy change (must be in kJ K$^{-1}$ mol$^{-1}$ — divide J values by 1000)

The Four Combinations of $\Delta H$ and $\Delta S$

$\Delta H$	$\Delta S$	$\Delta G = \Delta H - T\Delta S$	Spontaneous?
Negative (exothermic)	Positive (more disorder)	Always negative	Always spontaneous at all temperatures
Negative (exothermic)	Negative (less disorder)	Depends on $T$	Spontaneous at low $T$ (enthalpy wins)
Positive (endothermic)	Positive (more disorder)	Depends on $T$	Spontaneous at high $T$ (entropy wins)
Positive (endothermic)	Negative (less disorder)	Always positive	Never spontaneous at any temperature

Spontaneity vs. Rate

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8. Quick Reference — Key Formulas and Sign Conventions

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Mixed Practice — Exam Style

1. [Exothermic/Endothermic] The dissolution of ammonium nitrate ($\text{NH}_4\text{NO}_3$) in water causes the temperature of the solution to decrease. Which row correctly describes this process?

   	$\Delta H$	Energy transfer
   A	Negative	Energy released to surroundings
   B	Positive	Energy absorbed from surroundings
   C	Negative	Energy absorbed from surroundings
   D	Positive	Energy released to surroundings

2. [Calorimetry] When 0.500 g of ethanol ($\text{C}_2\text{H}_5\text{OH}$, $M_r = 46.08$) is burned and the heat is used to warm 200.0 g of water from 20.0$^\circ$C to 33.2$^\circ$C, the experimental enthalpy of combustion of ethanol is approximately:

   A. $-1020 \text{ kJ mol}^{-1}$

   B. $-509 \text{ kJ mol}^{-1}$

   C. $+1020 \text{ kJ mol}^{-1}$

   D. $-11.0 \text{ kJ mol}^{-1}$

3. [Hess’s Law — Formation Data] Given the following standard enthalpies of formation:

   $\Delta H_f^\circ[\text{SO}_3\text{(g)}] = -395.7 \text{ kJ mol}^{-1}$, $\Delta H_f^\circ[\text{SO}_2\text{(g)}] = -296.8 \text{ kJ mol}^{-1}$

   Calculate $\Delta H^\circ$ for: $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{SO}_3\text{(g)}$

   A. $-98.9 \text{ kJ}$

   B. $-197.8 \text{ kJ}$

   C. $+197.8 \text{ kJ}$

   D. $-692.5 \text{ kJ}$

4. [Bond Enthalpies] The bond enthalpy of H-H is 436 kJ mol$^{-1}$ and Cl-Cl is 242 kJ mol$^{-1}$. The bond enthalpy of H-Cl is 431 kJ mol$^{-1}$. What is the approximate $\Delta H$ for $\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)}$?

   A. $-184 \text{ kJ}$

   B. $+184 \text{ kJ}$

   C. $-247 \text{ kJ}$

   D. $-862 \text{ kJ}$

5. [Entropy — Predicting Sign] For which reaction is $\Delta S$ most likely to be large and positive?

   A. $\text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{O(s)}$

   B. $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}$

   C. $\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$

   D. $2\text{CO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)}$

6. [Gibbs Free Energy — Calculation] A reaction has $\Delta H = -120 \text{ kJ mol}^{-1}$ and $\Delta S = +45 \text{ J K}^{-1}\text{mol}^{-1}$ at 298 K. What is $\Delta G$?

   A. $-133.4 \text{ kJ mol}^{-1}$

   B. $-106.6 \text{ kJ mol}^{-1}$

   C. $+106.6 \text{ kJ mol}^{-1}$

   D. $-13410 \text{ kJ mol}^{-1}$

7. [Gibbs — Spontaneity at Temperature] A reaction has $\Delta H = +50.0 \text{ kJ mol}^{-1}$ and $\Delta S = +120 \text{ J K}^{-1}\text{mol}^{-1}$. Above what temperature is the reaction spontaneous?

   A. 417 K

   B. 240 K

   C. 0.42 K

   D. 6000 K

8. [Enthalpy Diagrams] In an enthalpy level diagram for an exothermic reaction with a catalyst, which statement is correct?

   A. The catalyst lowers both the activation energy and $\Delta H$

   B. The catalyst lowers the activation energy but does not change $\Delta H$

   C. The catalyst raises the activation energy but makes the reaction more exothermic

   D. The catalyst has no effect on the activation energy

9. [Calorimetry — Sources of Error] A student measures the enthalpy of combustion of propan-1-ol using a spirit burner and gets a value significantly less negative than the literature value. The most likely reason is:

   A. The thermometer was reading too high

   B. Heat was lost to the surroundings, so not all energy was transferred to the water

   C. The student used too much water

   D. The specific heat capacity of water was too low

10. [Gibbs — Four Combinations] A reaction has $\Delta H > 0$ and $\Delta S < 0$. Which statement about spontaneity is correct?

    A. The reaction is spontaneous at all temperatures

    B. The reaction is spontaneous at high temperatures

    C. The reaction is spontaneous at low temperatures

    D. The reaction is never spontaneous at any temperature

11. [Hess’s Law — Combustion Data] Given:

    $\Delta H_c^\circ[\text{C(s)}] = -393.5 \text{ kJ mol}^{-1}$, $\Delta H_c^\circ[\text{H}_2\text{(g)}] = -285.8 \text{ kJ mol}^{-1}$, $\Delta H_c^\circ[\text{C}_2\text{H}_6\text{(g)}] = -1560.7 \text{ kJ mol}^{-1}$

    Calculate $\Delta H_f^\circ$ for ethane: $2\text{C(s)} + 3\text{H}_2\text{(g)} \rightarrow \text{C}_2\text{H}_6\text{(g)}$

    A. $-84.7 \text{ kJ mol}^{-1}$

    B. $+84.7 \text{ kJ mol}^{-1}$

    C. $-881.4 \text{ kJ mol}^{-1}$

    D. $-1560.7 \text{ kJ mol}^{-1}$

12. [Entropy Calculation] Given: $S^\circ[\text{N}_2\text{(g)}] = 191.6$, $S^\circ[\text{H}_2\text{(g)}] = 130.7$, $S^\circ[\text{NH}_3\text{(g)}] = 192.5$ (all in J K$^{-1}$ mol$^{-1}$). What is $\Delta S^\circ$ for $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}$?

    A. $-198.7 \text{ J K}^{-1}\text{mol}^{-1}$

    B. $+198.7 \text{ J K}^{-1}\text{mol}^{-1}$

    C. $-129.8 \text{ J K}^{-1}\text{mol}^{-1}$

    D. $+61.9 \text{ J K}^{-1}\text{mol}^{-1}$

Show Answers

1. B — The temperature decreases, so energy is absorbed from the surroundings (the water cools). This is endothermic, so $\Delta H$ is positive. A and C both have negative $\Delta H$ which would mean exothermic. D has positive $\Delta H$ but says energy is released, which is contradictory.

2. A — Moles of ethanol = $\frac{0.500}{46.08} = 0.01085$ mol. $q = mc\Delta T = 200.0 \times 4.18 \times 13.2 = 11035$ J $= 11.04$ kJ. $\Delta H_c = -\frac{11.04}{0.01085} = -1018 \approx -1020$ kJ mol$^{-1}$. B divides by twice the moles. C has the wrong sign. D forgets to divide by moles.

3. B — $\Delta H^\circ = [2(-395.7)] - [2(-296.8) + 0] = -791.4 - (-593.6) = -791.4 + 593.6 = -197.8$ kJ. Remember $\Delta H_f^\circ[\text{O}_2] = 0$. A halves the answer (forgets the coefficient of 2). C has the wrong sign. D adds instead of subtracting.

4. A — Bonds broken: 1(H-H) + 1(Cl-Cl) = 436 + 242 = 678 kJ. Bonds formed: 2(H-Cl) = 2(431) = 862 kJ. $\Delta H = 678 - 862 = -184$ kJ. B has the wrong sign (subtracted the wrong way). C uses incorrect bond counts. D forgot to subtract.

5. C — A solid decomposes to a solid and a gas. The production of a gas from a solid gives a large positive $\Delta S$. A is freezing (decrease in disorder, negative $\Delta S$). B goes from 4 moles of gas to 2 moles (negative $\Delta S$). D goes from 3 moles of gas to 2 moles (negative $\Delta S$).

6. A — Convert $\Delta S$ to kJ: $45 \div 1000 = 0.045$ kJ K$^{-1}$ mol$^{-1}$. $\Delta G = -120 - (298 \times 0.045) = -120 - 13.4 = -133.4$ kJ mol$^{-1}$. B forgets to subtract the $T\Delta S$ term properly (adds instead of subtracts). D fails to convert J to kJ ($298 \times 45 = 13410$).

7. A — Set $\Delta G = 0$: $T = \frac{\Delta H}{\Delta S} = \frac{50.0}{0.120} = 417$ K. Above 417 K, the $T\Delta S$ term exceeds $\Delta H$, making $\Delta G < 0$. B inverts the fraction. C fails to convert J to kJ. D multiplies instead of dividing.

8. B — A catalyst provides an alternative pathway with lower activation energy. It does NOT change $\Delta H$ — the enthalpy difference between reactants and products is determined by the chemical bonds, not the pathway. A is wrong because $\Delta H$ is unchanged. C and D are wrong about the effect on $E_a$.

9. B — Heat loss to surroundings is the primary source of error in spirit burner experiments. Less heat reaches the water, so $\Delta T$ is smaller than expected, giving a less negative (smaller magnitude) $\Delta H_c$. C is not a source of systematic error in the right direction. D is irrelevant — the accepted value of specific heat capacity is used.

10. D — When $\Delta H > 0$ and $\Delta S < 0$: $\Delta G = \Delta H - T\Delta S = (\text{positive}) - T(\text{negative}) = (\text{positive}) + (\text{positive}) = \text{always positive}$. $\Delta G$ is positive at all temperatures, so the reaction is never spontaneous. A describes $\Delta H < 0, \Delta S > 0$. B describes $\Delta H > 0, \Delta S > 0$. C describes $\Delta H < 0, \Delta S < 0$.

11. A — Using combustion data: $\Delta H_f^\circ = \sum \Delta H_c^\circ(\text{reactants}) - \sum \Delta H_c^\circ(\text{products})$. $= [2(-393.5) + 3(-285.8)] - [-1560.7] = [-787.0 + (-857.4)] - [-1560.7] = -1644.4 + 1560.7 = -83.7 \approx -84.7$ kJ mol$^{-1}$. B has the wrong sign. C and D use incorrect formulas.

12. A — $\Delta S^\circ = [2(192.5)] - [191.6 + 3(130.7)] = 385.0 - [191.6 + 392.1] = 385.0 - 583.7 = -198.7$ J K$^{-1}$ mol$^{-1}$. The negative value is expected: 4 moles of gas become 2 moles, a decrease in disorder. B has the wrong sign. C uses incorrect coefficients. D forgets to multiply $S^\circ(\text{H}_2)$ by 3.