Thermodynamics | AP - Wyatt's Notes

The First Law of Thermodynamics (CED Unit 6)

Energy cannot be created or destroyed. The change in internal energy of a system is:

\Delta U = q + w

Where $q$ is heat and $w$ is work. The first law is a statement of energy conservation: any change In the internal energy of a system must be accounted for by heat flow and work done.

Sign Convention (Chemistry)

$q \gt 0$ : heat absorbed by the system (endothermic)
$q \lt 0$ : heat released by the system (exothermic)
$w \gt 0$ : work done on the system (compression)
$w \lt 0$ : work done by the system (expansion)

This sign convention is used by chemists (IUPAC convention). Some physics texts use the opposite Sign for work.

Pressure-Volume Work

For a gas expanding/contracting against constant external pressure:

W = -P\Delta V

The negative sign reflects the chemistry convention: when a gas expands ( $\Delta V \gt 0$ ), it does Work on the surroundings ( $w \lt 0$ ).

Work at Constant Pressure: Enthalpy

At constant pressure, $w = -P\Delta V$ So:

\Delta U = q_P - P\Delta V \implies q_P = \Delta U + P\Delta V = \Delta H

Enthalpy ( $H$ ) is defined as $H = U + PV$ . At constant pressure, $\Delta H = q_P$ . Enthalpy is Convenient because most chemical reactions occur at constant (atmospheric) pressure.

Derivation: Why $\Delta H = q_P$

Starting from the first law at constant pressure:

$\Delta U = q_P + w = q_P - P\Delta V$

$q_P = \Delta U + P\Delta V$

$q_P = (U_2 - U_1) + P(V_2 - V_1) = (U_2 + PV_2) - (U_1 + PV_1) = H_2 - H_1 = \Delta H$

This derivation shows that enthalpy change equals heat at constant pressure because the $P\Delta V$ Work term is absorbed into the enthalpy definition.

Enthalpy of Reaction (CED Unit 6)

Standard Enthalpy of Formation ( $\Delta H_f^\circ$ )

The enthalpy change when 1 mole of a compound forms from its elements in their standard states.

\Delta H_{\mathrm{rxn}^\circ = \sum n\Delta H_f^\circ(\mathrm{products) - \sum m\Delta H_f^\circ(\mathrm{reactants)

Standard conditions: $1 \mathrm{ atm$ , $298 \mathrm{ K$ ( $25^\circ\mathrm{C$ ), pure substances in Their most stable form. By convention, $\Delta H_f^\circ = 0$ for elements in their standard state.

The standard state of an element is its most stable form at $1 \mathrm{ atm$ and $25^\circ\mathrm{C$ : e.g., $\mathrm{O_2(g)$ Not $\mathrm{O_3(g)$ or $\mathrm{O_2(l)$ ; $\mathrm{C(graphite)$ Not $\mathrm{C(diamond)$ .

Hess”s Law

The total enthalpy change for a reaction is the same regardless of the pathway. If a reaction can be Written as the sum of several steps:

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

Hess’s law is a direct consequence of enthalpy being a state function: only on the Initial and final states, not on the path between them. This allows us to calculate enthalpy changes That cannot be measured directly.

Worked Example: Hess’s Law

Calculate $\Delta H^\circ$ for the reaction $\mathrm{C(\mathrm{graphite) + 2\mathrm{H_2(g) \to \mathrm{CH_4(g)$ using the following data:

$\mathrm{C(\mathrm{graphite) + \mathrm{O_2(g) \to \mathrm{CO_2(g)$ $\Delta H_1 = -393.5 \mathrm{ kJ/mol$

$\mathrm{H_2(g) + \frac{1}{2}\mathrm{O_2(g) \to \mathrm{H_2\mathrm{O(l)$ $\Delta H_2 = -285.8 \mathrm{ kJ/mol$

$\mathrm{CH_4(g) + 2\mathrm{O_2(g) \to \mathrm{CO_2(g) + 2\mathrm{H_2\mathrm{O(l)$ $\Delta H_3 = -890.3 \mathrm{ kJ/mol$

Using Hess’s law: $\Delta H_f(\mathrm{CH_4) = \Delta H_1 + 2\Delta H_2 - \Delta H_3$

$= -393.5 + 2(-285.8) - (-890.3) = -393.5 - 571.6 + 890.3 = -74.8 \mathrm{ kJ/mol$

Worked Example: Hess’s Law with Multiple Steps

Calculate $\Delta H^\circ$ for $\mathrm{C(s) + 2\mathrm{H_2(g) + \frac{1}{2}\mathrm{O_2(g) \to \mathrm{CH_3\mathrm{OH(l)$ .

Given: $\mathrm{C(s) + \mathrm{O_2(g) \to \mathrm{CO_2(g)$ , $\Delta H = -393.5 \mathrm{ kJ/mol$ $\mathrm{H_2(g) + \frac{1}{2}\mathrm{O_2(g) \to \mathrm{H_2\mathrm{O(l)$ $\Delta H = -285.8 \mathrm{ kJ/mol$ $\mathrm{CH_3\mathrm{OH(l) + \frac{3}{2}\mathrm{O_2(g) \to \mathrm{CO_2(g) + 2\mathrm{H_2\mathrm{O(l)$ $\Delta H = -726.4 \mathrm{ kJ/mol$

Target: $\mathrm{C(s) + 2\mathrm{H_2(g) + \frac{1}{2}\mathrm{O_2(g) \to \mathrm{CH_3\mathrm{OH(l)$

Reverse equation 3 and add equations 1 and 2:

$\mathrm{C(s) + \mathrm{O_2(g) \to \mathrm{CO_2(g)$ , $\Delta H = -393.5$

$2\mathrm{H_2(g) + \mathrm{O_2(g) \to 2\mathrm{H_2\mathrm{O(l)$ , $\Delta H = 2(-285.8) = -571.6$

$\mathrm{CO_2(g) + 2\mathrm{H_2\mathrm{O(l) \to \mathrm{CH_3\mathrm{OH(l) + \frac{3}{2}\mathrm{O_2(g)$ $\Delta H = +726.4$

Sum: $\Delta H = -393.5 - 571.6 + 726.4 = -238.7 \mathrm{ kJ/mol$

Worked Example: Bond Enthalpy Calculation

Estimate $\Delta H$ for $\mathrm{N_2(g) + 3\mathrm{H_2(g) \to 2\mathrm{NH_3(g)$ using bond Enthalpies.

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

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

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

(The exact value is $-92.2 \mathrm{ kJ/mol$ Showing that bond enthalpies give a good Approximation.)

Example Calculate $\Delta H^\circ$ 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)

Using standard enthalpies of formation (kJ/mol):

Substance	$\Delta H_f^\circ$ (kJ/mol)
$\mathrm{C_3\mathrm{H_8(g)$	-103.8
$\mathrm{O_2(g)$	0
$\mathrm{CO_2(g)$	-393.5
$\mathrm{H_2\mathrm{O(l)$	-285.8

\Delta H^\circ = [3(-393.5) + 4(-285.8)] - [-103.8 + 5(0)]

= [-1180.5 - 1143.2] - [-103.8] = -2323.7 + 103.8 = -2219.9 \mathrm{ kJ

Bond Enthalpies

The average enthalpy change when a bond is broken in the gas phase:

\Delta H \approx \sum D(\mathrm{bonds broken) - \sum D(\mathrm{bonds formed)

This is an approximation because bond energies are average values that depend on the molecular Environment. The approximation is most accurate when all species are in the gas phase.

Example Estimate $\Delta H$ for $\mathrm{CH_4(g) + 2\mathrm{O_2(g) \to \mathrm{CO_2(g) + 2\mathrm{H_2\mathrm{O(g)$ using bond Enthalpies.

Bonds broken: $4 \mathrm{ C--H (413) + 2 \mathrm{ O=O (495) = 1648 + 990 = 2642 \mathrm{ kJ/mol$

Bonds formed: $2 \mathrm{ C=O (799) + 4 \mathrm{ O--H (463) = 1598 + 1852 = 3450 \mathrm{ kJ/mol$

\Delta H \approx 2642 - 3450 = -808 \mathrm{ kJ/mol

(The exact value using formation enthalpies is about $-802 \mathrm{ kJ/mol$ for the gaseous Products.)

Calorimetry

Specific Heat Capacity

Q = mc\Delta T

Where $m$ is mass, $c$ is specific heat capacity (J/g $\cdot$ K), and $\Delta T$ is the temperature Change.

The specific heat capacity of water is $4.18 \mathrm{ J/g\cdot\mathrm{K$ . Water has an unusually High specific heat capacity because hydrogen bonds must be broken to increase its temperature.

Coffee-Cup Calorimetry (Constant Pressure)

Measures $\Delta H$ of reactions in solution. The calorimeter is open to the atmosphere, so $P$ is Constant.

Q_{\mathrm{rxn} = -q_{\mathrm{solution} = -(m_{\mathrm{solution} \cdot c_{\mathrm{solution} \cdot \Delta T)

The negative sign ensures that if the solution temperature increases ( $\Delta T \gt 0$ ), the Reaction is exothermic ( $q_{\mathrm{rxn} \lt 0$ ).

Bomb Calorimetry (Constant Volume)

Measures $\Delta U$ of combustion reactions. The calorimeter has a fixed volume.

Q_{\mathrm{rxn} = -C_{\mathrm{calorimeter} \cdot \Delta T

Where $C_{\mathrm{calorimeter}$ is the heat capacity of the entire calorimeter (including the water Bomb, stirrer, etc.).

Derivation: Converting $\Delta U$ to $\Delta H$ in Bomb Calorimetry

For a bomb calorimetry experiment:

$\Delta U = q_V = -C_{\mathrm{cal} \cdot \Delta T$

To convert to $\Delta H$ :

$\Delta H = \Delta U + \Delta n_g RT$

Where $\Delta n_g$ is the change in moles of gas. For combustion reactions, $\Delta n_g$ is often Negative (fewer gas moles of products than reactants), making $\Delta H$ slightly more negative than $\Delta U$ .

Worked Example: Bomb Calorimetry with $\Delta H$ Conversion

A bomb calorimeter with $C_{\mathrm{cal} = 894 \mathrm{ J/K$ is used to determine the enthalpy of Combustion of glucose ( $\mathrm{C_6\mathrm{H_{12}\mathrm{O_6$ ). Burning $1.00 \mathrm{ g$ raises The temperature by $3.02 \mathrm{ K$ . Calculate $\Delta U$ and $\Delta H$ per mole.

Molar mass of glucose: $180.16 \mathrm{ g/mol$ .

$\Delta U = -894 \times 3.02 = -2700 \mathrm{ J = -2.70 \mathrm{ kJ per gram$

$\Delta U_{\mathrm{per mol} = -2.70 \times 180.16 = -486 \mathrm{ kJ/mol$

For $\mathrm{C_6\mathrm{H_{12}\mathrm{O_6(s) + 6\mathrm{O_2(g) \to 6\mathrm{CO_2(g) + 6\mathrm{H_2\mathrm{O(l)$ :

$\Delta n_g = 6 - 6 = 0$ (no net change in gas moles).

$\Delta H = \Delta U + \Delta n_g RT = -486 + 0 = -486 \mathrm{ kJ/mol$

(Literature value: $-2803 \mathrm{ kJ/mol$ . The difference is due to the calorimeter containing a Solution rather than pure water.)

Example When $50.0 \mathrm{ mL$ of $1.00 \mathrm{ M \mathrm{HCl$ is mixed with $50.0 \mathrm{ mL$ of $1.00 \mathrm{ M$ $\mathrm{NaOH$ in a coffee-cup calorimeter, the temperature rises from $25.0^\circ\mathrm{C$ to $31.6^\circ\mathrm{C$ . Calculate $\Delta H$ per mole of $\mathrm{H_2\mathrm{O$ formed.

Total mass: $50.0 + 50.0 = 100.0 \mathrm{ g$ (assume density = $1.00 \mathrm{ g/mL$ $c = 4.18 \mathrm{ J/g\cdot\mathrm{K$ ).

Q_{\mathrm{solution} = (100.0)(4.18)(31.6 - 25.0) = 100.0 \times 4.18 \times 6.6 = 2759 \mathrm{ J = 2.76 \mathrm{ kJ

Q_{\mathrm{rxn} = -2.76 \mathrm{ kJ

Moles of $\mathrm{H_2\mathrm{O$ formed: $1.00 \times 0.0500 = 0.0500 \mathrm{ mol$ .

\Delta H = \frac{-2.76}{0.0500} = -55.2 \mathrm{ kJ/mol

Entropy and the Second Law (CED Unit 9)

Entropy ( $S$ )

Entropy is a measure of disorder or randomness at the molecular level. More precisely, it is a Measure of the number of microstates available to a system.

The second law states that the entropy of the universe increases for any spontaneous process:

\Delta S_{\mathrm{universe} = \Delta S_{\mathrm{system} + \Delta S_{\mathrm{surroundings} \gt 0

Factors Affecting Entropy

Phase changes: $S_{\mathrm{gas} \gg S_{\mathrm{liquid} \gt S_{\mathrm{solid}$ (gases have many more microstates because molecules are free to move in three dimensions)
Temperature: higher $T$ means higher $S$ (more energy is distributed among more microstates)
Number of particles: more particles means higher $S$ (more microstates for more particles)
Dissolution: dissolving a solid increases $S$ (ions are dispersed in solution)
Complexity: more complex molecules have higher $S$ (more vibrational modes)

Worked Example: Predicting Entropy Changes

Predict the sign of $\Delta S^\circ$ for each reaction:

(a) $2\mathrm{Na(s) + \mathrm{Cl_2(g) \to 2\mathrm{NaCl(s)$ : Negative (gas consumed, solid Formed)

(b) $\mathrm{CaCO_3(s) \to \mathrm{CaO(s) + \mathrm{CO_2(g)$ : Positive (gas produced)

(d) $\mathrm{NH_4\mathrm{NO_3(s) \to \mathrm{N_2\mathrm{O(g) + 2\mathrm{H_2\mathrm{O(g)$ : Positive (solid to 3 moles gas)

Standard Entropy Change

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

Note: unlike $\Delta H_f^\circ$ , $S^\circ$ is not zero for elements (elements have nonzero absolute Entropy because they are not perfect crystals at absolute zero under standard conditions).

The Third Law of Thermodynamics

The entropy of a perfect crystal at absolute zero ( $0 \mathrm{ K$ ) is zero. This provides an Absolute reference point for entropy, unlike enthalpy.

Worked Example: Standard Entropy Calculation

Calculate $\Delta S^\circ$ for the reaction $\mathrm{CaCO_3(s) \to \mathrm{CaO(s) + \mathrm{CO_2(g)$ .

Given: $S^\circ(\mathrm{CaCO_3, s) = 92.9$ , $S^\circ(\mathrm{CaO, s) = 39.7$ $S^\circ(\mathrm{CO_2, g) = 213.7 \mathrm{ J/(mol\cdot\mathrm{K)$ .

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

The positive $\Delta S^\circ$ is expected because a solid decomposes to produce a gas.

Worked Example: Entropy and Phase Changes

Calculate $\Delta S$ for the vaporisation of $1.00 \mathrm{ mol$ of water at $100^{\circ}\mathrm{C$ .

$\Delta H_{\mathrm{vap} = 40.7 \mathrm{ kJ/mol$

At the boiling point, $\Delta G = 0$ So $\Delta H = T\Delta S$ :

$\Delta S = \frac{\Delta H_{\mathrm{vap}}{T} = \frac{40700}{373.15} = 109.1 \mathrm{ J/(mol\cdot\mathrm{K)$

This positive entropy change confirms that the gas phase has more disorder than the liquid phase.

Worked Example: Entropy of Surroundings

For the reaction $\mathrm{NH_4\mathrm{NO_3(s) \to \mathrm{N_2\mathrm{O(g) + 2\mathrm{H_2\mathrm{O(g)$ at $298 \mathrm{ K$ :

$\Delta H^\circ = -36.0 \mathrm{ kJ/mol$ (exothermic, heat released to surroundings).

$\Delta S_{\mathrm{surroundings} = \frac{-\Delta H^\circ}{T} = \frac{36000}{298} = 120.8 \mathrm{ J/(mol\cdot\mathrm{K)$

$\Delta S_{\mathrm{system} = 439 \mathrm{ J/(mol\cdot\mathrm{K)$ (given)

$\Delta S_{\mathrm{universe} = 439 + 120.8 = 559.8 \mathrm{ J/(mol\cdot\mathrm{K) \gt 0$

The reaction is spontaneous because $\Delta S_{\mathrm{universe} \gt 0$ . The positive $\Delta S_{\mathrm{system}$ (more gas molecules produced) and the positive $\Delta S_{\mathrm{surroundings}$ (exothermic) both contribute.

Gibbs Free Energy (CED Unit 9)

Definition

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

The Gibbs free energy combines enthalpy and entropy into a single criterion for spontaneity. A Negative $\Delta G$ means the process is spontaneous at constant temperature and pressure.

Spontaneity at Constant Temperature and Pressure

Condition	Spontaneity
$\Delta G \lt 0$	Spontaneous
$\Delta G = 0$	At equilibrium
$\Delta G \gt 0$	Nonspontaneous

Standard Gibbs Free Energy of Formation

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

Relationship to the Equilibrium Constant

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

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

$K$	$\Delta G^\circ$	Favorability
$K \gg 1$	$\Delta G^\circ \ll 0$	Products favored
$K = 1$	$\Delta G^\circ = 0$	Neither
$K \ll 1$	$\Delta G^\circ \gg 0$	Reactants favored

This equation bridges thermodynamics and equilibrium: the equilibrium constant is determined by the Standard free energy change.

Derivation: Non-Standard Gibbs Free Energy

\Delta G = \Delta G^\circ + RT\ln Q

Where $Q$ is the reaction quotient. At equilibrium, $\Delta G = 0$ and $Q = K$ Giving $\Delta G^\circ = -RT\ln K$ .

Temperature Dependence of Spontaneity

$\Delta H$	$\Delta S$	Spontaneous?	Temperature Effect
Negative	Positive	Always	Spontaneous at all $T$
Positive	Negative	Never	Nonspontaneous at all $T$
Negative	Negative	Low $T$	Becomes nonspontaneous above $T = \frac{\Delta H}{\Delta S}$
Positive	Positive	High $T$	Becomes spontaneous above $T = \frac{\Delta H}{\Delta S}$

Temperature at Which Spontaneity Changes

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

At this temperature, $\Delta G = 0$ and $K = 1$ .

Worked Example: Gibbs Free Energy Calculation

For the reaction $\mathrm{N_2(g) + 3\mathrm{H_2(g) \to 2\mathrm{NH_3(g)$ at $298 \mathrm{ K$ :

$\Delta H^\circ = -92.2 \mathrm{ kJ/mol$ $\Delta S^\circ = -198.8 \mathrm{ J/(mol\cdot\mathrm{K)$ .

Calculate $\Delta G^\circ$ and $K$ .

$\Delta G^\circ = -92200 - 298(-198.8) = -92200 + 59242 = -32958 \mathrm{ J/mol = -33.0 \mathrm{ kJ/mol$

$K = e^{-\Delta G^\circ/(RT)} = e^{32958/(8.314 \times 298)} = e^{13.29} = 5.9 \times 10^5$

$K \gg 1$ Confirming the reaction strongly favours products at $298 \mathrm{ K$ .

Worked Example: Non-Standard Gibbs Free Energy

Calculate $\Delta G$ for the reaction $\mathrm{N_2(g) + 3\mathrm{H_2(g) \to 2\mathrm{NH_3(g)$ at $298 \mathrm{ K$ when $P(\mathrm{N_2) = 10.0 \mathrm{ atm$ $P(\mathrm{H_2) = 30.0 \mathrm{ atm$ , $P(\mathrm{NH_3) = 0.500 \mathrm{ atm$ .

$Q = \frac{(0.500)^2}{(10.0)(30.0)^3} = \frac{0.250}{270000} = 9.26 \times 10^{-7}$

$\Delta G = \Delta G^\circ + RT\ln Q = -33000 + (8.314)(298)\ln(9.26 \times 10^{-7})$

$= -33000 + 2478 \times (-13.89) = -33000 - 34420 = -67420 \mathrm{ J/mol = -67.4 \mathrm{ kJ/mol$

$\Delta G \lt 0$ So the reaction is spontaneous under these conditions. The high pressure of Reactants and low pressure of product drive the reaction forward.

flowchart TD
    A[4_Thermodynamics] --> 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 Table: Thermodynamic Quantities

Quantity	Symbol	Units	State Function?	Zero Reference
Internal energy	$U$	kJ	Yes	None (arbitrary)
Enthalpy	$H$	kJ	Yes	Elements in standard state
Entropy	$S$	J/(mol $\cdot$ K)	Yes	Perfect crystal at 0 K
Gibbs free energy	$G$	kJ	Yes	Elements in standard state

Summary Table: Spontaneity Criteria

$\Delta H$	$\Delta S$	Low $T$ ( $\Delta G$ )	High $T$ ( $\Delta G$ )	Spontaneous at
$-$	$+$	$-$ (spontaneous)	$-$ (spontaneous)	All $T$
$+$	$-$	$+$ (nonspontaneous)	$+$ (nonspontaneous)	Never
$-$	$-$	$-$ (spontaneous)	$+$ (nonspontaneous)	Low $T$ only
$+$	$+$	$+$ (nonspontaneous)	$-$ (spontaneous)	High $T$ only

Common Pitfalls

Confusing $\Delta U$ and $\Delta H$ . $\Delta H = \Delta U + P\Delta V$ . They are equal only when there is no gas produced/consumed or when $\Delta V = 0$ .
Wrong sign for work. In chemistry, $w = -P\Delta V$ . When a gas expands ( $\Delta V \gt 0$ ), the system does work on the surroundings ( $w \lt 0$ ).
Forgetting that $\Delta H_f^\circ = 0$ for elements in their standard state. This is a convention.
Using the wrong sign convention for calorimetry. $q_{\mathrm{rxn} = -q_{\mathrm{surroundings}$ .
Confusing entropy of the system with entropy of the universe. Spontaneity requires $\Delta S_{\mathrm{universe} \gt 0$ Not just $\Delta S_{\mathrm{system} \gt 0$ .
Incorrect units in the Gibbs equation. $\Delta H$ is in kJ/mol; $\Delta S$ is in J/(mol $\cdot$ K). Convert one of them before combining.
Using $R = 0.08206$ in $\Delta G^\circ = -RT\ln K$ . Use $R = 8.314 \mathrm{ J/(mol\cdot\mathrm{K)$ because $\Delta G^\circ$ is in J/mol.
Assuming a negative $\Delta H$ guarantees spontaneity. If $\Delta S$ is sufficiently negative, $\Delta G$ can be positive even when $\Delta H$ is negative.
Forgetting that standard conditions are $298 \mathrm{ K$ and $1 \mathrm{ atm$ Not STP.

Practice Questions

Calculate $\Delta H^\circ$ for $2\mathrm{Fe_2\mathrm{O_3(s) + 3\mathrm{C(s) \to 4\mathrm{Fe(s) + 3\mathrm{CO_2(g)$ using standard enthalpies of formation.
When $3.50 \mathrm{ g$ of $\mathrm{NaOH$ is dissolved in $100.0 \mathrm{ g$ of water in a calorimeter, the temperature rises from $23.0^\circ\mathrm{C$ to $36.5^\circ\mathrm{C$ . Calculate $\Delta H$ per mole of $\mathrm{NaOH$ .
For a reaction with $\Delta H = 125 \mathrm{ kJ/mol$ and $\Delta S = 200 \mathrm{ J/(mol\cdot\mathrm{K)$ Find the temperature range where the reaction is spontaneous.
Given $\Delta G_f^\circ$ values: $\mathrm{NO_2(g) = 51.3 \mathrm{ kJ/mol$ $\mathrm{N_2\mathrm{O_4(g) = 97.8 \mathrm{ kJ/mol$ . Find $\Delta G^\circ$ and $K$ for $2\mathrm{NO_2(g) \rightleftharpoons \mathrm{N_2\mathrm{O_4(g)$ at $298 \mathrm{ K$ .
Estimate the enthalpy of combustion of $\mathrm{CH_4$ using bond enthalpies. Compare with the value calculated from standard enthalpies of formation.
A reaction has $\Delta G^\circ = -20.0 \mathrm{ kJ/mol$ at $298 \mathrm{ K$ . Calculate $K$ .
Explain why the melting of ice is spontaneous above $0^\circ\mathrm{C$ but not below, using $\Delta G = \Delta H - T\Delta S$ .
Calculate $\Delta S^\circ$ for the reaction $2\mathrm{H_2(g) + \mathrm{O_2(g) \to 2\mathrm{H_2\mathrm{O(l)$ given: $S^\circ(\mathrm{H_2) = 130.7$ , $S^\circ(\mathrm{O_2) = 205.1$ $S^\circ(\mathrm{H_2\mathrm{O, l) = 69.9 \mathrm{ J/(mol\cdot\mathrm{K)$ .
A bomb calorimeter has $C_{\mathrm{cal} = 850 \mathrm{ J/K$ . Burning $1.00 \mathrm{ g$ of naphthalene ( $\mathrm{C_{10}\mathrm{H_8$ ) raises the temperature by $2.46 \mathrm{ K$ . Calculate the enthalpy of combustion per mole of naphthalene.
For the reaction $\mathrm{NH_4\mathrm{NO_3(s) \to \mathrm{N_2\mathrm{O(g) + 2\mathrm{H_2\mathrm{O(g)$ $\Delta H^\circ = -36.0 \mathrm{ kJ/mol$ and $\Delta S^\circ = 439 \mathrm{ J/(mol\cdot\mathrm{K)$ . Is the reaction spontaneous at $298 \mathrm{ K$ ? At what temperature does it become nonspontaneous?
Explain why the dissolution of $\mathrm{NH_4\mathrm{NO_3$ in water is endothermic yet spontaneous at room temperature.
Calculate $\Delta G$ (not $\Delta G^\circ$ ) for the reaction $\mathrm{N_2\mathrm{O_4(g) \rightleftharpoons 2\mathrm{NO_2(g)$ at $298 \mathrm{ K$ when $P(\mathrm{N_2\mathrm{O_4) = 0.50 \mathrm{ atm$ and $P(\mathrm{NO_2) = 0.10 \mathrm{ atm$ . Given: $\Delta G^\circ = 4.72 \mathrm{ kJ/mol$ .
Predict the sign of $\Delta S^\circ$ for each reaction and explain your reasoning: (a) $2\mathrm{Na(s) + \mathrm{Cl_2(g) \to 2\mathrm{NaCl(s)$ (b) $\mathrm{CaCO_3(s) \to \mathrm{CaO(s) + \mathrm{CO_2(g)$
Calculate $\Delta H^\circ$ for the reaction $\mathrm{N_2(g) + 2\mathrm{O_2(g) \to 2\mathrm{NO_2(g)$ using the following data: $\frac{1}{2}\mathrm{N_2(g) + \mathrm{O_2(g) \to \mathrm{NO_2(g)$ $\Delta H^\circ = 33.2 \mathrm{ kJ/mol$ .
A student calculates $\Delta G^\circ = -15 \mathrm{ kJ/mol$ for a reaction at $298 \mathrm{ K$ and concludes that the reaction will reach completion. Explain why this conclusion may not be justified.
Calculate $\Delta G^\circ$ at $500 \mathrm{ K$ for a reaction with $\Delta H^\circ = -50 \mathrm{ kJ/mol$ and $\Delta S^\circ = -80 \mathrm{ J/(mol\cdot\mathrm{K)$ . Is the reaction spontaneous at this temperature?
Using the data below, calculate the standard enthalpy change for the reaction $\mathrm{C_2\mathrm{H_5\mathrm{OH(l) + 3\mathrm{O_2(g) \to 2\mathrm{CO_2(g) + 3\mathrm{H_2\mathrm{O(l)$ . $\Delta H_f^\circ(\mathrm{C_2\mathrm{H_5\mathrm{OH, l) = -277.7 \mathrm{ kJ/mol$ .
Explain why a reaction with $\Delta H \gt 0$ and $\Delta S \gt 0$ is nonspontaneous at low temperatures but becomes spontaneous at high temperatures.
Calculate the boiling point of $\mathrm{Br_2$ given that $\mathrm{Br_2(l) \to \mathrm{Br_2(g)$ has $\Delta H^\circ = 30.9 \mathrm{ kJ/mol$ and $\Delta S^\circ = 93.2 \mathrm{ J/(mol\cdot\mathrm{K)$ .
A calorimeter contains $200 \mathrm{ g$ of water at $25.0^\circ\mathrm{C$ . When $5.00 \mathrm{ g$ of $\mathrm{KOH$ is dissolved, the temperature rises to $35.0^\circ\mathrm{C$ . Calculate the enthalpy of solution of $\mathrm{KOH$ in kJ/mol.
For the reaction $2\mathrm{NO(g) + \mathrm{O_2(g) \to 2\mathrm{NO_2(g)$ Given $\Delta H^\circ = -114.1 \mathrm{ kJ/mol$ and $\Delta S^\circ = -146.5 \mathrm{ J/(mol\cdot\mathrm{K)$ Calculate the temperature above which the reaction is no longer spontaneous.
Using the following data, calculate $\Delta S^\circ$ for the reaction $4\mathrm{Fe(s) + 3\mathrm{O_2(g) \to 2\mathrm{Fe_2\mathrm{O_3(s)$ : $S^\circ(\mathrm{Fe, s) = 27.3$ , $S^\circ(\mathrm{O_2, g) = 205.1$ $S^\circ(\mathrm{Fe_2\mathrm{O_3, s) = 87.4 \mathrm{ J/(mol\cdot\mathrm{K)$ .
Calculate the normal boiling point of chloroform ( $\mathrm{CHCl_3$ ) given that $\mathrm{CHCl_3(l) \to \mathrm{CHCl_3(g)$ has $\Delta H^\circ = 31.4 \mathrm{ kJ/mol$ and $\Delta S^\circ = 94.2 \mathrm{ J/(mol\cdot\mathrm{K)$ .
For the reaction $\mathrm{C(s) + \mathrm{H_2\mathrm{O(g) \to \mathrm{CO(g) + \mathrm{H_2(g)$ $\Delta H^\circ = 131.3 \mathrm{ kJ/mol$ and $\Delta S^\circ = 133.7 \mathrm{ J/(mol\cdot\mathrm{K)$ . Calculate the minimum temperature at which this reaction becomes spontaneous.
Explain why the following statement is incorrect: “An exothermic reaction is always spontaneous.”
A bomb calorimeter with $C_{\mathrm{cal} = 950 \mathrm{ J/K$ is used to determine the enthalpy of combustion of benzoic acid ( $\mathrm{C_7\mathrm{H_6\mathrm{O_2$ ). Burning $1.00 \mathrm{ g$ raises the temperature by $3.24 \mathrm{ K$ . Calculate the enthalpy of combustion per mole.
Calculate $\Delta G^\circ$ and $K$ at $298 \mathrm{ K$ for the reaction $\mathrm{H_2(g) + \mathrm{I_2(g) \to 2\mathrm{HI(g)$ given: $\Delta G_f^\circ(\mathrm{HI, g) = 1.7 \mathrm{ kJ/mol$ .
For a certain reaction, $\Delta G^\circ = -5.4 \mathrm{ kJ/mol$ at $300 \mathrm{ K$ . Calculate $K$ at this temperature and determine whether products or reactants are favoured.
Calculate the work done when $2.00 \mathrm{ mol$ of a gas expands from $5.0 \mathrm{ L$ to $15.0 \mathrm{ L$ against a constant external pressure of $1.00 \mathrm{ atm$ .
Explain, using thermodynamic principles, why ice melts spontaneously at temperatures above $0^{\circ}\mathrm{C$ even though the process is endothermic.
Calculate the work done by the system and $\Delta U$ when $3.00 \mathrm{ L$ of gas at $2.00 \mathrm{ atm$ expands against a constant external pressure of $0.50 \mathrm{ atm$ to a final volume of $8.00 \mathrm{ L$ .
Given $\Delta H_f^\circ(\mathrm{NH_3, g) = -46.1 \mathrm{ kJ/mol$ $\Delta H_f^\circ(\mathrm{NO, g) = 90.3 \mathrm{ kJ/mol$ And $\Delta H_f^\circ(\mathrm{H_2\mathrm{O, g) = -241.8 \mathrm{ kJ/mol$ Calculate $\Delta H^\circ$ for the reaction $4\mathrm{NH_3(g) + 5\mathrm{O_2(g) \to 4\mathrm{NO(g) + 6\mathrm{H_2\mathrm{O(g)$ .
The standard entropy values are: $S^\circ(\mathrm{C, s) = 5.7$ $S^\circ(\mathrm{CO_2, g) = 213.7$ $S^\circ(\mathrm{CO, g) = 197.7 \mathrm{ J/(mol\cdot\mathrm{K)$ . Calculate $\Delta S^\circ$ for the reaction $\mathrm{C(s) + \mathrm{CO_2(g) \to 2\mathrm{CO(g)$ and comment on the sign.

Practice Problems

Question 1: Hess's law and enthalpy of formation

Given the following data, calculate the standard enthalpy of formation of $\mathrm{CH_3\mathrm{OH(l)$ :

$\mathrm{C(s) + \mathrm{O_2(g) \to \mathrm{CO_2(g)$ , $\Delta H^\circ = -393.5 \mathrm{ kJ/mol$
$\mathrm{H_2(g) + \frac{1}{2}\mathrm{O_2(g) \to \mathrm{H_2\mathrm{O(l)$ $\Delta H^\circ = -285.8 \mathrm{ kJ/mol$
$\mathrm{CH_3\mathrm{OH(l) + \frac{3}{2}\mathrm{O_2(g) \to \mathrm{CO_2(g) + 2\mathrm{H_2\mathrm{O(l)$ $\Delta H^\circ = -726.4 \mathrm{ kJ/mol$

Answer

Target: $\mathrm{C(s) + 2\mathrm{H_2(g) + \frac{1}{2}\mathrm{O_2(g) \to \mathrm{CH_3\mathrm{OH(l)$

Manipulate the given equations:

(1) $\mathrm{C(s) + \mathrm{O_2(g) \to \mathrm{CO_2(g)$ $\Delta H^\circ = -393.5 \mathrm{ kJ/mol$ — keep

(2) $2\mathrm{H_2(g) + \mathrm{O_2(g) \to 2\mathrm{H_2\mathrm{O(l)$ $\Delta H^\circ = 2(-285.8) = -571.6 \mathrm{ kJ/mol$ — multiply by 2

(3) $\mathrm{CO_2(g) + 2\mathrm{H_2\mathrm{O(l) \to \mathrm{CH_3\mathrm{OH(l) + \frac{3}{2}\mathrm{O_2(g)$ $\Delta H^\circ = +726.4 \mathrm{ kJ/mol$ — reverse

Add (1) + (2) + (3):

$\mathrm{C(s) + 2\mathrm{H_2(g) + \frac{1}{2}\mathrm{O_2(g) \to \mathrm{CH_3\mathrm{OH(l)$

$\Delta H_f^\circ = -393.5 + (-571.6) + 726.4 = -238.7 \mathrm{ kJ/mol$ .

Question 2: Gibbs free energy and spontaneity

For the reaction $\mathrm{NH_4\mathrm{NO_3(s) \to \mathrm{N_2\mathrm{O(g) + 2\mathrm{H_2\mathrm{O(g)$ $\Delta H^\circ = -36.0 \mathrm{ kJ/mol$ and $\Delta S^\circ = 347 \mathrm{ J/(mol\cdot K)$ . Calculate $\Delta G^\circ$ at $298 \mathrm{ K$ and determine the temperature range over which the Reaction is spontaneous.

Answer

$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -36,000 - 298 \times 347 = -36,000 - 103,406 = -139,406 \mathrm{ J/mol = -139.4 \mathrm{ kJ/mol$ .

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

The reaction is spontaneous when $\Delta G \lt 0$ :

$\Delta H - T\Delta S \lt 0$

Since $\Delta H = -36.0 \mathrm{ kJ/mol$ (negative) and $\Delta S = +347 \mathrm{ J/(mol\cdot K)$ (positive), both terms favour spontaneity. The reaction is spontaneous at all temperatures. There is No upper temperature limit because the $-T\Delta S$ term always contributes negatively to $\Delta G$ When $\Delta S$ is positive.

Question 3: Calorimetry and specific heat

A $50.0 \mathrm{ g$ sample of an unknown metal is heated to $100.0^\circ\mathrm{C$ and then placed In $100.0 \mathrm{ g$ of water at $25.0^\circ\mathrm{C$ in a coffee-cup calorimeter. The final Temperature of the mixture is $28.8^\circ\mathrm{C$ . Calculate the specific heat capacity of the Metal. Assume no heat loss to the calorimeter.

Answer

Heat gained by water = heat lost by metal.

$q_{\mathrm{water} = m_{\mathrm{water} \times c_{\mathrm{water} \times \Delta T_{\mathrm{water} = 100.0 \times 4.184 \times (28.8 - 25.0) = 100.0 \times 4.184 \times 3.8 = 1589.9 \mathrm{ J$ .

$q_{\mathrm{metal} = m_{\mathrm{metal} \times c_{\mathrm{metal} \times \Delta T_{\mathrm{metal} = 50.0 \times c_{\mathrm{metal} \times (100.0 - 28.8) = 50.0 \times c_{\mathrm{metal} \times 71.2$ .

Setting equal: $1589.9 = 50.0 \times c_{\mathrm{metal} \times 71.2$ .

$c_{\mathrm{metal} = 1589.9 / 3560 = 0.447 \mathrm{ J/(g\cdot^\circ\mathrm{C)}$ .

This value is close to that of iron ( $0.449 \mathrm{ J/(g\cdot^\circ\mathrm{C)}$ ), suggesting the Unknown metal may be iron.

Question 4: Entropy change of surroundings

For the vaporisation of water at $100^\circ\mathrm{C$ and $1 \mathrm{ atm$ : $\mathrm{H_2\mathrm{O(l) \to \mathrm{H_2\mathrm{O(g)$ $\Delta H_{\mathrm{vap} = 40.7 \mathrm{ kJ/mol$ . Calculate $\Delta S_{\mathrm{system}$ $\Delta S_{\mathrm{surroundings}$ And $\Delta S_{\mathrm{universe}$ . Is the process spontaneous At this temperature?

Answer

$\Delta S_{\mathrm{system} = \Delta H_{\mathrm{vap} / T = 40,700 / 373 = 109.1 \mathrm{ J/(mol\cdot K)$ .

$\Delta S_{\mathrm{surroundings} = -\Delta H_{\mathrm{vap} / T = -40,700 / 373 = -109.1 \mathrm{ J/(mol\cdot K)$ .

$\Delta S_{\mathrm{universe} = \Delta S_{\mathrm{system} + \Delta S_{\mathrm{surroundings} = 109.1 + (-109.1) = 0 \mathrm{ J/(mol\cdot K)$ .

At $100^\circ\mathrm{C$ and $1 \mathrm{ atm$ Liquid and gaseous water are in equilibrium, so $\Delta G = 0$ and $\Delta S_{\mathrm{universe} = 0$ . The process is at equilibrium, not Spontaneous in either direction. Above $100^\circ\mathrm{C$ Vaporisation becomes spontaneous ( $\Delta S_{\mathrm{universe} \gt 0$ ).

Question 5: Bond enthalpy calculation

Using the following average bond enthalpies, estimate $\Delta H$ for the reaction $\mathrm{CH_4(g) + 2\mathrm{Cl_2(g) \to \mathrm{CH_2\mathrm{Cl_2(g) + 2\mathrm{HCl(g)$ :

C-H: $413 \mathrm{ kJ/mol$ Cl-Cl: $242 \mathrm{ kJ/mol$ C-Cl: $339 \mathrm{ kJ/mol$ H-Cl: $431 \mathrm{ kJ/mol$ .

Answer

Bonds broken (reactants):

4 C-H bonds in $\mathrm{CH_4$ : $4 \times 413 = 1652 \mathrm{ kJ/mol$
2 Cl-Cl bonds: $2 \times 242 = 484 \mathrm{ kJ/mol$

Wait — not all C-H bonds break. In $\mathrm{CH_2\mathrm{Cl_2$ Two C-H bonds remain. So only 2 C-H bonds break.

Corrected bonds broken:

2 C-H bonds: $2 \times 413 = 826 \mathrm{ kJ/mol$
2 Cl-Cl bonds: $2 \times 242 = 484 \mathrm{ kJ/mol$
Total broken: $826 + 484 = 1310 \mathrm{ kJ/mol$

Bonds formed (products):

2 C-Cl bonds: $2 \times 339 = 678 \mathrm{ kJ/mol$
2 H-Cl bonds: $2 \times 431 = 862 \mathrm{ kJ/mol$
Total formed: $678 + 862 = 1540 \mathrm{ kJ/mol$

$\Delta H = \mathrm{bonds broken - \mathrm{bonds formed = 1310 - 1540 = -230 \mathrm{ kJ/mol$ .

The reaction is exothermic because stronger bonds (H-Cl, C-Cl) are formed than are broken (C-H, Cl-Cl).

Worked Examples

Example 1: Conservation of energy

A $0.50\,\text{kg}$ ball is dropped from a height of $20\,\text{m}$ . Calculate its speed just before it hits the ground (ignore air resistance).

Solution:

Using conservation of energy: $mgh = \frac{1}{2}mv^2$

v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 20} = \sqrt{392.4} \approx 19.8\,\text{m\,s}^{-1}

Intuition

Physics explores the fundamental rules governing matter, energy, space, and time. At its heart lies the principle that complex phenomena emerge from simple interactions - gravity shapes orbits, electromagnetism binds atoms, and quantum mechanics governs the subatomic realm. Understanding these laws allows us to build technologies from smartphones to spacecraft and to comprehend our place in the cosmos.

Cross-References

Mechanics
Waves
Electricity
Fields
AP Physics — Work, Energy, and Power: The work-energy theorem from physics underpins enthalpy and internal energy calculations in thermodynamics.

Mark this page as reviewed

Thermodynamics | AP - Wyatt's Notes

The First Law of Thermodynamics (CED Unit 6)

Sign Convention (Chemistry)

Pressure-Volume Work

Work at Constant Pressure: Enthalpy

Derivation: Why ΔH=qP\Delta H = q_PΔH=qP​

Enthalpy of Reaction (CED Unit 6)

Standard Enthalpy of Formation (ΔHf∘\Delta H_f^\circΔHf∘​)

Hess”s Law

Worked Example: Hess’s Law

Worked Example: Hess’s Law with Multiple Steps

Worked Example: Bond Enthalpy Calculation

Worked Example: Gibbs Free Energy Calculation

Worked Example: Non-Standard Gibbs Free Energy

Summary Table: Thermodynamic Quantities

Summary Table: Spontaneity Criteria

Common Pitfalls

Practice Questions

Practice Problems

Worked Examples

Intuition

Cross-References

Derivation: Why $\Delta H = q_P$

Standard Enthalpy of Formation ( $\Delta H_f^\circ$ )