Thermodynamics & Energetics

Fundamental Concepts

System, Surroundings, and the Universe

A thermodynamic system is the part of the universe under study. The surroundings are everything else. The universe is the system plus the surroundings.

• Open system: exchanges both matter and energy with surroundings.
• Closed system: exchanges energy but not matter.
• Isolated system: exchanges neither matter nor energy.

Enthalpy ($H$)

Enthalpy is a thermodynamic state function defined as:

$$H = U + pV$$

Where $U$ is internal energy, $p$ is pressure, and $V$ is volume. For constant-pressure processes (the usual condition in chemistry):

$$\Delta H = q_p$$

The enthalpy change $\Delta H$ is the heat exchanged at constant pressure. It cannot be measured absolutely; only changes in enthalpy are measurable.

Internal Energy ($U$)

The internal energy of a system is the total kinetic and potential energy of all particles within the system. At constant volume:

$$\Delta U = q_V$$

The relationship between $\Delta H$ and $\Delta U$:

$$\Delta H = \Delta U + \Delta(pV)$$

For reactions involving only solids and liquids, $\Delta(pV) \approx 0$So $\Delta H \approx \Delta U$. For reactions involving gases:

$$\Delta(pV) = \Delta n_g RT$$

Where $\Delta n_g$ is the change in moles of gas.

Standard Conditions

Standard enthalpy changes are measured under the following conditions:

• Pressure: $100\,\mathrm{kPa}$ ($1\,\mathrm{bar}$Approximately $1\,\mathrm{atm}$).
• Temperature: $298\,\mathrm{K}$ ($25^\circ\mathrm{C}$), unless otherwise stated.
• Concentration: $1\,\mathrm{mol/dm}^3$ for solutions.
• All substances in their standard states (most stable form at the specified conditions).

The standard symbol is $\Delta H^\circ$ with the superscript circle.

Exothermic and Endothermic Reactions

• Exothermic: $\Delta H \lt 0$; heat is released to the surroundings.
• Endothermic: $\Delta H \gt 0$; heat is absorbed from the surroundings.

Hess’s Law

Statement

Hess’s Law states that the enthalpy change for a chemical reaction is the same regardless of the route by which the reaction occurs, provided the initial and final conditions are the same.

This is a direct consequence of enthalpy being a state function — its value depends only on the current state of the system, not on the path taken to reach it.

Application: Indirect Determination of Enthalpy Changes

Worked Example 1. Calculate $\Delta H_f^\circ$ of $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}(l)$.

Given:

• $\Delta H_c^\circ(\mathrm{C}) = -394\,\mathrm{kJ/mol}$
• $\Delta H_c^\circ(\mathrm{H}_2) = -286\,\mathrm{kJ/mol}$
• $\Delta H_c^\circ(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}) = -1367\,\mathrm{kJ/mol}$

Formation: $2\mathrm{C}(s) + 3\mathrm{H}_2(g) + \tfrac{1}{2}\mathrm{O}_2(g) \to \mathrm{C}_2\mathrm{H}_5\mathrm{OH}(l)$

Combustion route: $2\mathrm{C}(s) + 3\mathrm{H}_2(g) + 3\frac{1}{2}\mathrm{O}_2(g) \to 2\mathrm{CO}_2(g) + 3\mathrm{H}_2\mathrm{O}(l)$

By Hess’s Law:

$$\Delta H_f^\circ + \Delta H_c^\circ(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}) = 2\Delta H_c^\circ(\mathrm{C}) + 3\Delta H_c^\circ(\mathrm{H}_2)$$ $$\Delta H_f^\circ = 2(-394) + 3(-286) - (-1367) = -788 - 858 + 1367 = -279\,\mathrm{kJ/mol}$$

Standard Enthalpy Changes

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

The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions.

• $\Delta H_f^\circ$ of an element in its standard state is zero by definition.
• Example: $\Delta H_f^\circ(\mathrm{H}_2\mathrm{O}(l)) = -286\,\mathrm{kJ/mol}$.

Standard Enthalpy of Combustion ($\Delta H_c^\circ$)

The enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions.

• Combustion reactions are always exothermic ($\Delta H_c^\circ \lt 0$).
• The products must be in their standard states ($\mathrm{CO}_2(g)$, $\mathrm{H}_2\mathrm{O}(l)$Etc.).

Standard Enthalpy of Atomisation ($\Delta H_\mathrm{at}^\circ$)

The enthalpy change when one mole of gaseous atoms is formed from an element in its standard state.

For a diatomic molecule $\mathrm{X}_2$:

$$\Delta H_\mathrm{at}^\circ = \frac{1}{2} \times \mathrm{bond dissociation enthalpy of } \mathrm{X}_2$$

Example: $\Delta H_\mathrm{at}^\circ(\mathrm{Cl}_2) = \tfrac{1}{2}(242) = 121\,\mathrm{kJ/mol}$.

For a solid element (e.g. Na), atomisation includes sublimation and bond breaking:

$$\mathrm{Na}(s) \to \mathrm{Na}(g) \quad \Delta H_\mathrm{at}^\circ = +108\,\mathrm{kJ/mol}$$

Bond Dissociation Enthalpy

The bond dissociation enthalpy is the enthalpy change required to break one mole of a specific bond in a specific molecule in the gaseous phase.

For a diatomic molecule, this is unambiguous. For polyatomic molecules, each bond-breaking step may have a different enthalpy. Mean bond enthalpies are averages across a range of compounds.

Note: $\mathrm{C-H}$ bond energy in $\mathrm{CH}_4$ is not the same as in $\mathrm{CH}_3\mathrm{Cl}$. Mean bond enthalpies introduce systematic error in thermochemical calculations.

Lattice Enthalpy

Definition

The lattice enthalpy $\Delta H_\mathrm{lat}^\circ$ is the enthalpy change when one mole of an ionic solid is formed from its constituent gaseous ions under standard conditions. This is always exothermic ($\Delta H_\mathrm{lat}^\circ \lt 0$).

The reverse process (separating one mole of solid into gaseous ions) is the lattice dissociation enthalpy and is always endothermic.

Born-Haber Cycle

The Born-Haber cycle for an ionic compound $\mathrm{M}_a\mathrm{X}_b$ constructs a thermochemical cycle:

$$\Delta H_f^\circ = a\Delta H_\mathrm{at}^\circ(\mathrm{M}) + b\Delta H_\mathrm{at}^\circ(\mathrm{X}) + \sum \mathrm{IE}(\mathrm{M}) + \sum \mathrm{EA}(\mathrm{X}) + \Delta H_\mathrm{lat}^\circ$$

Rearranging gives the lattice enthalpy.

Worked Example. Calculate the lattice enthalpy of $\mathrm{MgO}$ given:

• $\Delta H_f^\circ(\mathrm{MgO}) = -602\,\mathrm{kJ/mol}$
• $\Delta H_\mathrm{at}^\circ(\mathrm{Mg}) = +148\,\mathrm{kJ/mol}$
• $\Delta H_\mathrm{at}^\circ(\mathrm{O}) = +248\,\mathrm{kJ/mol}$ (per atom; for $\tfrac{1}{2}\mathrm{O}_2$)
• $\mathrm{IE}_1(\mathrm{Mg}) = +738\,\mathrm{kJ/mol}$
• $\mathrm{IE}_2(\mathrm{Mg}) = +1451\,\mathrm{kJ/mol}$
• $\mathrm{EA}_1(\mathrm{O}) = -141\,\mathrm{kJ/mol}$
• $\mathrm{EA}_2(\mathrm{O}) = +798\,\mathrm{kJ/mol}$

$$\Delta H_\mathrm{lat}^\circ = \Delta H_f^\circ - \Delta H_\mathrm{at}^\circ(\mathrm{Mg}) - \Delta H_\mathrm{at}^\circ(\mathrm{O}) - \mathrm{IE}_1 - \mathrm{IE}_2 - \mathrm{EA}_1 - \mathrm{EA}_2$$ $$\Delta H_\mathrm{lat}^\circ = -602 - 148 - 248 - 738 - 1451 - (-141) - 798$$ $$\Delta H_\mathrm{lat}^\circ = -602 - 148 - 248 - 738 - 1451 + 141 - 798 = -3844\,\mathrm{kJ/mol}$$

Theoretical Lattice Enthalpy: Born-Lande Equation

The theoretical lattice enthalpy is calculated from the electrostatic model:

$$\Delta H_\mathrm{lat}^\circ \approx -\frac◆LB◆N_A M z^+ z^- e^2◆RB◆◆LB◆4\pi\varepsilon_0 r_0◆RB◆\left(1 - \frac{1}{n}\right)$$

Where:

• $N_A$ is Avogadro’s constant
• $M$ is the Madelung constant (depends on the lattice geometry)
• $z^+$, $z^-$ are the ion charges
• $e$ is the elementary charge
• $\varepsilon_0$ is the permittivity of free space
• $r_0$ is the sum of the ionic radii
• $n$ is the Born exponent (related to the electron configuration of the ions)

This equation predicts purely ionic bonding. Differences between theoretical and experimental (Born-Haber) lattice enthalpies indicate the degree of covalent character in the ionic bond.

Polarisation and Covalent Character (Fajans’ Rules)

A small, highly charged cation (high charge density) can polarise a large, deformable anion, drawing electron density towards itself and introducing covalent character.

Fajans’ Rules predict increasing covalent character when:

1. The cation is small and/or highly charged (e.g. $\mathrm{Al}^{3+}$).
2. The anion is large and/or highly charged (e.g. $\mathrm{I}^-$).
3. The cation has an electronic configuration resembling a noble gas with no inert pair (e.g. $\mathrm{Li}^+$ vs $\mathrm{Tl}^+$).

Example: $\mathrm{AgI}$ has significant covalent character because $\mathrm{Ag}^+$ is a relatively soft cation and $\mathrm{I}^-$ is a large, polarisable anion. The experimental lattice enthalpy is less exothermic than the theoretical value.

Entropy ($S$)

Definition

Entropy is a measure of the number of ways energy can be distributed among the particles of a system — a measure of disorder or randomness.

The second law of thermodynamics states that the total entropy of the universe increases in any spontaneous process:

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

Factors Affecting Entropy

Factor	Effect on $S$
More particles (gaseous products)	Increase
Higher temperature	Increase
Change from solid to liquid to gas	Increase
Dissolution of a solid (if ions are dispersed)	Increase
Fewer moles of gas	Decrease

Standard Entropy Change ($\Delta S^\circ$)

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

Standard entropies are always positive (absolute values, not relative like enthalpy).

Worked Example. Calculate $\Delta S^\circ$ for:

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

Given: S^\circ(\mathrm{CaCO}_3) = 92.9\,\mathrm{J\,mol^{-1}\,K^{-1}}$$S^\circ(\mathrm{CaO}) = 38.1\,\mathrm{J\,mol^{-1}\,K^{-1}}$$S^\circ(\mathrm{CO}_2) = 213.7\,\mathrm{J\,mol^{-1}\,K^{-1}}.

$$\Delta S^\circ = (38.1 + 213.7) - 92.9 = 251.8 - 92.9 = +158.9\,\mathrm{J\,mol^{-1}\,K^{-1}}$$

The entropy increases because a gas is produced from a solid (more microstates).

Entropy Change of the Surroundings

$$\Delta S_\mathrm{surroundings} = -\frac◆LB◆\Delta H◆RB◆◆LB◆T◆RB◆$$

For an exothermic reaction ($\Delta H \lt 0$), $\Delta S_\mathrm{surroundings} \gt 0$ (the surroundings gain heat, increasing their disorder).

Gibbs Free Energy ($G$)

Definition

The Gibbs free energy combines enthalpy and entropy into a single criterion for spontaneity:

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

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

This is valid at constant temperature and pressure (standard laboratory conditions).

Standard Gibbs Free Energy Change ($\Delta G^\circ$)

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

The relationship between $\Delta G^\circ$ and the equilibrium constant:

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

Where $R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ and $T$ is in Kelvin.

When $\Delta G^\circ = 0$: $K = 1$ (equal amounts of reactants and products at equilibrium).

When $\Delta G^\circ \lt 0$: $K \gt 1$ (products favoured).

When $\Delta G^\circ \gt 0$: $K \lt 1$ (reactants favoured).

Temperature Dependence of Spontaneity

$\Delta H$	$\Delta S$	Spontaneity
$-$ (exothermic)	$+$ (entropy increases)	Spontaneous at all temperatures
$-$ (exothermic)	$-$ (entropy decreases)	Spontaneous at low temperatures
$+$ (endothermic)	$+$ (entropy increases)	Spontaneous at high temperatures
$+$ (endothermic)	$-$ (entropy decreases)	Non-spontaneous at all temperatures

Worked Example. For the thermal decomposition of $\mathrm{CaCO}_3$:

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

\Delta H^\circ = +178\,\mathrm{kJ/mol}$$\Delta S^\circ = +160\,\mathrm{J\,mol^{-1}\,K^{-1}}.

At what temperature does the reaction become spontaneous?

$$\Delta G^\circ = 0 \implies T = \frac◆LB◆\Delta H^\circ◆RB◆◆LB◆\Delta S^\circ◆RB◆ = \frac◆LB◆178 \times 10^3◆RB◆◆LB◆160◆RB◆ = 1113\,\mathrm{K} = 840^\circ\mathrm{C}$$

Above 1113\,\mathrm{K}$$\Delta G^\circ \lt 0 and the decomposition is spontaneous.

Gibbs-Helmholtz Equation

$$\left(\frac◆LB◆\partial (\Delta G/T)◆RB◆◆LB◆\partial T◆RB◆\right)_p = -\frac◆LB◆\Delta H◆RB◆◆LB◆T^2◆RB◆$$

This shows how $\Delta G/T$ varies with temperature. In the simplified form used at A-Level, it justifies the linear relationship:

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

A plot of $\Delta G$ vs $T$ is linear with gradient $= -\Delta S$ and y-intercept $= \Delta H$.

Gibbs Free Energy and Equilibrium

The relationship between $\Delta G^\circ$ and the equilibrium constant $K$ is one of the most important equations in physical chemistry:

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

This equation allows prediction of the equilibrium position from thermodynamic data:

• When $\Delta G^\circ \lt 0$: $K \gt 1$Products are favoured at equilibrium.
• When $\Delta G^\circ = 0$: $K = 1$Reactants and products are present in equal amounts.
• When $\Delta G^\circ \gt 0$: $K \lt 1$Reactants are favoured at equilibrium.

Worked Example. Calculate $K$ at $298\,\mathrm{K}$ for the reaction $\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2\mathrm{NO}_2(g)$ given $\Delta G^\circ = +4.72\,\mathrm{kJ/mol}$.

$$K = \exp\left(\frac◆LB◆-\Delta G^\circ◆RB◆◆LB◆RT◆RB◆\right) = \exp\left(\frac◆LB◆-4720◆RB◆◆LB◆8.314 \times 298◆RB◆\right) = \exp(-1.904) = 0.149$$

$K = 0.149$So reactants ($\mathrm{N}_2\mathrm{O}_4$) are favoured at equilibrium.

Using $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ with the equilibrium expression:

Combining the two equations:

$$-RT\ln K = \Delta H^\circ - T\Delta S^\circ$$ $$\ln K = -\frac◆LB◆\Delta H^\circ◆RB◆◆LB◆RT◆RB◆ + \frac◆LB◆\Delta S^\circ◆RB◆◆LB◆R◆RB◆$$

This is the van ‘t Hoff equation (linear form), which is identical in structure to the Arrhenius equation.

Calorimetry

Experimental Determination of Enthalpy Changes

Coffee-cup calorimetry: A polystyrene cup minimises heat exchange with the surroundings. The heat absorbed by the solution equals the heat released by the reaction:

$$Q = mc\Delta T$$

Where $m$ is the mass of solution (g), $c$ is the specific heat capacity ($4.18\,\mathrm{J\,g^{-1}\,K^{-1}}$ for water), and $\Delta T$ is the temperature change.

Sources of error in calorimetry:

1. Heat loss to the surroundings (exothermic reactions underestimate $\Delta H$; endothermic reactions overestimate it).
2. The calorimeter itself absorbs heat ($q_\mathrm{calorimeter} = C_\mathrm{cal}\Delta T$). For accurate work, include this.
3. Incomplete reaction (if the reaction does not go to completion).
4. Non-standard conditions (reactions are not at $298\,\mathrm{K}$ or $100\,\mathrm{kPa}$).

Hess’s Law Cycles with Mean Bond Enthalpies

Mean bond enthalpies are averages across different molecules. They introduce systematic error because the actual bond enthalpy in a specific molecule may differ from the mean.

Example of systematic error: The C—O bond enthalpy in $\mathrm{CH}_3\mathrm{OH}$ differs from the mean C—O bond enthalpy (which averages C—O bonds in alcohols, ethers, esters, carboxylic acids). Using mean values for a specific molecule gives only an estimate of $\Delta H$.

Rule: Calculations using mean bond enthalpies give estimated values. Calculations using standard enthalpies of formation give accurate values (provided the data are reliable). Always use formation enthalpies in preference to bond enthalpies when both are available.

Enthalpy of Neutralisation

The enthalpy of neutralisation is the enthalpy change when one mole of water is formed from the reaction of an acid and a base:

$$\mathrm{H}^+(aq) + \mathrm{OH}^-(aq) \to \mathrm{H}_2\mathrm{O}(l) \quad \Delta H^\circ_\mathrm{neut} = -57.9\,\mathrm{kJ/mol}$$

This value is approximately constant for all strong acid-strong base reactions because the net ionic equation is always the same. Weak acid-strong base or strong acid-weak base neutralisations have less exothermic values because some energy is consumed in dissociating the weak acid or weak base.

Reaction type	$\Delta H_\mathrm{neut}$ ($\mathrm{kJ/mol}$)	Explanation
Strong acid + strong base	$-57.9$	Full ionisation of both
Weak acid + strong base	$-56.1$ (e.g. $\mathrm{CH}_3\mathrm{COOH} + \mathrm{NaOH}$)	Energy needed to dissociate weak acid
Strong acid + weak base	$-57.3$ (e.g. $\mathrm{HCl} + \mathrm{NH}_3$)	Energy needed to dissociate weak base
Weak acid + weak base	Variable	Depends on relative strengths

Worked Example: Hess’s Law with Indirect Routes

Calculate $\Delta H_f^\circ$ of $\mathrm{CH}_4(g)$ given:

• $\mathrm{C}(s) + \mathrm{O}_2(g) \to \mathrm{CO}_2(g)$; $\Delta H_1 = -394\,\mathrm{kJ/mol}$
• $\mathrm{H}_2(g) + \tfrac{1}{2}\mathrm{O}_2(g) \to \mathrm{H}_2\mathrm{O}(l)$; $\Delta H_2 = -286\,\mathrm{kJ/mol}$
• $\mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \to \mathrm{CO}_2(g) + 2\mathrm{H}_2\mathrm{O}(l)$; $\Delta H_3 = -890\,\mathrm{kJ/mol}$

Formation: $\mathrm{C}(s) + 2\mathrm{H}_2(g) \to \mathrm{CH}_4(g)$

By Hess’s Law, route the formation through combustion:

$$\Delta H_f^\circ + \Delta H_3 = \Delta H_1 + 2\Delta H_2$$ $$\Delta H_f^\circ = -394 + 2(-286) - (-890) = -394 - 572 + 890 = -76\,\mathrm{kJ/mol}$$

Common Pitfalls

1. Sign errors in Hess’s Law cycles. When constructing a cycle, ensure all arrows point in the correct direction. If a step is reversed, change the sign of the enthalpy.

2. Confusing $\Delta S^\circ$ with $S^\circ$. $\Delta S^\circ$ is the change in entropy; $S^\circ$ is the absolute entropy of a single species. $S^\circ$ is always positive; $\Delta S^\circ$ can be positive or negative.

3. Using $\Delta H$ instead of $\Delta H^\circ$ in Gibbs calculations. Standard values must be used for standard free energy calculations.

4. Forgetting units in entropy calculations. Entropy is in $\mathrm{J\,mol^{-1}\,K^{-1}}$Not $\mathrm{kJ}$. Always convert $\Delta H$ to $\mathrm{J}$ (or $\Delta S$ to $\mathrm{kJ}$) before combining in $\Delta G = \Delta H - T\Delta S$.

5. Misidentifying the theoretical vs experimental lattice enthalpy in Fajans’ Rules discussions. The experimental (Born-Haber) value is less exothermic when covalent character is present; the theoretical (Born-Lande) value assumes perfect ionic bonding.

6. Assuming $\Delta G = 0$ means the reaction has stopped. $\Delta G = 0$ means the reaction is at equilibrium, not that it has stopped. Both forward and reverse reactions continue at equal rates.

7. Using the wrong formula for the surroundings entropy. $\Delta S_\mathrm{surroundings} = -\Delta H / T$. The negative sign is critical: an exothermic reaction ($\Delta H \lt 0$) increases the entropy of the surroundings.

Bond Enthalpy Calculations

Standard Enthalpy of Formation vs Mean Bond Enthalpy

Standard enthalpies of formation ($\Delta H_f^\circ$) are measured experimentally. Mean bond enthalpies are calculated as averages. When both are available, $\Delta H_f^\circ$ data give more accurate results.

When to use bond enthalpies: Use mean bond enthalpies when $\Delta H_f^\circ$ data are not available for all species in the reaction (e.g. For gaseous atoms or free radicals).

Worked Example: Bond Enthalpy Calculation for an Unknown Reaction

Calculate $\Delta H$ for the reaction:

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

Using mean bond enthalpies: \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}.

$$\text{Bonds broken} = 1 \times \mathrm{N}\equiv\mathrm{N} + 3 \times \mathrm{H}-\mathrm{H} = 945 + 3(436) = 2253\,\mathrm{kJ/mol}$$ $$\text{Bonds formed} = 6 \times \mathrm{N}-\mathrm{H} = 6(391) = 2346\,\mathrm{kJ/mol}$$ $$\Delta H = 2253 - 2346 = -93\,\mathrm{kJ/mol}$$

The experimental value is $-92\,\mathrm{kJ/mol}$Showing good agreement.

Born-Haber Cycles and Lattice Enthalpy (Summary)

Lattice Enthalpy Definitions

• Lattice dissociation enthalpy: Enthalpy change when one mole of an ionic lattice is separated into its gaseous ions (endothermic, always positive).
• Lattice formation enthalpy: Enthalpy change when one mole of an ionic lattice is formed from its gaseous ions (exothermic, always negative).

$$\Delta H_\mathrm{latt}(\mathrm{dissociation}) = -\Delta H_\mathrm{latt}(\mathrm{formation})$$

Factors Affecting Lattice Enthalpy

1. Ionic charge: Higher charge $\to$ stronger electrostatic attraction $\to$ more exothermic lattice enthalpy. $\mathrm{MgO}$ (\mathrm{Mg}^{2+}$$\mathrm{O}^{2-}) has a much more exothermic lattice enthalpy than $\mathrm{NaCl}$ (\mathrm{Na}^+$$\mathrm{Cl}^-).
2. Ionic radius: Smaller ions $\to$ shorter internuclear distance $\to$ stronger attraction $\to$ more exothermic lattice enthalpy. $\mathrm{LiF}$ has a more exothermic lattice enthalpy than $\mathrm{NaF}$ because $\mathrm{Li}^+$ is smaller than $\mathrm{Na}^+$.

Fajans’ Rules and Covalent Character

When the cation is small and highly charged (e.g. $\mathrm{Al}^{3+}$) and the anion is large and highly charged (e.g. $\mathrm{I}^-$), the electron cloud of the anion is distorted (polarised) towards the cation. This introduces covalent character, making the experimental lattice enthalpy less exothermic than the theoretical (purely ionic) value.

Cation	Anion	Covalent character	Explanation
$\mathrm{Na}^+$ (large, low charge)	$\mathrm{Cl}^-$ (moderate)	Low	Minimal polarisation
$\mathrm{Al}^{3+}$ (small, high charge)	$\mathrm{I}^-$ (large)	High	Strong polarisation

Worked Example: Born-Haber Cycle for $\mathrm{CaCl}_2$

Construct a Born-Haber cycle for $\mathrm{CaCl}_2(s)$ and calculate the lattice enthalpy.

Steps (values are illustrative):

Step	Description	$\Delta H$ ($\mathrm{kJ/mol}$)
1	$\mathrm{Ca}(s) \to \mathrm{Ca}(g)$ (atomisation)	$+178$
2	$\mathrm{Ca}(g) \to \mathrm{Ca}^+(g) + e^-$ (1st IE)	$+590$
3	$\mathrm{Ca}^+(g) \to \mathrm{Ca}^{2+}(g) + e^-$ (2nd IE)	$+1145$
4	$\mathrm{Cl}_2(g) \to 2\mathrm{Cl}(g)$ (atomisation)	$+244$
5	$2\mathrm{Cl}(g) + 2e^- \to 2\mathrm{Cl}^-(g)$ (1st EA $\times 2$)	$-728$
6	$\mathrm{Ca}(s) + \mathrm{Cl}_2(g) \to \mathrm{CaCl}_2(s)$ ($\Delta H_f^\circ$)	$-796$

Lattice enthalpy (formation):

$$\Delta H_\mathrm{latt} = \Delta H_f^\circ - (\text{Steps 1--5}) = -796 - (178 + 590 + 1145 + 244 - 728)$$ $$\Delta H_\mathrm{latt} = -796 - 1435 = -2231\,\mathrm{kJ/mol}$$

Applications of Thermodynamics

Predicting the Feasibility of Industrial Processes

Haber process: $\mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightleftharpoons 2\mathrm{NH}_3(g)$

• $\Delta H^\circ = -92\,\mathrm{kJ/mol}$ (exothermic)
• $\Delta S^\circ = -199\,\mathrm{J\,mol^{-1}\,K^{-1}}$ (4 moles gas $\to$ 2 moles gas)
• $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$

At $298\,\mathrm{K}$: $\Delta G^\circ = -92 - 298(-0.199) = -92 + 59.3 = -32.7\,\mathrm{kJ/mol}$ (spontaneous, $K \approx 5.6 \times 10^5$).

At $700\,\mathrm{K}$: $\Delta G^\circ = -92 - 700(-0.199) = -92 + 139.3 = +47.3\,\mathrm{kJ/mol}$ (non-spontaneous).

The Haber process uses high temperature for kinetic reasons (faster rate), despite the thermodynamic penalty. High pressure shifts equilibrium towards products (fewer moles of gas).

Contact process: $2\mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{SO}_3(g)$

• $\Delta H^\circ = -198\,\mathrm{kJ/mol}$ (exothermic)
• $\Delta S^\circ = -190\,\mathrm{J\,mol^{-1}\,K^{-1}}$
• Optimal temperature is a compromise: $400$—$450^\circ\mathrm{C}$ gives acceptable rate and reasonable equilibrium yield. $\mathrm{V}_2\mathrm{O}_5$ catalyst lowers the activation energy.

Practice Problems

Problem 1

Calculate $\Delta G^\circ$ at $298\,\mathrm{K}$ for the reaction:

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

Given: \Delta H^\circ = -114\,\mathrm{kJ/mol}$$\Delta S^\circ = -146\,\mathrm{J\,mol^{-1}\,K^{-1}}.

Is the reaction spontaneous?

Solution:

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -114 \times 10^3 - 298 \times (-146) = -114000 + 43508 = -70492\,\mathrm{J/mol} = -70.5\,\mathrm{kJ/mol}$$

Since $\Delta G^\circ \lt 0$The reaction is spontaneous at $298\,\mathrm{K}$. The negative $\Delta H$ dominates over the unfavourable negative $\Delta S$.

Problem 2

The melting of ice: $\mathrm{H}_2\mathrm{O}(s) \to \mathrm{H}_2\mathrm{O}(l)$ has $\Delta H^\circ = +6.01\,\mathrm{kJ/mol}$ and $\Delta S^\circ = +22.0\,\mathrm{J\,mol^{-1}\,K^{-1}}$. Calculate the melting point of ice.

Solution:

At the melting point, $\Delta G = 0$:

$$T = \frac◆LB◆\Delta H^\circ◆RB◆◆LB◆\Delta S^\circ◆RB◆ = \frac{6010}{22.0} = 273\,\mathrm{K} = 0^\circ\mathrm{C}$$

Problem 3

Use bond enthalpy data to estimate $\Delta H$ for the hydrogenation of ethyne to ethane:

$$\mathrm{HC}\equiv\mathrm{CH}(g) + 2\mathrm{H}_2(g) \to \mathrm{CH}_3\mathrm{CH}_3(g)$$

Bond enthalpies: \mathrm{C}\equiv\mathrm{C} = 839\,\mathrm{kJ/mol}$$\mathrm{C}-\mathrm{C} = 347\,\mathrm{kJ/mol}$$\mathrm{C}-\mathrm{H} = 413\,\mathrm{kJ/mol}$$\mathrm{H}-\mathrm{H} = 436\,\mathrm{kJ/mol}.

Solution:

Bonds broken: $1 \times \mathrm{C}\equiv\mathrm{C} + 2 \times \mathrm{C}-\mathrm{H} + 2 \times \mathrm{H}-\mathrm{H} = 839 + 2(413) + 2(436) = 839 + 826 + 872 = 2537\,\mathrm{kJ/mol}$

Bonds formed: $1 \times \mathrm{C}-\mathrm{C} + 6 \times \mathrm{C}-\mathrm{H} = 347 + 6(413) = 347 + 2478 = 2825\,\mathrm{kJ/mol}$

$$\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds formed}) = 2537 - 2825 = -288\,\mathrm{kJ/mol}$$

The reaction is exothermic, as expected for hydrogenation. Note that mean bond enthalpies are used, so this is an estimate. The experimental value is approximately $-312\,\mathrm{kJ/mol}$; the discrepancy arises because the $\mathrm{C}-\mathrm{H}$ bond enthalpy in $\mathrm{HC}\equiv\mathrm{CH}$ differs from the mean value used.

Problem 4

The decomposition of ammonium chloride:

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

Has $\Delta H^\circ = +176\,\mathrm{kJ/mol}$ and $\Delta S^\circ = +285\,\mathrm{J\,mol^{-1}\,K^{-1}}$.

(a) Explain why $\Delta S^\circ$ is positive. (b) Calculate the minimum temperature at which the decomposition becomes spontaneous. (c) State two assumptions made in the calculation.

Solution:

(a) $\Delta S^\circ$ is positive because one mole of solid produces two moles of gas. Gases have much higher entropy than solids due to the large number of accessible microstates. The increase in the number of gas molecules and the change from a highly ordered solid to freely moving gas molecules both contribute to a large positive entropy change.

(b) At the threshold of spontaneity, $\Delta G^\circ = 0$:

$$T = \frac◆LB◆\Delta H^\circ◆RB◆◆LB◆\Delta S^\circ◆RB◆ = \frac◆LB◆176 \times 10^3◆RB◆◆LB◆285◆RB◆ = 618\,\mathrm{K} = 345^\circ\mathrm{C}$$

Above 618\,\mathrm{K}$$\Delta G^\circ \lt 0 and the decomposition is spontaneous.

(c) Assumptions: (i) $\Delta H^\circ$ and $\Delta S^\circ$ are constant over the temperature range (they are not strictly constant but vary little for most reactions). (ii) The reaction is at standard pressure ($100\,\mathrm{kPa}$).

Problem 5

Use the following standard enthalpies of formation to calculate $\Delta H_f^\circ$ for ethanol:

• $\Delta H_f^\circ(\mathrm{C}(\text{graphite})) = 0\,\mathrm{kJ/mol}$
• $\Delta H_f^\circ(\mathrm{H}_2(g)) = 0\,\mathrm{kJ/mol}$
• $\Delta H_f^\circ(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}(l)) = -277\,\mathrm{kJ/mol}$
• $\Delta H_c^\circ(\mathrm{C}(\text{graphite})) = -394\,\mathrm{kJ/mol}$
• $\Delta H_c^\circ(\mathrm{H}_2(g)) = -286\,\mathrm{kJ/mol}$
• $\Delta H_c^\circ(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}(l)) = -1367\,\mathrm{kJ/mol}$

Verify the consistency of the data by constructing a Hess’s Law cycle.

Solution:

The combustion of ethanol:

$$\mathrm{C}_2\mathrm{H}_5\mathrm{OH}(l) + 3\mathrm{O}_2(g) \to 2\mathrm{CO}_2(g) + 3\mathrm{H}_2\mathrm{O}(l) \quad \Delta H_c = -1367\,\mathrm{kJ/mol}$$

Route via elements:

$$2\mathrm{C}(\text{graphite}) + 3\mathrm{H}_2(g) + \tfrac{1}{2}\mathrm{O}_2(g) \to \mathrm{C}_2\mathrm{H}_5\mathrm{OH}(l) \quad \Delta H_f^\circ$$

Combustion of elements:

$$2\mathrm{C}(\text{graphite}) + 2\mathrm{O}_2(g) \to 2\mathrm{CO}_2(g) \quad 2 \times (-394) = -788\,\mathrm{kJ/mol}$$ $$3\mathrm{H}_2(g) + \tfrac{3}{2}\mathrm{O}_2(g) \to 3\mathrm{H}_2\mathrm{O}(l) \quad 3 \times (-286) = -858\,\mathrm{kJ/mol}$$

By Hess’s Law: $\Delta H_f^\circ + \Delta H_c^\circ(\text{ethanol}) = 2\Delta H_c^\circ(\mathrm{C}) + 3\Delta H_c^\circ(\mathrm{H}_2)$

$$\Delta H_f^\circ = (-788) + (-858) - (-1367) = -1646 + 1367 = -279\,\mathrm{kJ/mol}$$

The calculated value ($-279\,\mathrm{kJ/mol}$) is close to the literature value ($-277\,\mathrm{kJ/mol}$), confirming consistency. The small discrepancy is within experimental uncertainty.

Problem 6

The enthalpy of neutralisation of $\mathrm{NaOH}(aq)$ and $\mathrm{CH}_3\mathrm{COOH}(aq)$ is $-56.1\,\mathrm{kJ/mol}$Whereas the enthalpy of neutralisation of $\mathrm{NaOH}(aq)$ and $\mathrm{HCl}(aq)$ is $-57.9\,\mathrm{kJ/mol}$. Explain the difference.

Solution:

The neutralisation of a strong acid ($\mathrm{HCl}$) with a strong base ($\mathrm{NaOH}$) always gives the same value ($-57.9\,\mathrm{kJ/mol}$) because the net ionic equation is:

$$\mathrm{H}^+(aq) + \mathrm{OH}^-(aq) \to \mathrm{H}_2\mathrm{O}(l) \quad \Delta H = -57.9\,\mathrm{kJ/mol}$$

With ethanoic acid (a weak acid), some energy is consumed in dissociating the acid:

$$\mathrm{CH}_3\mathrm{COOH}(aq) \rightleftharpoons \mathrm{CH}_3\mathrm{COO}^-(aq) + \mathrm{H}^+(aq) \quad \Delta H_\mathrm{diss} \approx +0.4\,\mathrm{kJ/mol}$$

The overall enthalpy is the sum of the dissociation and neutralisation:

$$\Delta H_\mathrm{neut}(\text{weak}) = \Delta H_\mathrm{diss} + \Delta H_\mathrm{neut}(\text{strong}) \approx +0.4 + (-57.9) = -57.5\,\mathrm{kJ/mol}$$

The measured value is $-56.1\,\mathrm{kJ/mol}$Suggesting additional endothermic contributions (the enthalpy of dissociation of ethanoic acid is endothermic, consuming some of the heat released by neutralisation). The reaction is less exothermic because the weak acid must first dissociate, which is an endothermic process.

Advanced Thermodynamic Calculations

Gibbs Free Energy and Equilibrium

The relationship between $\Delta G^\circ$ and the equilibrium constant:

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

This is one of the most important equations in A-Level chemistry. It connects thermodynamics (energetics) with equilibrium (composition).

Worked Example: Calculate $K_c$ at $298\,\mathrm{K}$ for a reaction with $\Delta G^\circ = -5.40\,\mathrm{kJ/mol}$.

$-5400 = -8.314 \times 298 \times \ln K_c$

$\ln K_c = \frac◆LB◆5400◆RB◆◆LB◆8.314 \times 298◆RB◆ = \frac{5400}{2478} = 2.179$

$K_c = e^{2.179} = 8.84$

Since \Delta G^\circ < 0$$K > 1Confirming the reaction is spontaneous and products are favoured at equilibrium.

Worked Example: Calculate the temperature at which $\Delta G = 0$ for the reaction $\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2\mathrm{NO}_2(g)$Given $\Delta H^\circ = +57.2\,\mathrm{kJ/mol}$ and \Delta S^\circ = +175.8\,\mathrm{J\,K^{-1}\,\mathrm{mol}^{-1}.

At $\Delta G = 0$: $0 = \Delta H^\circ - T\Delta S^\circ$

$T = \frac◆LB◆\Delta H^\circ◆RB◆◆LB◆\Delta S^\circ◆RB◆ = \frac{57200}{175.8} = 325\,\mathrm{K}$

Below $325\,\mathrm{K}$, $\Delta G < 0$ and the forward reaction is spontaneous. Above $325\,\mathrm{K}$, $\Delta G > 0$ and the reverse reaction is spontaneous. At $325\,\mathrm{K}$The system is at equilibrium ($K = 1$).

Entropy Calculations

Worked Example: Calculate $\Delta S^\circ$ for the reaction $\mathrm{CaCO}_3(s) \to \mathrm{CaO}(s) + \mathrm{CO}_2(g)$.

$S^\circ$ values: \mathrm{CaCO}_3(s) = 92.9\,\mathrm{J\,K^{-1}\,\mathrm{mol}^{-1}}$$\mathrm{CaO}(s) = 38.1\,\mathrm{J\,K^{-1}\mathrm{mol}^{-1}}$$\mathrm{CO}_2(g) = 213.7\,\mathrm{J\,K^{-1}\mathrm{mol}^{-1}}.

$\Delta S^\circ = S^\circ(\mathrm{CaO}) + S^\circ(\mathrm{CO}_2) - S^\circ(\mathrm{CaCO}_3)$

$= 38.1 + 213.7 - 92.9 = 158.9\,\mathrm{J\,K^{-1}\mathrm{mol}^{-1}}$

The entropy change is positive, as expected: a solid decomposes to give a gas (increased disorder).

Born-Haber and Hess’s Law Applications

Worked Example: Use Hess’s Law to calculate $\Delta H_f^\circ$ for $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}(l)$.

Given data:

• $\mathrm{C}(s) + \mathrm{O}_2(g) \to \mathrm{CO}_2(g)$, $\Delta H = -393.5\,\mathrm{kJ/mol}$
• $\mathrm{H}_2(g) + \frac{1}{2}\mathrm{O}_2(g) \to \mathrm{H}_2\mathrm{O}(l)$, $\Delta H = -285.8\,\mathrm{kJ/mol}$
• $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}(l) + 3\mathrm{O}_2(g) \to 2\mathrm{CO}_2(g) + 3\mathrm{H}_2\mathrm{O}(l)$, $\Delta H = -1367\,\mathrm{kJ/mol}$

By Hess’s Law:

$2(\mathrm{C} \to \mathrm{CO}_2) + 3(\mathrm{H}_2 + \tfrac{1}{2}\mathrm{O}_2 \to \mathrm{H}_2\mathrm{O}) - (\mathrm{C}_2\mathrm{H}_5\mathrm{OH} + 3\mathrm{O}_2 \to 2\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O}) = \Delta H_f^\circ(\mathrm{C}_2\mathrm{H}_5\mathrm{OH})$

$2(-393.5) + 3(-285.8) - (-1367) = -787.0 - 857.4 + 1367 = -277.4\,\mathrm{kJ/mol}$

$\Delta H_f^\circ(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}) = -277\,\mathrm{kJ/mol}$ (to 3 s.f.)

Gibbs Free Energy: Predicting Feasibility

Worked Example: Is the reduction of $\mathrm{TiO}_2$ to $\mathrm{Ti}$ by carbon thermodynamically feasible at $1500\,\mathrm{K}$?

$\mathrm{TiO}_2(s) + 2\mathrm{C}(s) \to \mathrm{Ti}(s) + 2\mathrm{CO}(g)$

$\Delta H^\circ = +877\,\mathrm{kJ/mol}$, \Delta S^\circ = +193\,\mathrm{J\,K^{-1}\mathrm{mol}^{-1}

$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 877000 - 1500 \times 193 = 877000 - 289500 = 587500\,\mathrm{J/mol} = +588\,\mathrm{kJ/mol}$

$\Delta G^\circ > 0$So the reaction is not thermodynamically feasible at $1500\,\mathrm{K}$.

At what temperature does it become feasible?

$T = \frac◆LB◆\Delta H^\circ◆RB◆◆LB◆\Delta S^\circ◆RB◆ = \frac{877000}{193} = 4544\,\mathrm{K}$

The reaction becomes feasible above approximately $4544\,\mathrm{K}$ (extremely high temperature, impractical). In practice, the Kroll process (reduction with $\mathrm{Mg}$ or $\mathrm{Cl}_2$) is used.

Exam-Style Questions with Full Mark Schemes

Q1 (5 marks)

Define the term standard enthalpy change of reaction, $\Delta H^\circ_\mathrm{r}$. Explain why the standard enthalpy change of neutralisation of a strong acid with a strong base is always approximately $-57\,\mathrm{kJ/mol}$ regardless of which strong acid and strong base are used.

Mark Scheme:

$\Delta H^\circ_\mathrm{r}$ is the enthalpy change when the reaction occurs under standard conditions with all reactants and products in their standard states (1 mark).

The neutralisation of any strong acid with any strong base has the same net ionic equation:

$\mathrm{H}^+(aq) + \mathrm{OH}^-(aq) \to \mathrm{H}_2\mathrm{O}(l)$ (1 mark).

The specific acid and base are irrelevant because strong acids and bases are fully dissociated in solution (1 mark). The enthalpy change depends only on the formation of the O—H bond in water, which is the same in every case (1 mark). Minor differences arise from the enthalpies of dilution of different ions (1 mark).

Q2 (6 marks)

For the reaction $\mathrm{A}(g) + 2\mathrm{B}(g) \to \mathrm{C}(g) + \mathrm{D}(g)$:

$\Delta H^\circ = -85\,\mathrm{kJ/mol}$, \Delta S^\circ = -120\,\mathrm{J\,K^{-1}\mathrm{mol}^{-1}

(a) Calculate $\Delta G^\circ$ at $298\,\mathrm{K}$. (2 marks)

(b) State whether the reaction is feasible at $298\,\mathrm{K}$Explaining your answer. (1 mark)

(c) Calculate the temperature above which the reaction becomes non-spontaneous. (2 marks)

(d) State the effect of increasing the pressure on the position of equilibrium. (1 mark)

Mark Scheme:

(a) $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -85000 - 298 \times (-120) = -85000 + 35760 = -49240\,\mathrm{J/mol} = -49.2\,\mathrm{kJ/mol}$ (1 mark for substitution, 1 mark for answer).

(b) $\Delta G^\circ < 0$So the reaction is feasible (spontaneous) at $298\,\mathrm{K}$ (1 mark).

(c) $\Delta G^\circ = 0$: $T = \frac◆LB◆\Delta H^\circ◆RB◆◆LB◆\Delta S^\circ◆RB◆ = \frac{-85000}{-120} = 708\,\mathrm{K}$ (1 mark). Above $708\,\mathrm{K}$, $\Delta G^\circ > 0$ and the reaction is non-spontaneous (1 mark).

(d) 3 moles of gas on the left, 2 moles on the right. Increasing pressure favours the side with fewer moles (products), shifting the equilibrium to the right (1 mark).

Q3 (4 marks)

Explain why the entropy change for the reaction $\mathrm{NH}_4\mathrm{Cl}(s) \to \mathrm{NH}_3(g) + \mathrm{HCl}(g)$ is positive.

Mark Scheme:

A solid is converted into two gases (2 marks). Gases have much higher entropy than solids because the particles are free to move in all directions (1 mark). The number of particles increases from 1 to 2, and the disorder increases (1 mark).

Q4 (5 marks)

Use the following data to calculate the lattice enthalpy of $\mathrm{KCl}$ using a Born-Haber cycle:

• $\Delta H_f^\circ(\mathrm{KCl}) = -437\,\mathrm{kJ/mol}$
• $\Delta H_\mathrm{at}^\circ(\mathrm{K}) = +89\,\mathrm{kJ/mol}$
• $\Delta H_\mathrm{at}^\circ(\mathrm{Cl}) = +122\,\mathrm{kJ/mol}$
• $\mathrm{IE}_1(\mathrm{K}) = +419\,\mathrm{kJ/mol}$
• $\mathrm{EA}_1(\mathrm{Cl}) = -349\,\mathrm{kJ/mol}$

Mark Scheme:

$\Delta H_\mathrm{lat}^\circ = \Delta H_f^\circ - \Delta H_\mathrm{at}^\circ(\mathrm{K}) - \Delta H_\mathrm{at}^\circ(\mathrm{Cl}) - \mathrm{IE}_1(\mathrm{K}) - \mathrm{EA}_1(\mathrm{Cl})$ (1 mark for equation)

$= -437 - 89 - 122 - 419 - (-349)$ (1 mark for substitution)

$= -437 - 89 - 122 - 419 + 349 = -718\,\mathrm{kJ/mol}$ (1 mark for arithmetic)

The lattice enthalpy of KCl is $-718\,\mathrm{kJ/mol}$ (1 mark for answer with sign).

The negative sign indicates the process of forming the ionic lattice from gaseous ions is exothermic (1 mark).

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:::tip Diagnostic Test Ready to test your understanding of Thermodynamics? The contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Thermodynamics with other chemistry topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

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: $mgh = \frac{1}{2}mv^2$

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

$$:::$$