Chemistry - Energetics / Thermochemistry

Enthalpy ($H$): The heat content of a system at constant pressure. It is a state function.

Enthalpy change ($\Delta H$): The heat exchanged with the surroundings during a process at Constant pressure.

$$\Delta H = H_{\mathrm{products}} - H_{\mathrm{reactants}}$$

Exothermic: $\Delta H \lt 0$. Heat is released to the surroundings. The products have less Enthalpy than the reactants.

Examples: combustion, neutralisation, respiration.

Endothermic: $\Delta H \gt 0$. Heat is absorbed from the surroundings. The products have more Enthalpy than the reactants.

Examples: thermal decomposition, photosynthesis, dissolving $\mathrm{NH}_4\mathrm{NO}_3$ in water.

Standard enthalpy changes are measured under standard conditions:

• Pressure: $1 \mathrm{ atm}$ ($101.3 \mathrm{ kPa}$)
• Concentration: $1 \mathrm{ mol/dm}^3$ for solutions
• Temperature: $25^\circ\mathrm{C}$ ($298 \mathrm{ K}$)
• All substances in their standard states

The standard state is the pure substance at 1 atm and the specified temperature.

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The enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions.

$$\mathrm{C}_{(s)} + \mathrm{O}_{2(g)} \to \mathrm{CO}_{2(g)} \quad \Delta H_f^\circ = -393.5 \mathrm{ kJ/mol}$$

By definition, $\Delta H_f^\circ$ of any element in its standard state = 0.

$$\Delta H_f^\circ(\mathrm{C}_{(s,\mathrm{ graphite})}) = 0 \quad ; \quad \Delta H_f^\circ(\mathrm{H}_{2(g)}) = 0$$

Caution: Warning Involves formation FROM elements, not the other way around.

The enthalpy change when 1 mole of a substance is completely burned in excess oxygen under standard Conditions.

$$\mathrm{CH}_{4(g)} + 2\mathrm{O}_{2(g)} \to \mathrm{CO}_{2(g)} + 2\mathrm{H}_2\mathrm{O}_{(l)} \quad \Delta H_c^\circ = -890.3 \mathrm{ kJ/mol}$$

Combustion is always exothermic, so $\Delta H_c^\circ \lt 0$.

The enthalpy change when 1 mole of water is formed from the reaction between an acid and an alkali under standard conditions.

$$\mathrm{H}^+_{(aq)} + \mathrm{OH}^-_{(aq)} \to \mathrm{H}_2\mathrm{O}_{(l)} \quad \Delta H_{\mathrm{neut}}^\circ = -57.3 \mathrm{ kJ/mol}$$

For strong acid-strong base reactions, $\Delta H_{\mathrm{neut}}^\circ$ is approximately constant at $-57.3 \mathrm{ kJ/mol}$ because the net ionic equation is always the same.

For weak acid-strong base reactions, $\Delta H_{\mathrm{neut}}^\circ$ is less exothermic (less negative) because energy is absorbed to dissociate the weak acid.

The enthalpy change to form 1 mole of gaseous atoms from the element in its standard state.

$$\frac{1}{2}\mathrm{Cl}_{2(g)} \to \mathrm{Cl}_{(g)} \quad \Delta H_{\mathrm{at}}^\circ = +122 \mathrm{ kJ/mol}$$

This is always endothermic (bonds must be broken).

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Hess’s Law states that the enthalpy change for a reaction is the same regardless of the route taken from reactants to products, provided the initial and final conditions are the same.

This is a consequence of enthalpy being a state function.

Using Enthalpies of Formation:

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

Using Enthalpies of Combustion:

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

Note the reversal of signs compared to formation.

Worked example 1: Calculate $\Delta H$ for the reaction:

$$\mathrm{C}_{3\mathrm{H}_{8(g)}} + 5\mathrm{O}_{2(g)} \to 3\mathrm{CO}_{2(g)} + 4\mathrm{H}_2\mathrm{O}_{(l)}$$

Given: $\Delta H_c^\circ(\mathrm{C}_{3\mathrm{H}_{8(g)}}) = -2220 \mathrm{ kJ/mol}$ $\Delta H_c^\circ(\mathrm{CO}_{2(g)}) = 0$ (it is already fully oxidised), $\Delta H_c^\circ(\mathrm{H}_2\mathrm{O}_{(l)}) = 0$.

Answer

Since $\mathrm{CO}_2$ and $\mathrm{H}_2\mathrm{O}$ are already combustion products, their $\Delta H_c^\circ = 0$.

$$\Delta H = \Delta H_c^\circ(\mathrm{C}_3\mathrm{H}_8) - [3\Delta H_c^\circ(\mathrm{CO}_2) + 4\Delta H_c^\circ(\mathrm{H}_2\mathrm{O})]$$ $$= -2220 - [3(0) + 4(0)] = -2220 \mathrm{ kJ/mol}$$

This makes sense: the enthalpy of combustion of propane equals the enthalpy change of its combustion Reaction.

Worked example 2: Calculate $\Delta H_f^\circ$ of ethanol given:

$$\mathrm{C}_2\mathrm{H}_{5}\mathrm{OH}_{(l)} + 3\mathrm{O}_{2(g)} \to 2\mathrm{CO}_{2(g)} + 3\mathrm{H}_2\mathrm{O}_{(l)} \quad \Delta H = -1367 \mathrm{ kJ/mol}$$

$\Delta H_f^\circ(\mathrm{CO}_{2(g)}) = -393.5 \mathrm{ kJ/mol}$ $\Delta H_f^\circ(\mathrm{H}_2\mathrm{O}_{(l)}) = -285.8 \mathrm{ kJ/mol}$

Answer

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

$$-1367 = [2(-393.5) + 3(-285.8)] - [\Delta H_f^\circ(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}) + 3(0)]$$ $$-1367 = [-787.0 + (-857.4)] - \Delta H_f^\circ(\mathrm{C}_2\mathrm{H}_5\mathrm{OH})$$ $$-1367 = -1644.4 - \Delta H_f^\circ(\mathrm{C}_2\mathrm{H}_5\mathrm{OH})$$ $$\Delta H_f^\circ(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}) = -1644.4 + 1367 = -277.4 \mathrm{ kJ/mol}$$

This agrees with the accepted value of $-277.7 \mathrm{ kJ/mol}$.

Worked example 3: Using enthalpies of combustion, calculate $\Delta H$ for:

$$3\mathrm{C}_{(s)} + 4\mathrm{H}_{2(g)} \to \mathrm{C}_3\mathrm{H}_{8(g)}$$

Given: $\Delta H_c^\circ(\mathrm{C}_{(s)}) = -393.5 \mathrm{ kJ/mol}$ (same as $\Delta H_c^\circ(\mathrm{CO}_2)$ since combustion of C gives $\mathrm{CO}_2$), $\Delta H_c^\circ(\mathrm{H}_{2(g)}) = -285.8 \mathrm{ kJ/mol}$ (gives $\mathrm{H}_2\mathrm{O}_{(l)}$), $\Delta H_c^\circ(\mathrm{C}_{3\mathrm{H}_{8(g)}}) = -2220 \mathrm{ kJ/mol}$.

Answer

$$\Delta H = \sum \Delta H_c^\circ(\mathrm{reactants}) - \sum \Delta H_c^\circ(\mathrm{products})$$ $$= [3(-393.5) + 4(-285.8)] - [(-2220)]$$ $$= [-1180.5 + (-1143.2)] - (-2220)$$ $$= -2323.7 + 2220 = -103.7 \mathrm{ kJ/mol}$$

This is $\Delta H_f^\circ(\mathrm{C}_{3\mathrm{H}_{8(g)}})$ Matching the reference value of $-103.8 \mathrm{ kJ/mol}$.

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The mean bond enthalpy is the average enthalpy change when 1 mole of a specified type of bond is broken in the gaseous state, averaged over a range of compounds.

Bond breaking is always endothermic ($\Delta H \gt 0$). Bond forming is always exothermic ($\Delta H \lt 0$).

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

Worked example 4: Estimate $\Delta H$ for the combustion of methane using bond enthalpies.

$$\mathrm{CH}_{4(g)} + 2\mathrm{O}_{2(g)} \to \mathrm{CO}_{2(g)} + 2\mathrm{H}_2\mathrm{O}_{(g)}$$

Answer

Bonds broken:

• 4 $\times$ C—H = $4 \times 413 = 1652 \mathrm{ kJ/mol}$
• 2 $\times$ O=O = $2 \times 498 = 996 \mathrm{ kJ/mol}$

Total bonds broken = $1652 + 996 = 2648 \mathrm{ kJ/mol}$

Bonds formed:

• 2 $\times$ C=O (in $\mathrm{CO}_2$) = $2 \times 805 = 1610 \mathrm{ kJ/mol}$
• 4 $\times$ O—H (in $2\mathrm{H}_2\mathrm{O}$) = $4 \times 464 = 1856 \mathrm{ kJ/mol}$

Total bonds formed = $1610 + 1856 = 3466 \mathrm{ kJ/mol}$

$\Delta H = 2648 - 3466 = -818 \mathrm{ kJ/mol}$

Note: the accepted value is $-890.3 \mathrm{ kJ/mol}$ (for $\mathrm{H}_2\mathrm{O}_{(l)}$). The discrepancy arises because bond enthalpies are averages and we used $\mathrm{H}_2\mathrm{O}_{(g)}$ rather than $\mathrm{H}_2\mathrm{O}_{(l)}$.

Worked example 5: Using bond enthalpies, estimate the enthalpy change for:

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

Answer

Bonds broken:

• 1 $\times$ N$\equiv$N = $945 \mathrm{ kJ/mol}$
• 3 $\times$ H—H = $3 \times 436 = 1308 \mathrm{ kJ/mol}$

Total bonds broken = $945 + 1308 = 2253 \mathrm{ kJ/mol}$

Bonds formed:

• 6 $\times$ N—H = $6 \times 391 = 2346 \mathrm{ kJ/mol}$

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

For 2 mol $\mathrm{NH}_3$: $\Delta H = -93 \mathrm{ kJ}$ So per mole of reaction: $-93 \mathrm{ kJ/mol}$.

Accepted $\Delta H_f^\circ(\mathrm{NH}_{3(g)}) = -46.0 \mathrm{ kJ/mol}$ So $\Delta H = 2 \times (-46.0) = -92.0 \mathrm{ kJ/mol}$. The estimate is close.

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Calorimetry measures the heat exchanged during a reaction by observing the temperature change of a known mass of water (or solution).

$$q = mc\Delta T$$

Where:

• $q$ = heat energy (J)
• $m$ = mass of water/solution (g)
• $c$ = specific heat capacity (4.18 J g$^{-1}$ K$^{-1}$ for water)
• $\Delta T$ = temperature change (K or $^\circ$ C)

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

The negative sign converts the perspective: if the solution temperature rises ($\Delta T \gt 0$), The reaction is exothermic ($\Delta H \lt 0$).

Used for reactions in solution (e.g., neutralisation, dissolution).

Assumptions:

1. The density of the solution is 1.00 g/cm$^3$ (so mass in g = volume in cm$^3$).
2. The specific heat capacity of the solution is the same as water (4.18 J g$^{-1}$ K$^{-1}$).
3. No heat is lost to the surroundings (or the calorimeter is well insulated).
4. The calorimeter itself absorbs negligible heat.

Worked example 6: 50.0 cm$^3$ of 1.00 mol/dm$^3$ HCl is mixed with 50.0 cm$^3$ of 1.00 Mol/dm$^3$ NaOH in a polystyrene cup. The temperature rises from $22.0^\circ\mathrm{C}$ to $28.8^\circ\mathrm{C}$. Calculate $\Delta H$ for the neutralisation per mole of water formed.

Answer

Total volume = $50.0 + 50.0 = 100.0 \mathrm{ cm}^3$

Mass of solution = $100.0 \mathrm{ g}$ (density = 1.00 g/cm$^3$)

$\Delta T = 28.8 - 22.0 = 6.8^\circ\mathrm{C}$

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

Moles of $\mathrm{H}_2\mathrm{O}$ formed = moles of $\mathrm{HCl}$ = $1.00 \times 50.0 / 1000 = 0.0500 \mathrm{ mol}$

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

This is close to the standard value of $-57.3 \mathrm{ kJ/mol}$.

Used to measure enthalpies of combustion. A known mass of fuel is burned, and the temperature rise of a known mass of water is measured.

Worked example 7: 1.50 g of ethanol is burned in a spirit burner. The heat produced raises the Temperature of 200 g of water from $20.0^\circ\mathrm{C}$ to $45.5^\circ\mathrm{C}$. Calculate the enthalpy of combustion of ethanol.

Answer

$q = mc\Delta T = 200 \times 4.18 \times (45.5 - 20.0) = 200 \times 4.18 \times 25.5 = 21318 \mathrm{ J} = 21.32 \mathrm{ kJ}$

Moles of $\mathrm{C}_2\mathrm{H}_5\mathrm{OH} = 1.50 / 46.1 = 0.0325 \mathrm{ mol}$

$\Delta H_c = -\dfrac{21.32}{0.0325} = -656 \mathrm{ kJ/mol}$

The accepted value is $-1367 \mathrm{ kJ/mol}$. The experimental value is much less exothermic due to heat losses to the surroundings and incomplete combustion.

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Born-Haber cycles calculate lattice energies of ionic compounds using Hess’s Law. The lattice energy is the enthalpy change when 1 mole of an ionic solid is formed from its gaseous ions.

1. Atomisation of sodium: $\mathrm{Na}_{(s)} \to \mathrm{Na}_{(g)}$ ($\Delta H_{\mathrm{at}}^\circ = +108 \mathrm{ kJ/mol}$)
2. Ionisation of sodium: $\mathrm{Na}_{(g)} \to \mathrm{Na}^+_{(g)} + e^-$ ($\mathrm{IE}_1 = +496 \mathrm{ kJ/mol}$)
3. Atomisation of chlorine: $\frac{1}{2}\mathrm{Cl}_{2(g)} \to \mathrm{Cl}_{(g)}$ ($\Delta H_{\mathrm{at}}^\circ = +122 \mathrm{ kJ/mol}$)
4. Electron affinity of chlorine: $\mathrm{Cl}_{(g)} + e^- \to \mathrm{Cl}^-_{(g)}$ ($\mathrm{EA} = -349 \mathrm{ kJ/mol}$)
5. Lattice energy: $\mathrm{Na}^+_{(g)} + \mathrm{Cl}^-_{(g)} \to \mathrm{NaCl}_{(s)}$ ($\Delta H_{\mathrm{latt}} = ?$)
6. Formation: $\mathrm{Na}_{(s)} + \frac{1}{2}\mathrm{Cl}_{2(g)} \to \mathrm{NaCl}_{(s)}$ ($\Delta H_f^\circ = -411 \mathrm{ kJ/mol}$)

By Hess’s Law:

$$\Delta H_f^\circ = \Delta H_{\mathrm{at}}(\mathrm{Na}) + \mathrm{IE}_1(\mathrm{Na}) + \Delta H_{\mathrm{at}}(\mathrm{Cl}) + \mathrm{EA}(\mathrm{Cl}) + \Delta H_{\mathrm{latt}}$$ $$-411 = 108 + 496 + 122 + (-349) + \Delta H_{\mathrm{latt}}$$ $$-411 = 377 + \Delta H_{\mathrm{latt}}$$ $$\Delta H_{\mathrm{latt}} = -411 - 377 = -788 \mathrm{ kJ/mol}$$

Worked example 8: Calculate the lattice energy of $\mathrm{MgO}$ given:

• $\Delta H_{\mathrm{at}}^\circ(\mathrm{Mg}) = +148 \mathrm{ kJ/mol}$
• $\mathrm{IE}_1(\mathrm{Mg}) = +738 \mathrm{ kJ/mol}$
• $\mathrm{IE}_2(\mathrm{Mg}) = +1451 \mathrm{ kJ/mol}$
• $\Delta H_{\mathrm{at}}^\circ(\mathrm{O}) = +248 \mathrm{ kJ/mol}$ (for $\frac{1}{2}\mathrm{O}_{2(g)} \to \mathrm{O}_{(g)}$)
• $\mathrm{EA}_1(\mathrm{O}) = -141 \mathrm{ kJ/mol}$
• $\mathrm{EA}_2(\mathrm{O}) = +798 \mathrm{ kJ/mol}$ (second electron affinity is endothermic)
• $\Delta H_f^\circ(\mathrm{MgO}) = -602 \mathrm{ kJ/mol}$

Answer

$$\Delta H_f^\circ = \Delta H_{\mathrm{at}}(\mathrm{Mg}) + \mathrm{IE}_1 + \mathrm{IE}_2 + \Delta H_{\mathrm{at}}(\mathrm{O}) + \mathrm{EA}_1 + \mathrm{EA}_2 + \Delta H_{\mathrm{latt}}$$ $$-602 = 148 + 738 + 1451 + 248 + (-141) + 798 + \Delta H_{\mathrm{latt}}$$ $$-602 = 3242 + \Delta H_{\mathrm{latt}}$$ $$\Delta H_{\mathrm{latt}} = -602 - 3242 = -3844 \mathrm{ kJ/mol}$$

The large magnitude reflects the high charges on $\mathrm{Mg}^{2+}$ and $\mathrm{O}^{2-}$.

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Entropy is a measure of the disorder or randomness of a system.

$$\Delta S = S_{\mathrm{products}} - S_{\mathrm{reactants}}$$

1. Physical state: Gas $\gt$ Liquid $\gt$ Solid (increasing disorder).
2. Temperature: Entropy increases with temperature.
3. Number of particles: More particles (especially gas molecules) means higher entropy.
4. Dissolution: Dissolving a solid in water generally increases entropy.

• Reactions producing more gas molecules: $\Delta S \gt 0$.
• Reactions consuming gas molecules: $\Delta S \lt 0$.
• Solid to liquid or liquid to gas transitions: $\Delta S \gt 0$.

$$\mathrm{CaCO}_{3(s)} \to \mathrm{CaO}_{(s)} + \mathrm{CO}_{2(g)} \quad \Delta S \gt 0 \mathrm{ (solid to solid + gas)}$$ $$\mathrm{N}_{2(g)} + 3\mathrm{H}_{2(g)} \to 2\mathrm{NH}_{3(g)} \quad \Delta S \lt 0 \mathrm{ (4 mol gas to 2 mol gas)}$$

Typical values (J mol$^{-1}$ K$^{-1}$):

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

Where:

• $\Delta G$ = Gibbs free energy change (kJ/mol)
• $\Delta H$ = enthalpy change (kJ/mol)
• $T$ = temperature (K)
• $\Delta S$ = entropy change (kJ mol$^{-1}$ K$^{-1}$)

• Chemical Kinetics And Energetics
• Chemistry - Chemical Kinetics
• DSE Chemistry