14.2 Calorimetry, Specific Heat & Enthalpy (ΔH)

14.2 Calorimetry, Specific Heat & Enthalpy (ΔH)

Quick Summary: Calorimetry experimentally measures heat transferred during physical and chemical processes. Thermal transfer is quantified by heat capacity ($C$), specific heat capacity ($c$), and molar heat capacity ($C_m$) using $q = m \cdot c \cdot \Delta T$. Constant-pressure calorimetry ("coffee-cup") measures aqueous solution reactions where $q_{\text{rxn}} = -q_{\text{calorimeter}} = \Delta H_{\text{rxn}}$. Constant-volume calorimetry ("bomb") suppresses expansion ($\Delta V = 0, w = 0$), measuring combustion heats as internal energy changes ($q_v = \Delta U$). Phase changes occur isothermally, governed by latent heats of fusion ($\Delta H_{\text{fus}}$), vaporization ($\Delta H_{\text{vap}}$), and sublimation ($\Delta H_{\text{sub}}$).

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1. Heat Capacity, Specific Heat & Molar Heat Capacity

When heat enters or leaves a substance without changing its phase or chemical composition, temperature changes proportionally to the quantity of thermal energy transferred:

Definitions & Distinctions

1. Heat Capacity ($C$): Heat required to raise sample temperature by $1\ ^\circ\text{C}$ (or $1\text{ K}$): $C = \frac{q}{\Delta T} \quad [\text{Units: J/}^\circ\text{C or J/K}]$ Heat capacity is an extensive property that scales directly with sample mass.
2. Specific Heat Capacity ($c$): Heat required to raise $1\text{ gram}$ of substance by $1\ ^\circ\text{C}$: $c = \frac{q}{m \cdot \Delta T} \quad [\text{Units: J/(g}\cdot^\circ\text{C)}]$ Specific heat is an intensive property characteristic of the pure material.
3. Molar Heat Capacity ($C_m$): Heat required to raise $1\text{ mole}$ of substance by $1\ ^\circ\text{C}$: $C_m = \frac{q}{n \cdot \Delta T} \quad [\text{Units: J/(mol}\cdot^\circ\text{C)}]$

The Fundamental Calorimetric Equation

$q = m \cdot c \cdot \Delta T = m \cdot c \cdot (T_{\text{final}} - T_{\text{initial}})$ where $\Delta T > 0$ denotes heat absorbed and $\Delta T < 0$ denotes heat released.

Reference Table: Specific Heat Capacities

Liquid water's remarkably high specific heat capacity allows large bodies of water to moderate climate extremes.

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2. Constant-Pressure ("Coffee-Cup") Calorimetry

Constant-pressure calorimetry measures enthalpy changes in aqueous reactions ($P_{\text{ext}} = \text{constant}$), meaning $q_{\text{rxn}} = q_p = \Delta H_{\text{rxn}}$.

Conservation of Energy in Solution

In an insulated polystyrene cup open to atmospheric pressure: $q_{\text{system}} + q_{\text{surroundings}} = 0 \implies q_{\text{rxn}} = -q_{\text{calorimeter}}$ $q_{\text{calorimeter}} = m_{\text{soln}} \cdot c_{\text{soln}} \cdot \Delta T + C_{\text{cal}} \cdot \Delta T$ Neglecting $C_{\text{cal}}$ and approximating dilute solutions as water ($c = 4.184\text{ J/(g}\cdot^\circ\text{C)}$, $\rho = 1.00\text{ g/mL}$): $q_{\text{rxn}} = -(m_{\text{soln}} \cdot 4.184 \cdot \Delta T)$ Per mole of limiting reactant: $\Delta H_{\text{rxn}} = \frac{q_{\text{rxn}}}{n_{\text{limiting}}}$

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3. Constant-Volume ("Bomb") Calorimetry

Combustion reactions producing high pressures are conducted in a constant-volume bomb calorimeter:

• The sample is ignited in excess oxygen inside a rigid steel vessel ("bomb") submerged in water.
• Because volume is fixed ($\Delta V = 0$), no expansion work occurs ($w = 0$): $\Delta U_{\text{comb}} = q_v = -C_{\text{bomb}} \cdot \Delta T$ where $C_{\text{bomb}}$ is the calibrated total heat capacity of the bomb, water, and vessel.
• Reaction enthalpy connects via $\Delta H = \Delta U + \Delta n_{\text{gas}} RT$.

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4. Latent Heats of Phase Transitions

Phase changes occur isothermally at transition temperatures as heat alters intermolecular attractions:

• Enthalpy of Fusion ($\Delta H_{\text{fus}}$): Heat needed to melt $1\text{ mole}$ of solid (for water, $+6.01\text{ kJ/mol}$).
• Enthalpy of Vaporization ($\Delta H_{\text{vap}}$): Heat needed to vaporize $1\text{ mole}$ of liquid (for water, $+40.67\text{ kJ/mol}$).
• Enthalpy of Sublimation ($\Delta H_{\text{sub}}$): Direct solid-to-gas conversion: $\Delta H_{\text{sub}} = \Delta H_{\text{fus}} + \Delta H_{\text{vap}}$ Vaporization requires far greater energy than fusion because molecules must separate entirely against intermolecular attractions.

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5. Thermochemical Equations & Stoichiometry

A thermochemical equation specifies stoichiometry alongside molar enthalpy ($\Delta H^\circ$): $2\text{ H}_2(g) + \text{O}_2(g) \longrightarrow 2\text{ H}_2\text{O}(l) \quad \Delta H^\circ = -571.6\text{ kJ}$ Enthalpy is proportional to moles reacted: $\frac{-571.6\text{ kJ}}{2\text{ mol H}_2} = \frac{-571.6\text{ kJ}}{1\text{ mol O}_2} = \frac{-571.6\text{ kJ}}{2\text{ mol H}_2\text{O}(l)}$ Reversing an equation changes the sign of $\Delta H^\circ$; halving coefficients divides $\Delta H^\circ$ by 2.

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6. Worked Quantitative Calorimetry Examples

Example 1: Coffee-Cup Neutralization Calorimetry

Problem: Mixing $50.0\text{ mL}$ of $1.00\text{ M HCl}$ with $50.0\text{ mL}$ of $1.00\text{ M NaOH}$ at $22.0\ ^\circ\text{C}$ in a calorimeter causes temperature to rise to $28.7\ ^\circ\text{C}$. With total mass $100.0\text{ g}$ and $c = 4.184\text{ J/(g}\cdot^\circ\text{C)}$, determine $\Delta H_{\text{rxn}}$ per mole.

Step 1: Calculate heat absorbed by solution $\Delta T = 28.7\ ^\circ\text{C} - 22.0\ ^\circ\text{C} = +6.7\ ^\circ\text{C}$ $q_{\text{soln}} = (100.0\text{ g})(4.184\text{ J/(g}\cdot^\circ\text{C)})(6.7\ ^\circ\text{C}) = +2803\text{ J} = +2.803\text{ kJ}$

Step 2: Determine molar enthalpy $q_{\text{rxn}} = -2.803\text{ kJ}, \quad n = (0.0500\text{ L})(1.00\text{ M}) = 0.0500\text{ mol}$ $\Delta H_{\text{rxn}} = \frac{-2.803\text{ kJ}}{0.0500\text{ mol}} = -56.1\text{ kJ/mol}$

Example 2: Bomb Calorimetry Combustion

Problem: Combusting $1.015\text{ g}$ sucrose ($342.3\text{ g/mol}$) in a bomb calorimeter with $C_{\text{bomb}} = 4.90\text{ kJ/}^\circ\text{C}$ raises temperature by $3.41\ ^\circ\text{C}$. Find $\Delta U_{\text{comb}}$ per mole. $q_{\text{calorimeter}} = (4.90\text{ kJ/}^\circ\text{C})(3.41\ ^\circ\text{C}) = +16.71\text{ kJ} \implies q_{\text{rxn}} = -16.71\text{ kJ}$ $n = \frac{1.015\text{ g}}{342.3\text{ g/mol}} = 0.002965\text{ mol}$ $\Delta U_{\text{comb}} = \frac{-16.71\text{ kJ}}{0.002965\text{ mol}} = -5.64 \times 10^3\text{ kJ/mol}$