1. Standard Enthalpy Definitions
The heat energy transferred in a chemical reaction at constant pressure. Standard conditions (\(^\circ\)): \(100\text{ kPa}\) (1 bar) pressure and a stated temperature (usually \(298\text{ K}\) / \(25^\circ\text{C}\)), with substances in their standard physical states.
The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions.
Rule: The standard enthalpy of formation of any element in its standard physical state is zero by definition (e.g. \(\Delta_f H^\circ[\text{O}_2(g)] = 0\)).
The enthalpy change when one mole of a substance burns completely in excess oxygen under standard conditions, with all reactants and products in their standard states.
Rule: Enthalpies of combustion are always negative (exothermic).
2. Calorimetry Principles
Calorimetry measures thermal energy changes by tracking temperature changes in an aqueous solution or calorimeter:
Heat Energy Transferred Formula
\[ q = mc\Delta T \]
- \(q\) = heat energy absorbed or released in joules (\(\text{J}\)).
- \(m\) = mass of solution being heated or cooled in grams (\(\text{g}\)). We assume aqueous solutions have a density of \(1.00\text{ g cm}^{-3}\) (so \(50\text{ cm}^3 = 50\text{ g}\)).
- \(c\) = specific heat capacity of water (\(4.18\text{ J g}^{-1}\text{K}^{-1}\)).
- \(\Delta T\) = temperature change in \(\text{K}\) or \(^\circ\text{C}\) (\(T_{\text{final}} - T_{\text{initial}}\)).
Molar Enthalpy Change Formula
\[ \Delta H = -\frac{q}{n \times 1000} \]
Where \(n\) is the amount of limiting reactant in moles (\(\text{mol}\)), dividing by 1000 converts joules to kilojoules (\(\text{kJ mol}^{-1}\)). The negative sign ensures that an exothermic temperature rise produces a negative \(\Delta H\).
3. Cooling Curve Temperature Extrapolation
In simple polystyrene cup calorimetry, heat is inevitably lost to the laboratory surroundings during the reaction. The recorded peak temperature is therefore lower than the true theoretical maximum.
Experimental Protocol (RP2 Alignment)
- Record the temperature of the initial solution every minute for 3 minutes to establish a stable baseline.
- At minute 4, add the second reactant and swirl thoroughly. Do not record a temperature at minute 4.
- Resume recording temperature at minute 5, taking readings every minute until minute 10 as the mixture cools.
- Plot temperature against time on graph paper.
- Draw a straight line of best fit through the cooling points (minutes 5 to 10) and extrapolate this line back to minute 4 (the exact moment of mixing).
- Determine \(\Delta T\) as the vertical difference between the extrapolated temperature at minute 4 and the initial baseline temperature.
4. Hess's Law Cycles
The total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final states are identical.
Enthalpies of Formation Route
\[ \Delta H_r^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants}) \]
Constituent elements in standard states form the bottom baseline. Arrows point upwards from elements to products and reactants.
Enthalpies of Combustion Route
\[ \Delta H_r^\circ = \sum \Delta_c H^\circ(\text{reactants}) - \sum \Delta_c H^\circ(\text{products}) \]
Combustion products (\(\text{CO}_2, \text{H}_2\text{O}\)) form the bottom baseline. Arrows point downwards from reactants and products to combustion products.
Formation equation: \[ 2\text{C}(s) + 3\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l) \]
Applying Hess combustion formula:
\[ \Delta_f H = \sum \Delta_c H(\text{reactants}) - \sum \Delta_c H(\text{products}) \]
\[ \Delta_f H = [2(-393.5) + 3(-285.8)] - [-1367.3] \]
\[ \Delta_f H = [-787.0 - 857.4] - [-1367.3] = -1644.4 + 1367.3 = \mathbf{-277.1\text{ kJ mol}^{-1}} \]
5. Mean Bond Enthalpies
The enthalpy change required to break one mole of a specified covalent bond in gaseous molecules, averaged across a wide range of chemical compounds.
\[ \Delta H_r = \sum (\text{Bonds Broken}) - \sum (\text{Bonds Formed}) \]
Why do bond enthalpy calculations differ from experimental Hess cycles?
- Mean bond enthalpies are averaged over many different molecules; the exact bond strength depends on the surrounding chemical environment.
- Bond enthalpies apply strictly to substances in the gaseous state. If reactants or products are liquids or solids, additional energy is involved in changes of state (e.g. enthalpy of vaporisation).
6. Practice Questions
Step 1: Total solution mass: \(m = 50.0 + 50.0 = 100.0\text{ g}\).
Step 2: Temperature rise: \(\Delta T = 26.2 - 19.5 = 6.7\text{ K}\).
Step 3: Heat energy released:
\[ q = mc\Delta T = 100.0 \times 4.18 \times 6.7 = 2800.6\text{ J} = 2.8006\text{ kJ} \]
Step 4: Moles of water formed: \(n = c \times V = 1.00 \times (50.0 / 1000) = 0.0500\text{ mol}\).
Step 5: Molar enthalpy change:
\[ \Delta H = -\frac{2.8006}{0.0500} = \mathbf{-56.0\text{ kJ mol}^{-1}} \]