REACTIVITY 1
What Drives Chemical Reactions?
Enthalpy • Calorimetry • Hess's Law • Born-Haber • Entropy • Gibbs Energy
1.1 Measuring Enthalpy
1. The Language of Thermodynamics
System: The reaction itself (reactants & products).
Surroundings: Everything else (water, thermometer, universe).
Temp ($T$): Measure of average kinetic energy.
Heat ($q$): Transfer of energy due to $\Delta T$.
2. Exothermic vs. Endothermic
Chemical reactions involve a trade-off: Breaking bonds (Endo) vs. Forming bonds (Exo).
| Feature | Exothermic ($\Delta H < 0$) | Endothermic ($\Delta H > 0$) |
|---|---|---|
| Bond Process | Bond Forming (Releases E) | Bond Breaking (Requires E) |
| Observation | Temp RISES | Temp FALLS |
| Stability | Products more stable | Products less stable |
Calorimetry (The Math)
We measure enthalpy changes indirectly by measuring the temperature change of water.
- $m$: Mass of the substance changing temp (usually water).
- $c$: Specific Heat Capacity ($4.18$).
- $\Delta T$: $T_{final} - T_{initial}$.
- $q$: Heat from Step 1.
- $n$: Moles of the limiting reactant.
- Negative Sign: Essential logic check. (Hot water = Exothermic).
Fatal Trap Alert
- The Unit Trap: $q$ is in Joules. $\Delta H$ is usually kJ/mol.
You MUST divide q by 1000 before finishing the calculation. - The Mass Trap: Do NOT use the mass of the fuel burned in $mc\Delta T$. Use the mass of the water that heated up.
1.2 Hess's Law & Bond Enthalpy
1. Hess's Law
The Law: The enthalpy change for a chemical reaction is independent of the pathway taken.
The "Cheat Code" Formulas
$\Delta H_f$ of elements ($O_2, Fe$) is ZERO.
Mnemonic: CRaP (Combustion: Reactants - Products).
2. Bond Enthalpy
Energy required to break 1 mole of bonds in gaseous state.
"Break Reactants (Endo) - Form Products (Exo)"
1.3 Energy from Fuels
Limited Oxygen. Produces CO (Toxic) or Soot. Releases less energy.
From biomass. Carbon Neutral (CO2 released was absorbed by plants).
Converts chemical $\to$ electrical directly.
2H2 + O2 → 2H2O
Advantage: Zero emissions.
Advanced Theory
The following section is for HL Students ONLY.
Born-Haber Cycles • Entropy • Gibbs Energy
Born-Haber Cycles
A special application of Hess's Law to calculate Lattice Enthalpy.
1. The Cycle Steps (Memorize Definitions)
Element $\rightarrow$ Gas Atom. (Endo)
Na(s) → Na(g)
Gas Atom $\rightarrow$ Cation. (Endo)
Na(g) → Na+(g) + e-
Gas Atom $\rightarrow$ Anion. (1st is Exo)
Cl(g) + e- → Cl-(g)
Gas Ions $\rightarrow$ Solid Lattice. (Highly Exo)
Na+(g) + Cl-(g) → NaCl(s)
2. Trends in Strength
- Ionic Charge: Higher charge = Stronger attraction. ($MgO > NaCl$).
- Ionic Radius: Smaller radius = Closer packing = Stronger attraction.
Entropy & Gibbs Energy
Measure of disorder (dispersal of energy).
The ultimate judge of spontaneity.
YES $\Delta G > 0$
NO $\Delta G = 0$
Equil.
The Fatal Unit Trap
$\Delta H$ is in kJ. $\Delta S$ is in J.
You MUST divide $\Delta S$ by 1000 to convert to kJ before plugging into the equation!
3. Equilibrium Link ($K$)
Negative $\Delta G \rightarrow$ Positive $\ln K \rightarrow$ $K > 1$ (Products Favored).
The Examiner's Vault
Strictly assessed on Reactivity 1 content.
(1) $A + 2B \rightarrow C \quad \Delta H = -100$
(2) $C \rightarrow D \quad \Delta H = +50$
Find $\Delta H$ for $A + 2B \rightarrow D$.
Add Eq 1 and Eq 2 directly. $-100 + 50 = -50$.
Bond E ($kJ/mol$): $H-H (436)$, $Cl-Cl (242)$, $H-Cl (431)$.
Calc $\Delta H$ for $H_2 + Cl_2 \rightarrow 2HCl$.
Broken: $436+242 = 678$. Formed: $2(431) = 862$.
$\Delta H = 678 - 862 = -184$.
Reaction is Endothermic ($\Delta H > 0$) and Order Increases ($\Delta S < 0$). Spontaneous?
$\Delta G = (+H) - T(-S) = (+H) + TS$. The result is always positive.
Combusting $0.50g$ Methanol ($M_r=32.04$) heats $200g$ water by $10.0^\circ C$. Calculate $\Delta H_c$.
- $q = 200 \times 4.18 \times 10.0 = 8360 J = 8.36 kJ$. [1]
- $n = 0.50 / 32.04 = 0.0156 mol$. [1]
- $\Delta H = -8.36 / 0.0156 = \mathbf{-536 \, kJ/mol}$. (Must have negative sign). [1]
Using standard data ($\Delta H_f = -796$, Lattice = -2253, etc.), calculate the Electron Affinity (EA) for Chlorine.
Equation: $\Delta H_f = \Delta H_{at}(Ca) + IE(Ca) + \Delta H_{at}(Cl_2) + 2(EA) + Lattice$
- $-796 = 178 + 1735 + 244 + 2(EA) - 2253$ [1]
- $2(EA) = -700 \rightarrow EA = \mathbf{-350 \, kJ/mol}$ [1]