TheIBAxolotl
System and surroundings
The system is the reacting chemicals; the surroundings are everything else (solvent, container, air, thermometer). Energy is conserved: whatever the system loses, the surroundings gain.
An open system exchanges matter and energy; a closed system exchanges only energy; an isolated system exchanges neither.
Exothermic and endothermic
| Exothermic | Endothermic | |
|---|---|---|
| Heat flow | system → surroundings | surroundings → system |
| Surroundings temperature | rises | falls |
| Sign of \(\Delta H\) | negative | positive |
| Examples | combustion, neutralisation, respiration | thermal decomposition, photosynthesis, dissolving NH\(_4\)NO\(_3\) |
Enthalpy, H, is the heat content of a system. We can only measure the change, \(\Delta H = H_\text{products} - H_\text{reactants}\).
Energy profiles and stability
An energy profile plots enthalpy against reaction progress.
- Exothermic: products lower than reactants; the products are more stable (lower energy).
- Endothermic: products higher than reactants.
- Both show a hump: the activation energy, \(E_a\), measured from reactants to the top.
Lower enthalpy means greater energetic stability, though a reaction may still be slow if \(E_a\) is high.
Standard enthalpy change
Standard enthalpy changes, \(\Delta H^\ominus\), refer to standard conditions: 100 kPa pressure, solutions at 1 mol dm\(^{-3}\), substances in their standard states, usually quoted at 298 K. Values are per mole of the reaction as written (kJ mol\(^{-1}\)).
Calorimetry
The heat absorbed or released by the surroundings (usually water) is:
\[Q = mc\Delta T\]
- m = mass of the substance being heated (for dilute solutions, take 1.00 g per cm\(^3\))
- c = specific heat capacity (water: 4.18 J g\(^{-1}\) K\(^{-1}\))
- \(\Delta T\) = temperature change (a change in °C equals a change in K)
Then:
\[\Delta H = \frac{-Q}{n}\]
where n is the moles of the limiting reactant (or of water formed, or fuel burned). The minus sign is added because heat gained by the water was lost by the system.
Worked example
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. The temperature rises by 6.70 °C. Find the enthalpy of neutralisation.
- \(m = 100.0\) g (total solution volume)
- \(Q = 100.0 \times 4.18 \times 6.70 = 2.80 \times 10^{3}\) J = 2.80 kJ
- \(n(\text{H}_2\text{O}) = 0.0500 \times 1.00 = 0.0500\) mol
- \(\Delta H = -2.80 / 0.0500 = -56.0\) kJ mol\(^{-1}\)
Errors in calorimetry
- Heat loss to the surroundings is the main systematic error: experimental values are usually less exothermic than literature values. Reduce it with a lid and an insulated (polystyrene) cup.
- For slow reactions, plot temperature against time and extrapolate the cooling line back to the moment of mixing to estimate the true maximum.
- In fuel combustion, incomplete combustion and heating the container also reduce the measured value.
- Assumptions: solution density and heat capacity equal to water's; heat capacity of the cup ignored.
Common mistakes
- Using only the acid's volume for m instead of the total.
- Using the mass of the fuel burned as m in \(Q = mc\Delta T\); m is the mass of water heated.
- Dropping the negative sign for exothermic reactions.
- Mixing J and kJ in the final answer.
- Adding 273 to \(\Delta T\) (a temperature difference is the same in K and °C).
Exam tips
- Always state the sign and the unit: "−56.0 kJ mol\(^{-1}\)".
- Match significant figures to the least precise data (often the temperature change).
- When asked why a value differs from literature, name heat loss first and suggest a specific improvement.
- On energy profiles, label axes (enthalpy, reaction progress), \(\Delta H\) and \(E_a\) with arrows in the correct direction.