Enthalpy and enthalpy change
Most reactions in a lab or kitchen happen in open vessels, at constant (air) pressure. For these we use a new state function, enthalpy: H = U + pV.
At constant pressure the first law gives q_p = ΔU + pΔV = ΔH. So ΔH is simply the heat taken in or given out at constant pressure. At constant volume, ΔV = 0 and q_V = ΔU.
Link between ΔH and ΔU
Solids and liquids hardly change volume, so for them ΔH ≈ ΔU. For gases, pV = nRT, so ΔH = ΔU + Δn_g RT, where Δn_g = moles of gaseous products − moles of gaseous reactants. Use R = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹ when ΔU is in kJ.
Extensive and intensive properties
- Extensive: depends on the amount of matter. Mass, volume, U, H, heat capacity. Double the sample, double the value.
- Intensive: does not depend on the amount. Temperature, pressure, density, molar heat capacity.
Heat capacity
Heat capacity C is the heat needed to raise the temperature by 1 K: q = C ΔT. Specific heat c is per gram (water: 4.18 J g⁻¹ K⁻¹), so q = m c ΔT. Molar heat capacity is per mole. For an ideal gas, C_p − C_v = R, because at constant pressure some heat is spent on expansion work.
Calorimetry: measuring ΔU and ΔH
Calorimetry means measuring heat. The reaction happens in or near a known mass of water, and we read the temperature change.
ΔU with a bomb calorimeter (constant volume)
A strong sealed steel vessel (the 'bomb') holds the sample and oxygen under pressure. It sits in water. An electric spark starts the burning. The volume cannot change, so no p–V work is done and q_V = ΔU. Heat released = C_calorimeter × ΔT. The sign for the reaction is opposite to the heat gained by the water.
ΔH with a cup calorimeter (constant pressure)
For reactions in solution (like acid + base), an insulated cup open to the air is enough. The pressure stays at atmospheric, so q_p = ΔH. Heat = m c ΔT of the solution; ΔH per mole = −q ÷ moles.
If ΔT rises, the reaction is exothermic (ΔH < 0). If it falls, the reaction is endothermic (ΔH > 0).
Reaction enthalpy, standard states and formation
The enthalpy of reaction ΔrH is the enthalpy change when the moles shown in the balanced equation react: ΔrH = ΣH(products) − ΣH(reactants).
Standard enthalpy
Values depend on conditions, so we compare them in the standard state: the pure substance at 1 bar and the stated temperature (usually 298 K). Standard values carry a ° sign: ΔrH°.
Thermochemical equations
A balanced equation with states and ΔrH, e.g. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); ΔrH° = −890.3 kJ mol⁻¹. Rules: (1) coefficients are moles; (2) double the equation → double ΔH (it is extensive); (3) reverse the equation → change the sign of ΔH; (4) states matter (H₂O(l) and H₂O(g) give different values).
Standard enthalpy of formation ΔfH°
The enthalpy change when 1 mol of a compound forms from its elements in their most stable (reference) states. For an element in its reference state (O₂(g), C graphite, H₂(g)), ΔfH° = 0. Then ΔrH° = Σ ai ΔfH°(products) − Σ bi ΔfH°(reactants).
Hess's law of constant heat summation
Hess's law: if a reaction takes place in several steps, its total enthalpy change is the sum of the enthalpy changes of the steps. The route does not matter, because H is a state function.
Example: C(graphite) + O₂ → CO₂, ΔH = −393.5 kJ. Or in two steps: C + ½O₂ → CO (−110.5 kJ), then CO + ½O₂ → CO₂ (−283.0 kJ). Sum = −393.5 kJ.
Why it is useful
Some enthalpies (like ΔfH of CO or CH₄) cannot be measured directly. We add, subtract, reverse or multiply known thermochemical equations until they give the wanted equation, and treat the ΔH values the same way.
In the exam
Hess's law numericals (3 marks), ΔH–ΔU conversion (2 marks) and calorimetry definitions are asked often.
Try it at home: a cup calorimeter
Take a steel tumbler inside a thermocol cup. Add 100 mL of tap water and note its temperature (a kitchen or fever thermometer is fine). Stir in two spoons of washing soda or a little detergent powder and read the new temperature. If it rose, the dissolving was exothermic. Now use q = m c ΔT = 100 × 4.18 × ΔT. Try the same numbers in the 3D free play.
Key formulas and definitions
- H = U + pV; at constant p: ΔH = q_p; at constant V: ΔU = q_V
- ΔH = ΔU + Δn_g RT (Δn_g = gas moles of products − reactants)
- q = C ΔT = m c ΔT (c of water = 4.18 J g⁻¹ K⁻¹)
- C_p − C_v = R (ideal gas, per mole)
- ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants)
- Reverse equation → change sign of ΔH; multiply equation → multiply ΔH
- Hess's law: ΔH(total) = ΔH₁ + ΔH₂ + ΔH₃ + …
Worked examples
1. How much heat is needed to warm 250 g of water from 20 °C to 80 °C?
q = m c ΔT = 250 × 4.18 × (80 − 20) = 250 × 4.18 × 60 = 62 700 J = 62.7 kJ.
2. 1.00 g of graphite is burnt in a bomb calorimeter of heat capacity 20.0 kJ K⁻¹. The temperature rises by 1.64 K. Find ΔU for burning 1 mol of carbon.
Heat gained by calorimeter = 20.0 × 1.64 = 32.8 kJ. The reaction gave this out, so q = −32.8 kJ per gram. 1 mol C = 12 g: ΔU = −32.8 × 12 = −393.6 kJ mol⁻¹.
3. 50 mL of 1 M HCl is mixed with 50 mL of 1 M NaOH in a cup calorimeter. The temperature rises by 6.8 K. Find the enthalpy of neutralisation (take 100 g solution, c = 4.18 J g⁻¹ K⁻¹).
q = 100 × 4.18 × 6.8 = 2842 J. Moles of water formed = 0.050 × 1 = 0.050 mol. ΔH = −2842 ÷ 0.050 = −56 840 J mol⁻¹ ≈ −56.8 kJ mol⁻¹.
4. For N₂(g) + 3H₂(g) → 2NH₃(g), ΔU = −87.0 kJ at 298 K. Find ΔH.
Δn_g = 2 − 4 = −2. ΔH = ΔU + Δn_g RT = −87.0 + (−2)(8.314 × 10⁻³)(298) = −87.0 − 4.96 = −91.96 ≈ −92.0 kJ.
5. Find ΔcH° of methane. ΔfH°: CH₄(g) = −74.8, CO₂(g) = −393.5, H₂O(l) = −285.8 kJ mol⁻¹.
CH₄ + 2O₂ → CO₂ + 2H₂O(l). ΔrH° = [−393.5 + 2(−285.8)] − [−74.8 + 0] = −965.1 + 74.8 = −890.3 kJ mol⁻¹.
6. Given C + O₂ → CO₂, ΔH = −393.5 kJ and CO + ½O₂ → CO₂, ΔH = −283.0 kJ. Find ΔfH° of CO.
Want: C + ½O₂ → CO. Take equation 1 as it is, and reverse equation 2: CO₂ → CO + ½O₂, ΔH = +283.0. Add: C + ½O₂ → CO, ΔH = −393.5 + 283.0 = −110.5 kJ mol⁻¹.
7. Use Hess's law to find ΔfH° of CH₄. ΔcH°: C(graphite) = −393.5, H₂(g) = −285.8, CH₄(g) = −890.3 kJ mol⁻¹.
Want: C + 2H₂ → CH₄. (i) C + O₂ → CO₂, −393.5. (ii) 2 × [H₂ + ½O₂ → H₂O], 2 × (−285.8) = −571.6. (iii) reverse CH₄ combustion: CO₂ + 2H₂O → CH₄ + 2O₂, +890.3. Add: −393.5 − 571.6 + 890.3 = −74.8 kJ mol⁻¹.
8. The molar heat capacity of an ideal gas at constant volume is 12.5 J K⁻¹ mol⁻¹. Find C_p and the heat needed to warm 2 mol by 10 K at constant pressure.
C_p = C_v + R = 12.5 + 8.3 = 20.8 J K⁻¹ mol⁻¹. q_p = n C_p ΔT = 2 × 20.8 × 10 = 416 J. This is ΔH.
Common mistakes
- Using R = 8.314 with ΔU in kJ. Convert R to 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹ first.
- Counting liquids and solids in Δn_g. Only gas moles count.
- Keeping the sign of ΔH when reversing an equation. Reverse the equation, flip the sign.
- Mixing the heat gained by water with ΔH of the reaction. If water gains heat, the reaction is exothermic: ΔH = −q/n.