Step 1. Calculate Heat Transferred (q)
Calculating Heat Energy Transferred (q)
\( q = mc\Delta T \)
- m = mass of the solution being heated/cooled (g), not the mass of solid added.
- c = specific heat capacity of water = 4.18 J g⁻¹ K⁻¹
- ΔT = temperature change (°C or K - identical numerical magnitude).
Step 2. Scale to Molar Enthalpy Change
Calculating Molar Enthalpy Change (ΔH)
\( \Delta H = -\dfrac{q}{n} \)
n = moles of the limiting reactant. The negative sign ensures the correct sign convention (temperature rise → exothermic → ΔH negative).
Always convert \(q\) from J to kJ by dividing by 1000 before reporting \(\Delta H\) in \(\text{kJ mol}^{-1}\).
Standard Assumptions (Solution Calorimetry)
| Assumption | Why |
|---|---|
| Density of solution = 1.00 g cm⁻³ | Treat volume (cm³) as mass (g) directly |
| Specific heat capacity = 4.18 J g⁻¹ K⁻¹ | Approximate dilute solution as pure water |
| No heat loss to surroundings | Polystyrene cup assumed to be a perfect insulator |
Five Common Calorimetry Traps
- Mass confusion: \(m\) = mass of the water/solution, NOT the mass of the solid reagent added.
- Mole error: Divide \(q\) by the moles of the limiting reactant only, not total moles.
- Units: \(q\) is in Joules. Divide by 1000 to get kJ mol⁻¹.
- Sign: If temperature rises, \(q\) is positive → ΔH must be negative (exothermic).
- °C to K: ΔT in °C ≡ ΔT in K. No conversion offset needed for temperature changes.
Step-by-Step Neutralisation Calorimetry Calculation
Question: 50.0 cm³ of 1.00 mol dm-3 HCl is mixed with 50.0 cm³ of 1.00 mol dm-3 NaOH. The temperature rises by 6.8 °C. Calculate ΔH for the neutralisation.
Step 1: Calculate heat released
Q = mcΔT = (100.0 g)(4.18 J g-1 K-1)(6.8 K) = 2842 J
Step 2: Calculate moles of limiting reactant
n(HCl) = 0.0500 dm³ × 1.00 mol dm-3 = 0.0500 mol
Step 3: Calculate ΔH
ΔH = −Q/n = −2842 / 0.0500 = −56,840 J mol-1 = −56.8 kJ mol-1
Note: The negative sign is applied because Q is positive (temperature rose), confirming the reaction is exothermic (ΔH < 0).
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