Step 1. Calculate Heat 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. Same magnitude)
Step 2. Scale to Molar Enthalpy Change
\( \Delta H = -\dfrac{q}{n} \)
n = moles of the limiting reactant. The negative sign ensures the correct sign convention (temperature ↑ = exothermic = ΔH negative).
Convert J → kJ by dividing by 1000.
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 |
⚠️ Examiner Traps
- Mass confusion: m = mass of the water/solution, NOT the solid reagent
- 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 temp rises, q is positive → ΔH must be negative (exothermic)
- °C to K: ΔT in °C ≡ ΔT in K. No conversion needed. Don't waste exam time!
📐 Worked Example: Neutralisation Calorimetry
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), but the reaction is exothermic (ΔH must be negative). Always divide by 1000 to convert J to kJ.
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