Students should be able to: use this equation to calculate the molar…

The idea

A molar enthalpy change needs two numbers: the heat, from q = mc∆T, and the amount in moles of the substance the question asks you to quote it for. Work out q in joules, divide by 1000, then divide by the moles of the limiting reactant, and attach the sign. So dissolving 4.00 g of ammonium nitrate, 0.0500 mol, in 50.0 cm³ of water and recording a fall of 6.2 K gives q = 50.0 × 4.18 × 6.2 = 1296 J, so ∆H = +1.296 ÷ 0.0500 = +25.9 kJ mol⁻¹, positive because the solution cooled. The same equation runs backwards: given ∆H and a temperature change you can find the mass or the concentration, and given the power of a heater, energy is power × time instead of mc∆T.

What earns the marks

• The amount in moles of the substance the enthalpy change is to be quoted per mole of.
• q calculated correctly from q = mc∆T, or from power × time when a heater is used.
• Dividing the energy in kilojoules by the moles to reach kJ mol⁻¹.
• The final value with the correct sign and units, allowing error carried forward from an earlier slip.

Common mistakes and how to put them right

• Dividing by the moles of the excess reagent instead of the limiting one.

• Dividing joules by moles and then quoting the answer as kJ mol⁻¹ without the factor of 1000.

• Losing the sign at the last step, so an exothermic result is written as positive.

• Rounding the moles to one significant figure early, which moves the final answer outside the accepted range.

• Using the mass of water only when the question tells you the total mass of the mixture changed temperature.

Exam tips

• Set the calculation out in three labelled lines, q, n, then ∆H = −q/n, so process marks survive an arithmetic slip.
• Quote the answer to the same number of significant figures as the least precise data, usually three.
• When a heater is specified in kW, energy in kJ is power in kW multiplied by time in seconds.