Calorimetry experiments for determining the enthalpy changes of reactions
2 marks · EasyA student dissolves 3.21 g of potassium nitrate in 50 g of water. The temperature of the water falls by 5.0°C. The specific heat capacity of water is 4.18 J/g°C. Calculate the molar enthalpy change (ΔH) of solution for potassium nitrate. (Relative formula mass of KNO₃ = 101)
Give your answer in kJ/mol to 2 significant figures. Show your working.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
–3.3 kJ/mol (±0.1)
Also accept: -3.3; –3.3 kJ/mol
Common mistakes: using positive instead of negative sign when temp falls; forgetting to divide Q by number of moles
Show mark scheme and worked solution
Marking points
M1 · 1 mark · Calculates Q = 50 × 4.18 × 5.0 = 1045 J; calculates moles of KNO₃ = 3.21 / 101 = 0.0318 mol. (Allow error carried forward.) (follow-through)
A1 · 1 mark · Correct calculation of ΔH = Q / moles, and converts to kJ/mol, final answer –3.3 kJ/mol (accept –3.2 to –3.4 kJ/mol), correct sign. (follow-through)
Worked solution
Q = mcΔT = 50 × 4.18 × 5.0 = 1045 J
Convert to kJ: 1045 J = 1.045 kJ
Number of moles of KNO₃ = 3.21 / 101 = 0.0318 mol
ΔH = Q / moles = 1.045 / 0.0318 = 32.8 kJ/mol
The temperature falls, so the process is endothermic. ΔH is positive: +33 kJ/mol. But for solution, convention is usually positive for endothermic (accept positive). If the question wants sign, accept positive as primary, but negative if heat is lost by water.
Final answer: +33 kJ/mol (to 2 significant figures).