Physical Chemistry
8 marks · NormalThe following graph shows the rate of a reaction (in g/s) for different concentrations of hydrochloric acid, and gives data for a calorimetry experiment involving the neutralisation of hydrochloric acid with sodium hydroxide.
| Concentration of HCl (mol/dm³) | Rate (g/s) |
|---|---|
| 0.20 | 0.013 |
| 0.40 | 0.026 |
| 0.60 | 0.039 |
| 0.80 | 0.052 |
Describe the relationship between concentration and rate as shown in the table.
Explain this relationship using particle (collision) theory.
A student mixes 25.0 cm³ of 1.00 mol/dm³ HCl with 25.0 cm³ of 1.00 mol/dm³ NaOH in a polystyrene cup. Initial temperature is 21.0°C and maximum temperature is 29.5°C. (i) Calculate the total volume of solution and the number of moles of HCl used. (ii) Calculate the heat evolved (Q) in joules. Assume density is 1.00 g/cm³ and c = 4.18 J/g°C. (iii) Calculate the molar enthalpy change (ΔH) for the reaction in kJ/mol.
Calculate the total volume of solution and the number of moles of HCl used.
Calculate Q, the heat evolved, in joules.
Calculate the molar enthalpy change (ΔH) for the reaction in kJ/mol. Give your answer to 2 significant figures and state if the reaction is exothermic or endothermic.
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
(a(i)) The rate is directly proportional to the concentration; when concentration doubles, the rate doubles.
(a(ii)) At higher concentration, there are more particles per unit volume so more frequent collisions, increasing the rate.
(b(i)) 50.0 cm³, 0.025 mol HCl mol
(b(ii)) Q = mcΔT = 50.0 × 4.18 × (29.5–21.0) = 50 × 4.18 × 8.5 = 1,776.5 J ≈ 1,780 J (3 sf) J
(b(iii)) ΔH = –71 kJ/mol. It is exothermic (negative sign). kJ/mol · must show correct sign and unit
Common mistakes: Ignores that heat is negative for exothermic; Confuses moles calculation
Show mark scheme and worked solution
Marking points
D1 (a(i)) · 1 mark · Describes direct proportionality between rate and concentration.
D2 (a(ii)) · 1 mark · Uses collision theory to explain increased rate at higher concentration (more collisions per second).
C1 (b(i)) · 1 mark · Correct moles of HCl (0.025 mol) and total volume (50.0 cm³).
C2 (b(ii)) · 2 marks · Accurate calculation of Q in J, all working shown.
C3 (b(iii)) · 2 marks · Correct calculation of ΔH in kJ/mol (correct sign and unit).
E1 (b(iii)) · 1 mark · States that the reaction is exothermic (as ΔH is negative).
Worked solution
a(i) The table shows that as concentration doubles, the rate doubles—this is a directly proportional relationship.
a(ii) Collision theory: Higher concentration means more HCl particles per unit volume, so there are more successful collisions per second, increasing the rate.
b(i) Total volume = 25.0 + 25.0 = 50.0 cm³; moles of HCl = 1.00 mol/dm³ × 25.0/1000 = 0.025 mol.
b(ii) Q = mcΔT = 50.0 g × 4.18 J/g°C × (29.5-21.0)°C = 50 × 4.18 × 8.5 = 1,776.5 J ≈ 1,780 J.
b(iii) ΔH = Q / moles = 1,780 J / 0.025 mol = 71,200 J/mol = 71 kJ/mol (rounded, 2 SF). As this reaction is exothermic, ΔH = –71 kJ/mol.