Calorimetry and enthalpy change: Class 11 practice
Heat absorbed by a solution q = mcΔT (c ≈ 4.18 J g⁻¹ K⁻¹ for water). Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Specific heat
- Moles
The idea behind calorimetry and enthalpy change
Heat absorbed by a solution q = mcΔT (c ≈ 4.18 J g⁻¹ K⁻¹ for water). ΔH per mole = −q ÷ moles reacting (negative when the solution warms).
Step-by-step method
- q = m × 4.18 × ΔT.
- Find moles of the limiting reactant.
- ΔH = −q/n, in kJ/mol.
Worked example 1 (Basic)
200 g of solution warms by 1 °C. How much heat was released? (c = 4.18 J/g°C)
- q = 200 × 4.18 × 1
- Answer: 836 J
Worked example 2 (Standard)
100 mL of 0.5 M HCl and 100 mL of 0.5 M NaOH (total 200 g) warm by 5 °C. Find ΔH of neutralisation.
- q = 4180 J
- n = 0.05 mol
- ΔH = −4180/0.05 = -83.6 kJ/mol
- Answer: -83.6 kJ/mol
Worked example 3 (Challenge)
50 mL of 1 M HCl and 50 mL of 1 M NaOH (total 100 g) warm by 6 °C. Find ΔH of neutralisation.
- q = 2508 J
- n = 0.05 mol
- ΔH = −2508/0.05 = -50.16 kJ/mol
- Answer: -50.16 kJ/mol
Common mistakes and misconceptions
- Wrong sign for exothermic
- Using only the acid volume as mass
Practice questions
- (Basic) 400 g of solution warms by 2.5 °C. How much heat was released? (c = 4.18 J/g°C)
- (Basic) 200 g of solution warms by 5 °C. How much heat was released? (c = 4.18 J/g°C)
- (Standard) 200 mL of 1 M HCl and 200 mL of 1 M NaOH (total 400 g) warm by 6.5 °C. Find ΔH of neutralisation.
- (Standard) 100 mL of 0.5 M HCl and 100 mL of 0.5 M NaOH (total 200 g) warm by 1 °C. Find ΔH of neutralisation.
- (Challenge) 50 mL of 2 M HCl and 50 mL of 2 M NaOH (total 100 g) warm by 3 °C. Find ΔH of neutralisation.
- (Challenge) 50 mL of 0.5 M HCl and 50 mL of 0.5 M NaOH (total 100 g) warm by 1 °C. Find ΔH of neutralisation.
Answer key with full working
- Q1: q = 400 × 4.18 × 2.5 → Answer: 4180 J
- Q2: q = 200 × 4.18 × 5 → Answer: 4180 J
- Q3: q = 10868 J → n = 0.2 mol → ΔH = −10868/0.2 = -54.34 kJ/mol → Answer: -54.34 kJ/mol
- Q4: q = 836 J → n = 0.05 mol → ΔH = −836/0.05 = -16.72 kJ/mol → Answer: -16.72 kJ/mol
- Q5: q = 1254 J → n = 0.1 mol → ΔH = −1254/0.1 = -12.54 kJ/mol → Answer: -12.54 kJ/mol
- Q6: q = 418 J → n = 0.025 mol → ΔH = −418/0.025 = -16.72 kJ/mol → Answer: -16.72 kJ/mol
Syllabus note
This skill is listed under "Thermodynamics" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is calorimetry and enthalpy change in?
"Thermodynamics", Class 11 Chemistry (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.