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Hess's law and enthalpy of reaction: Class 11 practice

Enthalpy change depends only on start and end states, so equations can be added like algebra: reverse an equation → change the sign of ΔH; multiply an equation → multiply ΔH. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Balancing equations

The idea behind hess's law and enthalpy of reaction

Enthalpy change depends only on start and end states, so equations can be added like algebra: reverse an equation → change the sign of ΔH; multiply an equation → multiply ΔH. ΔH = ΣΔH_f(products) − ΣΔH_f(reactants).

Step-by-step method

  1. Arrange the given equations to add up to the target.
  2. Reverse/multiply ΔH accordingly.
  3. Add the ΔH values.

Worked example 1 (Basic)

ΔH_f: CO₂ = -238 kJ/mol, H₂O = -252 kJ/mol, CH₄ = -233 kJ/mol. Find ΔH for CH₄ + 2O₂ → CO₂ + 2H₂O.

  1. ΔH = [-238 + 2(-252)] − [-233 + 0]
  2. = -509 kJ
  3. Answer: -509 kJ

Worked example 2 (Standard)

A → B has ΔH = -149 kJ. What is ΔH for 2B → 2A?

  1. Reverse: +149
  2. Double: 298 kJ
  3. Answer: 298 kJ

Worked example 3 (Challenge)

C + O₂ → CO₂, ΔH = -160 kJ; CO + ½O₂ → CO₂, ΔH = -247 kJ. Find ΔH for C + ½O₂ → CO.

  1. Target = eq1 − eq2
  2. ΔH = -160 − (-247) = 87 kJ
  3. Answer: 87 kJ

Common mistakes and misconceptions

  • Forgetting to change sign on reversal
  • Not multiplying ΔH with coefficients

Practice questions

  1. (Basic) ΔH_f: CO₂ = -355 kJ/mol, H₂O = -165 kJ/mol, CH₄ = -139 kJ/mol. Find ΔH for CH₄ + 2O₂ → CO₂ + 2H₂O.
  2. (Basic) ΔH_f: CO₂ = -307 kJ/mol, H₂O = -141 kJ/mol, CH₄ = -235 kJ/mol. Find ΔH for CH₄ + 2O₂ → CO₂ + 2H₂O.
  3. (Standard) A → B has ΔH = -258 kJ. What is ΔH for 2B → 2A?
  4. (Standard) A → B has ΔH = -111 kJ. What is ΔH for 2B → 2A?
  5. (Challenge) C + O₂ → CO₂, ΔH = -341 kJ; CO + ½O₂ → CO₂, ΔH = -265 kJ. Find ΔH for C + ½O₂ → CO.
  6. (Challenge) C + O₂ → CO₂, ΔH = -200 kJ; CO + ½O₂ → CO₂, ΔH = -75 kJ. Find ΔH for C + ½O₂ → CO.

Answer key with full working

  • Q1: ΔH = [-355 + 2(-165)] − [-139 + 0] → = -546 kJ → Answer: -546 kJ
  • Q2: ΔH = [-307 + 2(-141)] − [-235 + 0] → = -354 kJ → Answer: -354 kJ
  • Q3: Reverse: +258 → Double: 516 kJ → Answer: 516 kJ
  • Q4: Reverse: +111 → Double: 222 kJ → Answer: 222 kJ
  • Q5: Target = eq1 − eq2 → ΔH = -341 − (-265) = -76 kJ → Answer: -76 kJ
  • Q6: Target = eq1 − eq2 → ΔH = -200 − (-75) = -125 kJ → Answer: -125 kJ

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 hess's law and enthalpy of reaction 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.

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