FSc Part 1 Chemistry chapter 7: Thermochemistry MCQs with answers

By Team Megalecture · Updated 4 Nov 2026

20 exam-style questions on Thermochemistry, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.

20 questions · 2 topics · 10 per test

Test this chapter freeAll FSc Part 1 Chemistry chapters

What chapter 7 covers

Ten Thermochemistry questions with answers

Try each one, then open the answer. These ten stay the same; the test draws new ones every time.

1. Hess's law of constant heat summation follows from the fact that

  1. A
    a catalyst does not change ΔH
  2. B
    mass is conserved in a reaction
  3. C
    all reactions are exothermic
  4. D
    enthalpy is a state function, so ΔH depends only on the initial and final states
Show answer

Answer: D. Whatever route is taken between the same reactants and products, the total enthalpy change is the same, because H is a state function.

2. Standard enthalpy changes (ΔH°) refer to which conditions?

  1. A
    25 °C and 1 atm
  2. B
    0 °C and 1 atm
  3. C
    25 °C and 2 atm
  4. D
    100 °C and 1 atm
Show answer

Answer: A. Standard thermochemical conditions are 25 °C (298 K) and 1 atm, with solutions at 1 mol dm−3. 0 °C and 1 atm is STP for gas volumes.

3. For CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) at 298 K, ΔE = −885.4 kJ mol−1. What is ΔH? (R = 8.314 J K−1 mol−1)

  1. A
    −880.4 kJ mol−1
  2. B
    −885.4 kJ mol−1
  3. C
    −887.9 kJ mol−1
  4. D
    −890.4 kJ mol−1
Show answer

Answer: D. Gas moles fall from 3 to 1 (water is liquid), so Δng = −2. ΔH = ΔE + ΔngRT = −885.4 + (−2 × 8.314 × 298 ÷ 1000) = −885.4 − 4.96 = −890.4 kJ mol−1.

4. Why is the enthalpy of neutralisation of any strong acid by any strong base almost the same, about −57.4 kJ mol−1?

  1. A
    in every case the reaction is simply H+(aq) + OH−(aq) → H2O(l)
  2. B
    all the salts formed have the same lattice energy
  3. C
    all strong acids have the same molar mass
  4. D
    heat is absorbed to ionise the acid in every case
Show answer

Answer: A. Strong acids and bases are fully ionised, so the other ions are spectators and the only change is the formation of 1 mol of water from H+ and OH−.

5. A reaction is carried out in a beaker, and the beaker becomes cold to the touch. Which statement is correct?

  1. A
    the reaction is endothermic; heat flows from the surroundings into the system
  2. B
    the reaction is exothermic; heat flows from the system to the surroundings
  3. C
    the reaction is endothermic and ΔH is negative
  4. D
    the reaction is exothermic and ΔH is positive
Show answer

Answer: A. The beaker (part of the surroundings) loses heat to the reacting system, so the reaction absorbs heat: endothermic, ΔH positive.

6. For which reaction is ΔH equal to ΔE?

  1. A
    N2(g) + 3H2(g) → 2NH3(g)
  2. B
    CaCO3(s) → CaO(s) + CO2(g)
  3. C
    H2(g) + Cl2(g) → 2HCl(g)
  4. D
    2H2(g) + O2(g) → 2H2O(l)
Show answer

Answer: C. ΔH = ΔE + ΔngRT. For H2 + Cl2 → 2HCl, 2 mol of gas give 2 mol of gas, so Δng = 0 and ΔH = ΔE.

7. ΔH°f: CH4(g) −74.8, CO2(g) −393.5, H2O(l) −285.8 kJ mol−1. What is ΔH° for CH4 + 2O2 → CO2 + 2H2O(l)?

  1. A
    −604.5 kJ mol−1
  2. B
    +890.3 kJ mol−1
  3. C
    −890.3 kJ mol−1
  4. D
    −1039.9 kJ mol−1
Show answer

Answer: C. ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants) = [−393.5 + 2(−285.8)] − (−74.8) = −965.1 + 74.8 = −890.3 kJ mol−1. Forgetting the 2 for water gives −604.5.

8. The standard enthalpy of atomisation of an element is the enthalpy change when

  1. A
    one mole of the element is formed from its gaseous atoms
  2. B
    one mole of gaseous atoms is formed from the element in its standard state
  3. C
    one mole of a compound is formed from its elements
  4. D
    one mole of the element burns completely in oxygen
Show answer

Answer: B. Atomisation produces 1 mol of gaseous atoms, e.g. ½Cl2(g) → Cl(g) or Na(s) → Na(g). It is always endothermic.

9. The standard enthalpy of combustion of a substance is

  1. A
    positive, because energy is needed to start burning
  2. B
    always negative, because combustion releases energy
  3. C
    zero for elements in their standard states
  4. D
    equal to its lattice energy
Show answer

Answer: B. ΔH°c is the enthalpy change when 1 mol burns completely in oxygen under standard conditions, and burning always gives out heat. Zero applies to ΔH°f of elements, not to combustion.

10. A system absorbs 50 kJ of heat and does 20 kJ of work on its surroundings. What is ΔE for the system?

  1. A
    +30 kJ
  2. B
    +70 kJ
  3. C
    −30 kJ
  4. D
    −70 kJ
Show answer

Answer: A. ΔE = q + w, with q = +50 kJ (heat absorbed) and w = −20 kJ (work done by the system). ΔE = 50 − 20 = +30 kJ. Adding both as positive gives 70 kJ.

Start the chapter testAbout the FSc Part 1 Chemistry test

Read the notes for this chapter

Ways to practise this chapter

Thermochemistry: questions students ask

How many Thermochemistry MCQs are there?

20, each with the answer and a short explanation, across 2 topics: Energy, the first law and enthalpy, Standard enthalpy changes, Hess's law and the Born-Haber cycle. Every chapter test draws ten of them, so you can take it many times.

Is the FSc Part 1 Chemistry chapter 7 test free?

Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.

How is the chapter test marked?

The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.

Can I read the explanations in Urdu?

Yes. In practice mode and on your results you can switch the explanations to Urdu.

Keep going

← Chapter 6: Chemical bondingChapter 8: Chemical equilibrium →All chaptersStuck? Ask a question