Energetics MCQs – 40 Practice Questions with Answers and Explanations
Energetics is one of the most calculation-heavy topics in chemistry – and one of the easiest to score in once the sign conventions click. Work through these 40 free original practice questions on exothermic and endothermic reactions, bond energies, calorimetry, Hess’s law, entropy and Gibbs free energy, then open the worked answer under each question to check your method.
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Section 1: Exothermic & Endothermic Reactions
Q1. A reaction mixture becomes warmer during the reaction and the temperature of the surroundings rises. The reaction is…
- (A) endothermic, because energy is taken in from the surroundings
- (B) exothermic, because energy is taken in from the surroundings
- (C) endothermic, because energy is transferred to the surroundings
- (D) exothermic, because energy is transferred to the surroundings
Show Answer
Answer: D – In an exothermic reaction the chemical system loses energy, so the surroundings gain it and warm up. Endothermic reactions do the opposite and cool the surroundings.
Q2. Which process is endothermic?
- (A) Combustion of methane
- (B) Respiration in muscle cells
- (C) Photosynthesis in a leaf
- (D) Neutralisation of an acid by an alkali
Show Answer
Answer: C – Photosynthesis takes in light energy to build glucose from carbon dioxide and water, so it is endothermic. Combustion, respiration and neutralisation all release energy.
Q3. An instant sports cold pack works when an inner bag breaks and ammonium nitrate dissolves in water. The pack feels cold because…
- (A) the dissolving process is exothermic and ΔH is negative
- (B) the dissolving process is exothermic and ΔH is positive
- (C) the dissolving process is endothermic and ΔH is negative
- (D) the dissolving process is endothermic and ΔH is positive
Show Answer
Answer: D – Energy is taken in from the water (and from the injured skin), so the temperature falls. Endothermic changes always have a positive ΔH value.
Q4. On a reaction profile diagram for an exothermic reaction, which statement is correct?
- (A) The products are drawn higher than the reactants and ΔH is positive
- (B) The products are drawn lower than the reactants and ΔH is negative
- (C) The activation energy is the difference in height between reactants and products
- (D) The curve has no peak because no activation energy is needed
Show Answer
Answer: B – Exothermic reactions lose energy overall, so the products sit at a lower energy than the reactants and ΔH is negative. The activation energy is the height from the reactants up to the peak of the curve.
Q5. Thermal decomposition of calcium carbonate, CaCO3 → CaO + CO2, needs continuous strong heating to keep it going. This shows the reaction is…
- (A) exothermic, because heat is produced
- (B) endothermic, because heat must be supplied
- (C) exothermic, because a gas is produced
- (D) endothermic, because it is a neutralisation
Show Answer
Answer: B – A reaction that only continues while energy is supplied from outside is taking energy in, so it is endothermic (ΔH positive).
Q6. During any chemical reaction, energy is needed to break bonds in the reactants and energy is released when new bonds form in the products. A reaction is exothermic overall when…
- (A) more energy is needed to break bonds than is released making bonds
- (B) exactly the same energy is involved in breaking and making bonds
- (C) more energy is released making bonds than is needed to break bonds
- (D) no bonds are broken at all
Show Answer
Answer: C – Bond breaking is endothermic (energy in) and bond making is exothermic (energy out). If bond making releases more than bond breaking absorbs, the excess is transferred to the surroundings.
Q7. The activation energy of a reaction is best defined as…
- (A) the total energy released by the reaction
- (B) the difference in energy between products and reactants
- (C) the minimum energy particles must have when they collide for a reaction to happen
- (D) the energy stored inside the products
Show Answer
Answer: C – Particles must collide with at least the activation energy for bonds to break and the reaction to start. It is the height of the barrier on the reaction profile, not the overall energy change.
Q8. A catalyst is added to an exothermic reaction. On the reaction profile, the catalyst…
- (A) lowers the peak of the curve but leaves the reactant and product levels unchanged
- (B) lowers the product level, making ΔH more negative
- (C) raises the activation energy so more collisions succeed
- (D) removes the activation energy completely
Show Answer
Answer: A – A catalyst provides an alternative route with a lower activation energy, so the peak drops. The energies of reactants and products, and therefore ΔH, stay exactly the same.
Q9. When magnesium reacts with dilute hydrochloric acid in an open test tube, the tube feels hot. Which energy transfer correctly describes the reaction?
- (A) Energy is transferred from the surroundings to the reacting chemicals
- (B) Energy is created by the reaction
- (C) No energy is transferred because the tube is open
- (D) Energy is transferred from the reacting chemicals to the surroundings
Show Answer
Answer: D – The reaction is exothermic: the chemical system loses energy to the surroundings, which is why the tube (part of the surroundings) warms up. Energy is transferred, never created.
Q10. Which pair of changes are both exothermic?
- (A) Melting ice and evaporating water
- (B) Freezing water and condensing steam
- (C) Thermal decomposition and photosynthesis
- (D) Dissolving ammonium nitrate and melting ice
Show Answer
Answer: B – Freezing and condensing are the reverse of melting and evaporating: particles lose energy as they become more ordered, so both release energy. Melting, evaporation, decomposition and photosynthesis all take energy in.
Section 2: Bond Energies & Calorimetry
Q11. Bond energies (kJ/mol): H−H = 436, Cl−Cl = 242, H−Cl = 431. Using H2 + Cl2 → 2HCl, the enthalpy change is…
- (A) +184 kJ/mol
- (B) −247 kJ/mol
- (C) +247 kJ/mol
- (D) −184 kJ/mol
Show Answer
Answer: D – Bonds broken: 436 + 242 = 678 kJ. Bonds formed: 2 × 431 = 862 kJ. ΔH = broken − formed = 678 − 862 = −184 kJ/mol.
Q12. Bond energies (kJ/mol): C−H = 413, O=O = 498, C=O = 805, O−H = 464. For CH4 + 2O2 → CO2 + 2H2O, ΔH is…
- (A) −818 kJ/mol
- (B) +818 kJ/mol
- (C) −3466 kJ/mol
- (D) +2648 kJ/mol
Show Answer
Answer: A – Broken: 4(413) + 2(498) = 1652 + 996 = 2648 kJ. Formed: 2(805) + 4(464) = 1610 + 1856 = 3466 kJ. ΔH = 2648 − 3466 = −818 kJ/mol. Remember CO2 contains two C=O bonds.
Q13. In a bond energy calculation, a student gets ΔH = +45 kJ/mol for a combustion reaction. The most likely mistake is that the student…
- (A) used bond energies in J instead of kJ
- (B) calculated bonds formed minus bonds broken instead of broken minus formed
- (C) forgot to include the activation energy
- (D) used the wrong temperature
Show Answer
Answer: B – Combustion is always exothermic, so a positive answer is impossible. Reversing the subtraction (formed − broken) flips the sign – the classic bond-energy trap.
Q14. A spirit burner heats 200 g of water from 21.0 °C to 38.5 °C. Taking the specific heat capacity of water as 4.18 J/g/°C, the energy transferred to the water is…
- (A) 1.46 kJ
- (B) 14.6 kJ
- (C) 146 kJ
- (D) 836 J
Show Answer
Answer: B – q = mcΔT = 200 × 4.18 × 17.5 = 14,630 J = 14.6 kJ. Watch the units: the formula gives joules, so divide by 1000 for kJ.
Q15. Burning 0.40 g of ethanol raised the temperature of 100 g of water by 11.5 °C. The experimental enthalpy change of combustion of ethanol (Mr = 46) is closest to…
- (A) −12.0 kJ/mol
- (B) +553 kJ/mol
- (C) −553 kJ/mol
- (D) −4807 kJ/mol
Show Answer
Answer: C – q = 100 × 4.18 × 11.5 = 4807 J = 4.81 kJ. Moles of ethanol = 0.40 / 46 = 0.00870 mol. ΔH = −4.81 / 0.00870 = −553 kJ/mol (negative because combustion is exothermic).
Q16. The data-book enthalpy of combustion of ethanol is −1367 kJ/mol, but a school calorimetry experiment gives about −550 kJ/mol. The main reason for the difference is…
- (A) the ethanol burned too quickly
- (B) much of the energy heated the air, the can and the thermometer instead of the water, and some ethanol may have burned incompletely
- (C) the water absorbed too much energy
- (D) bond energies are only approximate
Show Answer
Answer: B – In a simple school burner, a large fraction of the released energy never reaches the water, so the measured value is always much smaller in magnitude than the true value. Incomplete combustion to carbon monoxide or soot lowers it further.
Q17. When 50 cm3 of 1.0 mol/dm3 hydrochloric acid is mixed with 50 cm3 of 1.0 mol/dm3 sodium hydroxide, the temperature rises by 6.8 °C. The enthalpy change of neutralisation (total solution mass 100 g, c = 4.18 J/g/°C) is…
- (A) −2.84 kJ/mol
- (B) −28.4 kJ/mol
- (C) +56.8 kJ/mol
- (D) −56.8 kJ/mol
Show Answer
Answer: D – q = 100 × 4.18 × 6.8 = 2842 J = 2.84 kJ. Moles of water formed = 0.050 mol. ΔH = −2.84 / 0.050 = −56.8 kJ/mol – close to the standard value of about −57 kJ/mol.
Q18. A fuel is burned to heat 0.150 kg of water by 25.0 °C (c = 4.18 J/g/°C). If 0.0250 mol of fuel was used, the enthalpy change of combustion is…
- (A) −627 kJ/mol
- (B) −15.7 kJ/mol
- (C) −62.7 kJ/mol
- (D) +627 kJ/mol
Show Answer
Answer: A – q = 150 × 4.18 × 25.0 = 15,675 J = 15.7 kJ. ΔH = −15.7 / 0.0250 = −627 kJ/mol.
Q19. Which change would NOT improve the accuracy of a calorimetry experiment using a spirit burner?
- (A) Putting a lid and insulation around the metal can
- (B) Using a draught shield around the flame
- (C) Stirring the water and using a more sensitive thermometer
- (D) Using a larger volume of water so the temperature rise is smaller
Show Answer
Answer: D – A larger water volume gives a smaller temperature rise, which makes the percentage uncertainty in ΔT bigger, not smaller. Lids, insulation, draught shields and stirring all reduce heat losses or measurement errors.
Q20. Bond energies (kJ/mol): N≡N = 945, H−H = 436, N−H = 391. For N2 + 3H2 → 2NH3, ΔH is…
- (A) +93 kJ/mol
- (B) −93 kJ/mol
- (C) −2253 kJ/mol
- (D) −147 kJ/mol
Show Answer
Answer: B – Broken: 945 + 3(436) = 2253 kJ. Formed: 2 × 3(391) = 2346 kJ (each NH3 has three N−H bonds). ΔH = 2253 − 2346 = −93 kJ/mol.
Section 3: Hess’s Law & Enthalpy Cycles
Q21. The standard enthalpy change of formation of a compound is the enthalpy change when…
- (A) 1 mole of the compound burns completely in oxygen under standard conditions
- (B) 1 mole of the compound is formed from its elements in their standard states under standard conditions
- (C) 1 mole of the compound is broken into gaseous atoms
- (D) 1 mole of an acid is neutralised by a base
Show Answer
Answer: B – Formation always means making exactly 1 mole of the compound from its elements in their standard states. By definition, the enthalpy of formation of an element in its standard state is zero.
Q22. Using ΔHf values (kJ/mol): CH4 = −75, CO2 = −394, H2O = −286, the enthalpy change of combustion of methane, CH4 + 2O2 → CO2 + 2H2O, is…
- (A) −605 kJ/mol
- (B) −755 kJ/mol
- (C) −891 kJ/mol
- (D) −966 kJ/mol
Show Answer
Answer: C – With formation data: ΔH = ΣΔHf(products) − ΣΔHf(reactants) = [(−394) + 2(−286)] − [(−75) + 0] = (−966) + 75 = −891 kJ/mol. O2 contributes zero as an element.
Q23. ΔHc values (kJ/mol): carbon = −394, hydrogen (H2) = −286, ethanol = −1367. Using a Hess cycle, the enthalpy of formation of ethanol, 2C + 3H2 + ½O2 → C2H5OH, is…
- (A) −279 kJ/mol
- (B) +279 kJ/mol
- (C) −1646 kJ/mol
- (D) −3013 kJ/mol
Show Answer
Answer: A – With combustion data the cycle goes the other way round: ΔHf = ΔHc(elements) − ΔHc(compound) = [2(−394) + 3(−286)] − (−1367) = (−1646) + 1367 = −279 kJ/mol. Using the formation-data direction here is the classic trap.
Q24. The enthalpy change of neutralisation for a strong acid with a strong alkali is always about −57 kJ/mol because…
- (A) all strong acids contain the same anion
- (B) the only reaction happening is H+ + OH− → H2O; the other ions are spectators
- (C) strong acids are fully concentrated
- (D) the volume of solution is always the same
Show Answer
Answer: B – Strong acids and alkalis are fully ionised, so whichever pair you choose, the actual reaction is the formation of 1 mole of water from H+ and OH− ions – the same enthalpy change every time.
Q25. Hess’s law states that…
- (A) the enthalpy change of a reaction depends only on the initial and final states, provided conditions (e.g. temperature and pressure) stay the same, whatever route is taken
- (B) the enthalpy change equals the activation energy
- (C) enthalpy changes can only be measured directly, never calculated
- (D) the enthalpy change of the reverse reaction has the same sign and value
Show Answer
Answer: A – Enthalpy is a state function, so the overall change is route-independent under the same conditions. The reverse reaction has the same magnitude but the opposite sign, which is why cycle directions matter.
Q26. The enthalpy change for the reaction C(graphite) → C(diamond) cannot be measured directly, but ΔHc(graphite) = −394 kJ/mol and ΔHc(diamond) = −396 kJ/mol. The enthalpy change of the conversion is…
- (A) −2 kJ/mol
- (B) +790 kJ/mol
- (C) +2 kJ/mol
- (D) −790 kJ/mol
Show Answer
Answer: C – Both forms burn to the same product, CO2. ΔH = ΔHc(graphite) − ΔHc(diamond) = (−394) − (−396) = +2 kJ/mol. Diamond is very slightly higher in energy than graphite.
Q27. The standard enthalpy change of atomisation of an element is the enthalpy change when…
- (A) 1 mole of gaseous atoms is formed from the element in its standard state
- (B) 1 mole of the element burns in oxygen
- (C) 1 mole of gaseous atoms is formed from 1 mole of a compound
- (D) an atom gains an electron
Show Answer
Answer: A – Atomisation always produces exactly 1 mole of separate gaseous atoms from the element in its standard state – for example ½Cl2(g) → Cl(g). It is a key step in Born-Haber cycles.
Q28. ΔHf values (kJ/mol): MgO = −602, CO2 = −394, MgCO3 = −1096. For MgCO3 → MgO + CO2, ΔH is…
- (A) +100 kJ/mol
- (B) −100 kJ/mol
- (C) +208 kJ/mol
- (D) −2092 kJ/mol
Show Answer
Answer: A – ΔH = ΣΔHf(products) − ΣΔHf(reactants) = [(−602) + (−394)] − (−1096) = (−996) + 1096 = +100 kJ/mol. Positive, as expected for a thermal decomposition.
Q29. When 25 cm3 of 2.0 mol/dm3 HCl is neutralised by excess NaOH, the temperature of the 50 g of solution rises by 13.7 °C (c = 4.18 J/g/°C). The enthalpy change of neutralisation is…
- (A) −28.6 kJ/mol
- (B) +57.3 kJ/mol
- (C) −114.6 kJ/mol
- (D) −57.3 kJ/mol
Show Answer
Answer: D – q = 50 × 4.18 × 13.7 = 2863 J = 2.86 kJ. Moles of acid = 2.0 × 0.025 = 0.050 mol. ΔH = −2.86 / 0.050 = −57.3 kJ/mol, matching the accepted value.
Q30. In a Hess cycle using enthalpies of combustion, the arrows for the combustion steps point…
- (A) upward from the products to the reactants
- (B) in the same direction as the target reaction only
- (C) from the elements to the compound
- (D) down from both reactants and products to the same combustion products (oxides)
Show Answer
Answer: D – A combustion cycle works because reactants and products can both be burned to identical products (CO2 and H2O), so both arrows point down to that common level. A formation cycle instead builds everything up from the elements.
Section 4: Entropy & Gibbs Free Energy (A-Level / IB / AP Stretch)
Q31. Which change is accompanied by an increase in entropy?
- (A) Water freezing at 0 °C
- (B) A gas condensing to a liquid
- (C) Sublimation of solid carbon dioxide, CO2(s) → CO2(g)
- (D) Forming a crystal lattice from aqueous ions
Show Answer
Answer: C – Entropy measures disorder. A gas is far more disordered than a solid, so sublimation gives a large entropy increase. Freezing, condensing and crystallising all make the system more ordered, so entropy decreases.
Q32. Standard entropies, S (J/K/mol): CaCO3(s) = 93, CaO(s) = 40, CO2(g) = 214. For CaCO3(s) → CaO(s) + CO2(g), ΔS is…
- (A) +161 J/K/mol
- (B) −161 J/K/mol
- (C) +347 J/K/mol
- (D) +121 J/K/mol
Show Answer
Answer: A – ΔS = ΣS(products) − ΣS(reactants) = (40 + 214) − 93 = +161 J/K/mol. The positive sign makes sense: a gas is produced from a solid.
Q33. For the Haber process reaction N2 + 3H2 → 2NH3, the entropy change of the system is expected to be…
- (A) positive, because ammonia is a gas
- (B) zero, because entropy never changes in a reaction
- (C) negative, because 4 moles of gas become 2 moles of gas
- (D) positive, because the reaction is exothermic
Show Answer
Answer: C – Gas molecules carry most of the entropy. Going from 4 moles of gas to 2 moles of gas reduces disorder, so ΔS is negative – whatever the enthalpy change is doing.
Q34. A reaction has ΔH = −92 kJ/mol and ΔS = −199 J/K/mol. At 298 K, ΔG is…
- (A) −151 kJ/mol
- (B) −32.7 kJ/mol
- (C) +32.7 kJ/mol
- (D) −92.2 kJ/mol
Show Answer
Answer: B – ΔG = ΔH − TΔS = −92,000 − 298 × (−199) = −92,000 + 59,302 = −32,698 J = −32.7 kJ/mol. Convert ΔS to kJ (or ΔH to J) before combining – the units trap catches many candidates.
Q35. From the result pattern in Q34 (ΔH negative, ΔS negative), this reaction becomes feasible (spontaneous) only when…
- (A) the temperature is low enough, because the −TΔS term grows more positive as T rises
- (B) the temperature is as high as possible
- (C) it is feasible at all temperatures
- (D) it is never feasible
Show Answer
Answer: A – With ΔH < 0 and ΔS < 0, ΔG = ΔH − TΔS is negative only at low temperatures, before the positive T|ΔS| term outweighs ΔH. This is why the Haber process yield falls as temperature rises.
Q36. Decomposition of calcium carbonate has ΔH = +178 kJ/mol and ΔS = +161 J/K/mol. The minimum temperature at which it becomes feasible is approximately…
- (A) 905 K
- (B) 1.11 K
- (C) 1106 K
- (D) 178 K
Show Answer
Answer: C – At the feasibility limit ΔG = 0, so T = ΔH / ΔS = 178,000 / 161 = 1106 K (about 833 °C). Above this temperature TΔS exceeds ΔH and ΔG turns negative.
Q37. Standard entropies, S (J/K/mol): H2(g) = 131, O2(g) = 205, H2O(l) = 70. For 2H2(g) + O2(g) → 2H2O(l), ΔS is…
- (A) +163 J/K/mol
- (B) −163 J/K/mol
- (C) −327 J/K/mol
- (D) −266 J/K/mol
Show Answer
Answer: C – ΔS = 2(70) − [2(131) + 205] = 140 − 467 = −327 J/K/mol. Three moles of gas become a liquid, so a large negative value is expected.
Q38. A reaction has ΔH = +46 kJ/mol and ΔS = +108 J/K/mol. At 500 K, the reaction is…
- (A) feasible, because ΔG = −8 kJ/mol
- (B) not feasible, because ΔG = +8 kJ/mol
- (C) feasible, because all endothermic reactions are feasible at high temperature
- (D) not feasible, because endothermic reactions are never feasible
Show Answer
Answer: A – ΔG = 46,000 − 500 × 108 = 46,000 − 54,000 = −8000 J = −8 kJ/mol. Negative ΔG means feasible – an endothermic reaction can be spontaneous when the entropy term wins.
Q39. In the Born-Haber cycle for sodium chloride, which sequence of steps correctly leads from Na(s) and ½Cl2(g) to gaseous Na+ and Cl− ions?
- (A) Electron affinity of chlorine, then ionisation of sodium, then atomisation
- (B) Ionisation of chlorine, then electron affinity of sodium
- (C) Lattice enthalpy first, then atomisation of both elements
- (D) Atomisation of sodium, ionisation of sodium, atomisation of chlorine, electron affinity of chlorine
Show Answer
Answer: D – The gaseous-ion route is: sublime/atomise Na, ionise Na(g), atomise ½Cl2 to Cl(g), then add an electron to Cl(g). Only then do the ions come together, releasing the lattice enthalpy. Sodium is ionised (loses an electron); chlorine gains it.
Q40. Born-Haber data for NaCl (kJ/mol): ΔHf = −411, atomisation of Na = +108, first ionisation energy of Na = +496, atomisation of Cl (from ½Cl2) = +122, electron affinity of Cl = −349. The lattice enthalpy of NaCl is…
- (A) −34 kJ/mol
- (B) +788 kJ/mol
- (C) −788 kJ/mol
- (D) +34 kJ/mol
Show Answer
Answer: C – Hess: ΔHf = atomisation(Na) + IE(Na) + atomisation(Cl) + EA(Cl) + lattice. So lattice = −411 − (108 + 496 + 122 − 349) = −411 − 377 = −788 kJ/mol. The lattice formation step is strongly exothermic, which is what makes the overall cycle work.
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