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Energy changes worksheet answers for GCSE chemistry (free PDF download)

The problems are on the energy changes worksheet. Try them on your own first, then compare with the answers below.

Quick answer: An exothermic reaction releases energy to the surroundings and its temperature rises; an endothermic reaction absorbs energy and its temperature falls. These worked answers correct the 20 problems of the Chemistry 9-10 worksheet on energy changes, from classify the changes to spot the errors: each solution shows the formula, the substitution and the final answer with its unit.

Energy changes worksheet answers for GCSE chemistry (free PDF download)

Each answer shows the formula, the substitution and the result in bold. Check the sign of every ΔH, because exothermic reactions are negative and endothermic reactions are positive. For bond energy problems, check that each bond is multiplied by its coefficient.

Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: HS-PS1-4

Problem 2 Thermometer evidence

1. 17.5 − 24.0 = −6.5 °C. 2. The temperature falls, so energy is taken in from the surroundings: endothermic. 3. ΔH is positive. Endothermic, ΔH positive.

Problem 3 Hand warmers and cold packs

1. Hand warmer: exothermic; cold pack: endothermic. 2. Energy is transferred from the chemicals to the surroundings, including your hands. 3. Warmer: cold winter morning at a cricket match; cold pack: treating a bruise after football. Warmer exothermic, cold pack endothermic.

Problem 4 Labelling a profile

1. A is the energy of the reactants and D the energy of the products. 2. B is the activation energy. 3. C is the energy change ΔH, and it is negative because the products are lower. 4. Exothermic. Exothermic; ΔH negative.

Problem 5 Breaking and making

1. Endothermic. 2. Exothermic. 3. Endothermic. 4. Exothermic. Breaking absorbs energy; making releases energy.

Problem 6 Signs of ΔH

1. Rises; negative. 2. More. 3. Activation. Rises and negative; more; activation.

Problem 7 Heat from magnesium

1. q = 50 × 4.2 × 11.0 = 2310 J. 2. 2310 ÷ 1000 = 2.31 kJ. 3. Exothermic. 2310 J = 2.31 kJ.

Problem 8 Hydrogen and bromine

1. 436 + 193 = 629 kJ. 2. 2 × 366 = 732 kJ. 3. ΔH = 629 − 732 = −103 kJ/mol, exothermic. ΔH = −103 kJ/mol.

Problem 9 Making ammonia

1. 945 + 3 × 436 = 945 + 1308 = 2253 kJ. 2. 6 × 391 = 2346 kJ. 3. ΔH = 2253 − 2346 = −93 kJ/mol. ΔH = −93 kJ/mol.

Problem 10 Sketch an endothermic profile

1. Draw the reactants at a low energy level and the products at a higher level. 2. The activation energy is an upward arrow from the reactants to the top of the hump; the energy change is an arrow from the reactants level up to the products level. 3. ΔH is positive. Products higher than reactants; ΔH positive.

Problem 11 Reading numbers from a profile

1. 120 kJ/mol. 2. −85 kJ/mol. 3. The reverse reaction starts 85 kJ lower and climbs to the same peak: 120 + 85 = 205 kJ/mol. Ea(reverse) = 205 kJ/mol.

Problem 12 Why light a match?

1. The gas and oxygen particles collide with less than the activation energy, so bonds do not break. The spark gives enough energy. 2. The energy released by the first reactions supplies the activation energy for the following particles. The spark supplies the activation energy.

Problem 13 Reading a temperature graph

1. 38 °C. 2. 38 − 20 = 18 °C. 3. After the reaction finishes, the warm solution loses energy to the cooler surroundings. 4. Exothermic. Maximum 38 °C; rise 18 °C.

Problem 14 Energy per gram of fuel

1. Hydrogen: 286 ÷ 2 = 143 kJ/g; methane: 890 ÷ 16 = 55.6 kJ/g; ethanol: 1367 ÷ 46 = 29.7 kJ/g. 2. Hydrogen, methane, ethanol. 3. Hydrogen is a gas that is hard to store and needs high pressure or very low temperature. 143, 55.6 and 29.7 kJ/g.

Problem 15 Per mole from joules

1. 4500 ÷ 1000 = 4.5 kJ. 2. 4.5 ÷ 0.030 = 150 kJ/mol. 3. The reaction is exothermic, so ΔH = −150 kJ/mol. ΔH = −150 kJ/mol.

Problem 16 Burning ethanol under a can

1. q = 100 × 4.2 × 30 = 12 600 J = 12.6 kJ. 2. n = 0.50 ÷ 46 = 0.0109 mol. 3. 12.6 ÷ 0.01087 = 1159 kJ/mol, about 1.2 × 10³ kJ/mol, which is about 15% lower than 1367 kJ/mol. 4. Energy is lost to the air and the can, and combustion may be incomplete. ΔH ≈ −1.2 × 10³ kJ/mol.

Problem 17 Burning methane

1. 4 × 413 + 2 × 498 = 1652 + 996 = 2648 kJ. 2. 2 × 805 + 4 × 464 = 1610 + 1856 = 3466 kJ. 3. ΔH = 2648 − 3466 = −818 kJ/mol. 4. n = 8.0 ÷ 16 = 0.50 mol, so 0.50 × 818 = 409 kJ. ΔH = −818 kJ/mol; 409 kJ released.

Problem 18 Hydrogen iodide

1. 2 × 299 = 598 kJ. 2. 436 + 151 = 587 kJ. 3. ΔH = 598 − 587 = +11 kJ, endothermic. 4. The products are 11 kJ higher than the reactants, below the top of the activation hump. ΔH = +11 kJ/mol; endothermic.

Problem 19 Evaluate a claim

1. ΔH only tells the energy difference. The particles still need 150 kJ/mol to start, so many reactions are slow at room temperature. 2. The reverse reaction starts 200 kJ/mol lower, so its activation energy is 150 + 200 = 350 kJ/mol. 3. Supply energy by heating or a spark, or use particles that collide more energetically. Ea reverse = 350 kJ/mol.

Problem 20 Spot the errors

1. Bond making is exothermic. 2. The products have less energy than the reactants. 3. ΔH = bonds broken − bonds formed. 4. The temperature rises in an exothermic reaction. Four corrections.

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