Energetics & Thermodynamics

Which statement correctly defines the standard enthalpy of formation, ΔHf°, of a compound?

• AThe enthalpy change when one mole of the compound is formed from its elements in their standard states under standard conditions✓
• BThe enthalpy change when one mole of the compound is burned completely in excess oxygen under standard conditions
• CThe enthalpy change when one mole of gaseous atoms is formed from the element under standard conditions
• DThe enthalpy change when one mole of water is formed in a neutralisation reaction under standard conditions

Answer: Standard enthalpy of formation is the enthalpy change when one mole of a compound is made from its constituent elements, each in their standard states, under standard conditions (100 kPa, stated temperature, usually 298 K). Burning one mole completely defines enthalpy of combustion; forming one mole of gaseous atoms is atomisation; forming one mole of water from an acid and an alkali is neutralisation.

The standard enthalpy of combustion of carbon is −394 kJ mol⁻¹, of hydrogen is −286 kJ mol⁻¹, and of methane is −890 kJ mol⁻¹. Using Hess's law, calculate the standard enthalpy of formation of methane, CH₄.

• A+76 kJ mol⁻¹
• B−210 kJ mol⁻¹
• C−76 kJ mol⁻¹✓
• D−1570 kJ mol⁻¹

Answer: ΔHf(CH₄) = [ΔHc(C) + 2ΔHc(H₂)] − ΔHc(CH₄) = [(−394) + 2(−286)] − (−890) = (−966) − (−890) = −76 kJ mol⁻¹. The combustion-data cycle subtracts the combustion of the product from the sum of the combustions of the elements.

Using the mean bond enthalpies H–H = 436 kJ mol⁻¹, Cl–Cl = 242 kJ mol⁻¹ and H–Cl = 431 kJ mol⁻¹, calculate the enthalpy change for the reaction H₂(g) + Cl₂(g) → 2HCl(g).

• A+184 kJ mol⁻¹
• B−184 kJ mol⁻¹✓
• C−431 kJ mol⁻¹
• D−94 kJ mol⁻¹

Answer: ΔH = Σ(bonds broken) − Σ(bonds formed) = (436 + 242) − (2 × 431) = 678 − 862 = −184 kJ mol⁻¹. The reaction is exothermic, so the sign is negative; the +184 distractor reverses the convention.

Enthalpy changes calculated from mean bond enthalpies often differ slightly from experimentally determined values. What is the main reason for this discrepancy?

• AMean bond enthalpies are quoted at 0 K, whereas the experiments themselves are run at room temperature (298 K)
• BExperimental values ignore the energy needed to break the bonds, and record only the energy that is released
• CMean bond enthalpies apply just to ionic compounds (such as NaCl), and not to simple covalent molecules
• DMean bond enthalpies are averaged over many different compounds, so they do not exactly match the bonds in a specific molecule✓

Answer: A mean (average) bond enthalpy is averaged across the same bond type in a range of different molecules and gaseous environments. The actual bond enthalpy in one specific molecule differs from this average, so bond-enthalpy calculations are approximate.

In a calorimetry experiment, burning a fuel raised the temperature of 200 g of water by 25.0 °C. Using the specific heat capacity of water c = 4.18 J g⁻¹ °C⁻¹, calculate the heat energy released to the water (q = mcΔT).

• A5.23 kJ
• B209 kJ
• C2.09 kJ
• D20.9 kJ✓

Answer: q = mcΔT = 200 × 4.18 × 25.0 = 20 900 J = 20.9 kJ. Care with unit conversion: 20 900 J ÷ 1000 = 20.9 kJ, not 2.09 kJ.

When 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid is mixed with 50.0 cm³ of 1.00 mol dm⁻³ sodium hydroxide, the temperature of the 100 g mixture rises by 6.80 °C. Taking c = 4.18 J g⁻¹ °C⁻¹, calculate the enthalpy of neutralisation per mole of water formed.

• A−2.84 kJ mol⁻¹
• B−56.8 kJ mol⁻¹✓
• C−28.4 kJ mol⁻¹
• D−113.7 kJ mol⁻¹

Answer: q = mcΔT = 100 × 4.18 × 6.80 = 2842 J = 2.842 kJ. Moles of water = 0.0500 dm³ × 1.00 = 0.0500 mol. ΔH = −2.842 / 0.0500 = −56.8 kJ mol⁻¹ (close to the textbook −57 kJ mol⁻¹ for strong acid + strong base).

For the thermal decomposition CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹. Above approximately what temperature does the reaction become feasible (ΔG ≤ 0)?

• A1106 K✓
• B28 658 K
• C0.904 K
• D1 106 000 K

Answer: At the feasibility limit ΔG = 0, so T = ΔH/ΔS. Converting ΔS to kJ: T = 178 / 0.161 = 1106 K. Above this temperature the −TΔS term outweighs the positive ΔH, making ΔG negative.

Which set of conditions guarantees that a reaction is feasible (spontaneous, ΔG < 0) at all temperatures?

• AΔH positive and ΔS positive
• BΔH negative and ΔS negative
• CΔH negative and ΔS positive✓
• DΔH positive and ΔS negative

Answer: ΔG = ΔH − TΔS. If ΔH is negative and ΔS is positive, then −TΔS is also negative at all positive temperatures, so ΔG is always negative. ΔH positive with ΔS negative is never feasible, and the two combinations where both have the same sign are temperature-dependent.

What is an exothermic reaction?

• AOne that releases energy, giving a negative enthalpy change✓
• BOne that releases energy, giving a positive enthalpy change
• COne that absorbs energy, giving a negative enthalpy change
• DOne that absorbs energy, giving a positive enthalpy change

Answer: Energy leaves the system and the surroundings warm up, so ΔH is negative by convention. 'One that absorbs energy, giving a positive enthalpy change' describes an endothermic reaction, which is the mirror image.

What does Hess's law state?

• AEnthalpy change depends entirely on the route taken
• BEnthalpy change is always negative for spontaneous change
• CEnthalpy change is independent of the route taken✓
• DEnthalpy change equals the activation energy of the reaction

Answer: Because enthalpy is a state function, the total change depends only on the initial and final states. That is what allows a cycle to be built and an unmeasurable enthalpy change to be found indirectly.

Why do bond enthalpy calculations give only approximate answers?

• AMean bond enthalpies are measured only in the solid state
• BBond breaking is exothermic and bond making endothermic
• CMean bond enthalpies are averaged across many compounds✓
• DBond enthalpies change with the temperature of the reaction

Answer: The same bond has slightly different strength in different molecular environments, so tabulated values are averages. Note also that breaking bonds absorbs energy and making them releases it, which is the reverse of the statement offered as a distractor.

Why is the standard enthalpy of formation of an element in its standard state zero?

• ANo chemical change occurs when an element forms from itself✓
• BElements contain no bonds that can be broken or formed at all
• CAll elements hold the same energy under standard conditions
• DThe value is defined as zero only for the gaseous elements

Answer: Formation means making one mole of a substance from its elements in their standard states. For an element already in its standard state there is no chemical change at all, so the enthalpy change is zero by definition — which is what makes Hess's-law cycles work.

In calorimetry, why is the measured enthalpy change usually smaller than the true value?

• AHeat is lost to the surroundings and the apparatus✓
• BHeat is gained from the surroundings and the apparatus
• CThe specific heat capacity of water is underestimated
• DThe reaction always fails to go to completion

Answer: Simple calorimeters lose heat through the walls and to the air, and some warms the apparatus itself, so less appears in the measured temperature rise. Insulation and extrapolating a cooling curve back to the mixing time both reduce the error.

What does the equation q = mcΔT calculate?

• AThe enthalpy change per mole of the limiting reactant
• BThe heat energy transferred to or from a substance✓
• CThe activation energy required to start the reaction
• DThe entropy change of the system during the reaction

Answer: It gives the energy in joules for a given mass, specific heat capacity and temperature change. Dividing by the moles reacted then converts it to an enthalpy change per mole.

Why is the enthalpy of combustion of an alcohol measured experimentally often too small?

• AComplete combustion and heat gain together reduce the value
• BThe alcohol evaporates first and so reacts far more completely
• CIncomplete combustion and heat loss both reduce the value✓
• DWater produced absorbs energy as it condenses to liquid

Answer: A spirit burner in air commonly burns incompletely, releasing less energy, and much of what is released escapes to the surroundings rather than heating the water. Both errors push the result the same way.

What is activation energy?

• AThe maximum energy that colliding particles can have
• BThe energy released when new bonds are formed
• CThe minimum energy needed for a collision to react✓
• DThe average kinetic energy of the reacting particles

Answer: Colliding particles must have at least this much energy, and be correctly oriented, to reach the transition state. A catalyst provides an alternative route with a lower activation energy, so more collisions succeed.

Why is the enthalpy of neutralisation similar for all strong acid and strong alkali pairs?

• AThe reaction occurring is always a metal displacing hydrogen
• BStrong acids and alkalis have identical concentrations
• CThe reaction occurring is always H⁺ plus OH⁻ forming water✓
• DThe salt formed is always fully soluble in water

Answer: Both are fully dissociated, so the spectator ions take no part and the only chemical change is hydrogen ions combining with hydroxide ions. Weak acids give lower values because energy is absorbed completing their dissociation.

Two experiments were done, each using 50.0 g of water (c = 4.18 J g⁻¹ K⁻¹): Experiment 1: 4.00 g of anhydrous CuSO₄ (Mr 159.6) dissolved; temperature rose by 7.8 K. Experiment 2: 6.25 g of CuSO₄·5H₂O (Mr 249.6) dissolved; temperature fell by 1.3 K.

Use a Hess cycle to calculate ΔH for CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s).

• A−75.9 kJ mol⁻¹✓
• B−54.2 kJ mol⁻¹
• C+75.9 kJ mol⁻¹
• D−65.0 kJ mol⁻¹

Answer: Expt 1: q = 50.0 × 4.18 × 7.8 = 1630 J for 0.02506 mol, so ΔH₁ = −65.0 kJ mol⁻¹. Expt 2: q = 50.0 × 4.18 × 1.3 = 272 J for 0.02504 mol, so ΔH₂ = +10.9 kJ mol⁻¹. Both routes end at the same solution, so ΔH = ΔH₁ − ΔH₂ = −65.0 − 10.9 = −75.9 kJ mol⁻¹. Adding the two values instead of subtracting gives −54.2 kJ mol⁻¹.

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g) ΔfH⦵ (kJ mol⁻¹): Fe₂O₃(s) −824; CO(g) −111; CO₂(g) −394

Calculate the standard enthalpy change for this reaction.

• A+25 kJ mol⁻¹
• B+541 kJ mol⁻¹
• C−25 kJ mol⁻¹✓
• D−2339 kJ mol⁻¹

Answer: ΔH = ΣΔfH(products) − ΣΔfH(reactants) = 3(−394) − [(−824) + 3(−111)] = −1182 − (−1157) = −25 kJ mol⁻¹. Iron is an element in its standard state, so its ΔfH is zero. Doing reactants minus products gives +25; ignoring the coefficient of 3 gives +541.

CH₃OH(g) + 1½O₂(g) → CO₂(g) + 2H₂O(g) Mean bond enthalpies (kJ mol⁻¹): C–H 413; C–O 358; O–H 464; O=O 498; C=O 805

Calculate the enthalpy change for this reaction.

• A+658 kJ mol⁻¹
• B−1122 kJ mol⁻¹
• C−658 kJ mol⁻¹✓
• D−907 kJ mol⁻¹

Answer: Bonds broken: 3(C–H) + C–O + O–H + 1½(O=O) = 1239 + 358 + 464 + 747 = 2808 kJ. Bonds made: 2(C=O) + 4(O–H) = 1610 + 1856 = 3466 kJ. ΔH = 2808 − 3466 = −658 kJ mol⁻¹. Leaving out the O–H bond in methanol gives −1122, and counting one O=O instead of 1½ gives −907.

Burning 0.512 g of propan-1-ol (Mr 60.0) in a spirit burner raised the temperature of 150 g of water by 21.5 K (c = 4.18 J g⁻¹ K⁻¹). The data-book value of ΔcH for propan-1-ol is −2021 kJ mol⁻¹.

What percentage of the energy released by burning the fuel was not transferred to the water?

• A78.2%
• B21.8%✓
• C27.9%
• D99.3%

Answer: q = 150 × 4.18 × 21.5 = 13 480 J. Moles burned = 0.512 ÷ 60.0 = 8.53 × 10⁻³ mol, so the experimental ΔcH = −13.48 ÷ 8.53 × 10⁻³ = −1580 kJ mol⁻¹. The fraction not transferred = (2021 − 1580) ÷ 2021 × 100 = 21.8%. The figure of 78.2% is the fraction that did reach the water, and 27.9% wrongly divides by the experimental value.

In a coffee-cup calorimetry practical, a student records the temperature every 30 s, adds the reactant at 3 minutes, and then plots temperature against time.

Why is the line after mixing extrapolated back to the 3-minute mark?

• ATo find the mean temperature of the mixture across the whole of the time period of the experiment
• BTo estimate the temperature change that would occur with no heat lost while the reaction proceeds✓
• CTo work out the specific heat capacity of the reaction mixture from the gradient of the cooling line
• DTo check that the reactants were mixed at room temperature before the timing of the readings began

Answer: Heat is lost to the surroundings while the mixture reaches its maximum temperature, so the highest reading underestimates the true temperature change. Extrapolating the cooling line back to the moment of mixing estimates the temperature that would have been reached instantly with no heat loss. The gradient shows the rate of cooling, not the specific heat capacity.