Questions on enthalpy change, standard conditions and the definitions

Question 1

Evaluate the claim that a reaction with a large negative ΔH must happen quickly when its reactants are mixed at room temperature. Refer in your answer to an enthalpy profile diagram and to one example.

Accepted Answer

Indicative content

ΔH is the gap between the reactant and product levels on the profile, so it measures how much heat is released, not how fast it is released
the rate depends on the activation energy, which is the height of the peak above the reactant level on the same diagram
if the activation energy is high, only a tiny fraction of colliding particles have enough energy to react at room temperature, so the reaction is slow
a large negative ΔH and a high activation energy can occur together, because the depth of the products and the height of the peak are independent features of the profile
methane and oxygen, with ΔcH° = −890 kJ mol⁻¹, do not react until a spark or flame supplies the activation energy; accept paper in air or any kinetically stable mixture
the claim is therefore wrong: such a mixture is energetically unstable but kinetically stable, and ΔH alone cannot predict the rate of a reaction

Levels

Level 3 (5–6 marks) ΔH and activation energy are both located on the profile and shown to be independent, the rate is linked to the activation energy through the fraction of particles able to react, a kinetically stable example is given, and the claim is judged wrong with the distinction between energetic and kinetic stability
Level 2 (3–4 marks) rate is tied to activation energy rather than ΔH and an example is given; the profile is not used to show the two are independent, or the verdict is not put in terms of kinetic stability
Level 1 (1–2 marks) the claim is accepted because a large negative ΔH means a lot of energy is released quickly, or activation energy is named with no profile and no example

Question 2

Give the definition of the standard enthalpy of formation of a compound, and write the equation that represents the standard enthalpy of formation of propan-1-ol, C₃H₇OH(l), with every state symbol.

Accepted Answer

B1 the enthalpy change when one mole of the compound is formed B1 from its constituent elements B1 with all reactants and products in their standard states under standard conditions B1 3C(s) + 4H₂(g) + ½O₂(g) → C₃H₇OH(l), with the coefficients 3, 4 and ½ and one mole of propan-1-ol as the only product B1 state symbols (s) for carbon, (g) for hydrogen and oxygen and (l) for propan-1-ol; reject O(g) for oxygen

Question 3

State the conditions of pressure and temperature to which a standard enthalpy change refers, and state the standard state of carbon, of bromine and of water.

Accepted Answer

B1 a pressure of 100 kPa; allow 101 kPa or one atmosphere B1 a stated temperature, which is usually 298 K; accept 25 °C B1 carbon as solid graphite, written C(s); reject diamond B1 bromine as a liquid, Br₂(l), and water as a liquid, H₂O(l); both are needed for the mark

Question 4

Describe the enthalpy profile diagram for an exothermic reaction, saying how the activation energy and ΔH are shown.

Accepted Answer

B1 axes labelled enthalpy and progress of reaction, with the reactant level higher than the product level B1 the line rises from the reactants to a peak before falling to the products B1 the activation energy is an arrow from the reactant level up to the peak; reject an arrow drawn from the axis B1 ΔH is an arrow from the reactant level down to the product level

Question 5

Calculate the heat released per gram when butane and ethanol burn completely, using the data in the table, and the mass of butane that must burn to release 1.00 × 10⁴ kJ.

Accepted Answer

M1 heat released per gram = ΔcH° ÷ Mr for each fuel; ignore the sign if the answer is stated as heat released A1 butane 2877 ÷ 58.0 = 49.6 kJ g⁻¹ and ethanol 1367 ÷ 46.0 = 29.7 kJ g⁻¹; both needed M1 amount of butane = 1.00 × 10⁴ ÷ 2877 = 3.48 mol A1 mass of butane = 3.48 × 58.0 = 202 g to three significant figures; allow 201 g, and error carried forward from the amount

Question 6

Suggest why the standard enthalpy of formation of glucose, C₆H₁₂O₆(s), cannot be measured directly, and outline how its value can be found from enthalpies of combustion.

Accepted Answer

B1 carbon, hydrogen and oxygen do not combine to form glucose under any conditions that can be measured in a calorimeter B1 any reaction between the elements gives other products, such as carbon dioxide and water, not glucose alone B1 the enthalpies of combustion of carbon, hydrogen and glucose can all be measured, because each burns completely in oxygen B1 by Hess's law, ΔfH°(glucose) = 6ΔcH°(C) + 6ΔcH°(H₂) − ΔcH°(glucose), since both routes end at 6CO₂(g) + 6H₂O(l)

Question 7

Identify which one of these equations represents a standard enthalpy of formation, and give the reason each of the other three does not. A: 2C(s) + 2H₂(g) → C₂H₄(g). B: C(s, diamond) + O₂(g) → CO₂(g). C: 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s). D: H₂(g) + Br₂(g) → 2HBr(g).

Accepted Answer

B1 A represents ΔfH° of ethene: one mole of the compound from its elements in their standard states B1 B starts from diamond, which is not the standard state of carbon; graphite is B1 C forms two moles of Fe₂O₃, so its ΔH is twice ΔfH°; accept 'not one mole of product' B1 D forms two moles of HBr and shows bromine as a gas, where its standard state is liquid; either reason earns the mark

Question 8

Give the definition of the standard enthalpy of combustion of a substance.

Accepted Answer

B1 the enthalpy change when one mole of the substance is burned B1 burned completely in oxygen; accept 'in an excess of oxygen' B1 with all reactants and products in their standard states under standard conditions

Question 9

Explain, in terms of the enthalpy of the reactants and the products, why the temperature of the surroundings rises during an exothermic reaction and why ΔH is negative.

Accepted Answer

B1 the products have less enthalpy than the reactants B1 the difference is released as heat energy to the surroundings, so the surroundings warm up B1 ΔH describes the change in the enthalpy of the chemicals, which has decreased, so ΔH is negative; ignore references to the beaker gaining energy

Question 10

Explain why the equation 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = −572 kJ mol⁻¹, does not show the standard enthalpy of formation of water, and give the value that it does have.

Accepted Answer

B1 ΔfH° refers to the formation of one mole of the compound, and this equation forms two moles B1 ΔfH° of H₂O(l) is therefore half of the value, −286 kJ mol⁻¹ B1 the equation as written represents ΔrH° for the reaction, which is also twice the standard enthalpy of combustion of hydrogen

Question 11

Suggest the sign of ΔH when ammonium nitrate dissolves in the water of a sports cold pack, and explain why the pack feels cold.

Accepted Answer

B1 ΔH is positive; dissolving ammonium nitrate is endothermic B1 the dissolving takes in heat energy from its surroundings, the water in the pack B1 so the temperature of the pack falls, and heat then flows from the skin into the pack

Question 12

Compare the enthalpy profile diagrams of an exothermic reaction and an endothermic reaction.

Accepted Answer

B1 in an exothermic profile the products are below the reactants, whereas in an endothermic profile they are above B1 so the ΔH arrow points down for exothermic and up for endothermic, starting from the reactant level in both B1 both show a peak above the reactants for the activation energy; in the endothermic profile the activation energy is at least as large as ΔH

Question 13

Calculate the enthalpy change when 85.0 g of ammonia is formed from its elements. N₂(g) + 3H₂(g) → 2NH₃(g), ΔrH° = −92 kJ mol⁻¹; Mr of NH₃ = 17.0.

Accepted Answer

M1 the equation forms two moles of NH₃, so the enthalpy change per mole of ammonia is −92 ÷ 2 = −46 kJ mol⁻¹ M1 amount of ammonia = 85.0 ÷ 17.0 = 5.00 mol A1 ΔH = 5.00 × (−46) = −230 kJ, that is −2.3 × 10² kJ to two significant figures; −460 kJ, from using −92 per mole, scores the second mark only

Question 14

Explain why the standard enthalpy of combustion of methanol, −726 kJ mol⁻¹, is less exothermic than that of methane, −890 kJ mol⁻¹, although each molecule contains one carbon atom.

Accepted Answer

B1 both burn completely to the same products, one mole of CO₂ and two moles of H₂O(l) B1 methanol already contains an oxygen atom bonded to its carbon, so it is partly oxidised before it burns B1 methanol therefore starts with less enthalpy than methane relative to those products, so less heat is released when it burns

Question 15

Give the definition of the standard enthalpy of neutralisation, and explain why its value for sulfuric acid with sodium hydroxide is about the same as for hydrochloric acid with sodium hydroxide.

Accepted Answer

B1 the enthalpy change when an acid and an alkali react under standard conditions to form one mole of water; reject 'one mole of acid' B1 sulfuric acid is diprotic, so one mole of it forms two moles of water and releases about twice the heat B1 per mole of water the reaction in both is H⁺(aq) + OH⁻(aq) → H₂O(l), so the value is about −57 kJ mol⁻¹ for both

Question 16

Explain why the standard enthalpy of formation of oxygen, O₂(g), is zero.

Accepted Answer

B1 oxygen is an element and O₂(g) is its standard state B1 forming O₂(g) from O₂(g) involves no change, so there is no enthalpy change

Question 17

Explain why the enthalpy change for burning a mole of methane would be less exothermic than ΔcH°(CH₄) if the water formed remained as steam.

Accepted Answer

B1 the standard enthalpy of combustion forms water in its standard state, liquid B1 condensing steam to liquid water releases energy, about 44 kJ per mole, so this is not released if the water stays as steam

Question 18

Suggest why the equation for the standard enthalpy of combustion of butane is written with 6½O₂(g) rather than doubled to 13O₂(g).

Accepted Answer

B1 the standard enthalpy of combustion is defined for one mole of butane, which needs 6½ moles of O₂ to form 4CO₂ and 5H₂O B1 the doubled equation burns two moles of butane, so its enthalpy change would be twice ΔcH°

Fuel	ΔcH° / kJ mol⁻¹	Mr
Butane, C₄H₁₀(g)	−2877	58.0
Ethanol, C₂H₅OH(l)	−1367	46.0