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OCR A Level Chemistry

Topic questions Energy changes Enthalpy changes

Enthalpy changes

8 questions · 74 marks · about 1.5 hours

Enthalpy changes questions

OCR H432 Enthalpy changes 8 questions

Periodic table, elements and physical chemistry June 2025, Question 17 16 marks

This question is about carbon monoxide and carbon as reducing agents in industry.

(a)

Iron(III) oxide, Fe2O3, can be reduced by carbon monoxide to form iron.

Question 17, shown as an image

Some standard enthalpy change of formation, maths symbolfHmaths symbol, and standard entropy, Smaths symbol, values are shown in the table.

SubstanceFe2O3(s)CO(g)Fe(s)CO2(g)
Ξ”f H / kJ mol–1–824–1110–394
S / J K–1 mol–187198To calculate in part (a)(iv)214
(a)(i)

Explain why the standard enthalpy change of formation of Fe has a value of 0 kJ mol–1.

[1]
Reveal mark scheme Mark scheme for question 17(a)(i)
(a)(ii)

Calculate the enthalpy change, maths symbolrH, for the forward reaction in Equilibrium 17.1.

maths symbolH =........ kJ mol–1

[2]
Reveal mark scheme Mark scheme for question 17(a)(ii)
(a)(iii)

Explain why the standard entropy of CO2 is greater than the standard entropy of Fe2O3.

[2]
Reveal mark scheme Mark scheme for question 17(a)(iii)
(a)(iv)

The standard entropy change, maths symbolSmaths symbol, for the forward reaction in Equilibrium 17.1 has a value of +15 J K–1 mol–1.

Use data from the table to calculate the standard entropy, Smaths symbol, of Fe.

Smaths symbol =........ J K–1 mol–1

[2]
Reveal mark scheme Mark scheme for question 17(a)(iv)
(b)

Carbon can also reduce Fe2O3(s) to iron as shown in Equilibrium 17.2.

Question 17, shown as an image

The forward reaction is only feasible at high temperatures.

β€’ Show that the forward reaction is not feasible at 100 Β°C.

β€’ Calculate the minimum temperature, in K, for the forward reaction to be feasible.

Give the minimum temperature to the nearest whole number.

[3]
Reveal mark scheme Mark scheme for question 17(b)
(c)

Steam can be reduced by carbon monoxide in a reversible reaction.

Question 17, shown as an image

At 700 K the value of Kp is 8.13.

(c)(i)

Explain why Kp for Equilibrium 17.3 has no units.

[1]
Reveal mark scheme Mark scheme for question 17(c)(i)
(c)(ii)

An equal number of moles of steam and carbon monoxide are left at 700 K in a sealed container until equilibrium is reached.

At equilibrium, both H2 and CO2 have a partial pressure of 211 kPa.

Calculate the partial pressures of H2O and CO at equilibrium.

[3]
Reveal mark scheme Mark scheme for question 17(c)(ii)
(c)(iii)

At a second temperature the value of Kp for Equilibrium 17.3 is 129.

Explain whether the second temperature is higher or lower than 700 K.

[2]
Reveal mark scheme Mark scheme for question 17(c)(iii)

Periodic table, elements and physical chemistry June 2024, Question 7 1 mark

The equation for the reaction of sulfuric acid with potassium hydroxide is shown below.

Question 7, shown as an image

25 cm3 of 1.00 mol dm–3 H2SO4 is reacted with excess KOH. The energy given out is 2.8 kJ.

What is the enthalpy change of neutralisation, in kJ mol–1?

A –56

B –70

C –112

D –224

Your answer

[1]
Reveal mark scheme Mark scheme for question 7

Unified chemistry June 2024, Question 2 14 marks

This question is about different areas of chemistry.

(a)
Figure for question 2
(a)(i)

Why do all average bond enthalpies have a positive value?

[1]
Reveal mark scheme Mark scheme for question 2(a)(i)
(a)(ii)

Determine the C–H bond enthalpy, in kJ mol–1, using the information above.

C–H bond enthalpy =........ kJ mol–1

[3]
Reveal mark scheme Mark scheme for question 2(a)(ii)
(a)(iii)

Hydrogen gas is being considered as a household fuel to replace methane.

The enthalpy change of formation, maths symbolfH, for H2O(l) is –285.8 kJ mol–1.

Determine the energy released when 60.0 m3 of hydrogen is used as a household fuel at RTP.

Give your answer to 3 significant figures and in standard form.

[2]
Reveal mark scheme Mark scheme for question 2(a)(iii)
(b)

Compound A is a chloride of a Period 3 element.

A student carries out the 2 steps below to find the formula of compound A.

Step 1 The student adds 5.00 maths symbol 10–4 mol of compound A to water. A colourless solution is formed.

Step 2 The colourless solution reacts with exactly 60.0 cm3 of 2.50 maths symbol 10–2 mol dm–3 AgNO3(aq) to form a white precipitate.

(b)(i)

Write an ionic equation, with state symbols, for the reaction in Step 2.

[1]
Reveal mark scheme Mark scheme for question 2(b)(i)
(b)(ii)

Determine the formula of compound A.

[3]
Reveal mark scheme Mark scheme for question 2(b)(ii)
(c)

Compound B, shown below, is an antiviral medicine.

Figure for question 2
(c)(i)

What is the molecular formula of compound B ?

[1]
Reveal mark scheme Mark scheme for question 2(c)(i)
(c)(ii)

How many chiral carbon atoms are there in one molecule of compound B ?

[1]
Reveal mark scheme Mark scheme for question 2(c)(ii)
(c)(iii)
Figure for question 2
[2]
Reveal mark scheme Mark scheme for question 2(c)(iii)

Periodic table, elements and physical chemistry June 2022, Question 17 14 marks

This question is about energy changes.

(a)

Magnesium reacts with aqueous silver nitrate, AgNO3(aq) as shown below.

Question 17, shown as an image

A student adds an excess of magnesium to 100.0 cm3 of 0.400 mol dm–3 AgNO3(aq). The initial temperature is 20.0 Β°C.

(a)(i)

Determine the maximum temperature reached in this reaction. Give your answer to 3 significant figures.

Assume that the specific heat capacity and density of the solution are the same as for water, and that there are no heat losses.

[4]
Reveal mark scheme Mark scheme for question 17(a)(i)
(a)(ii)

The student wants to repeat the experiment, but there is not enough AgNO3(aq) left to use another 100.0 cm3 portion.

The student decides to modify the method by adding an excess of magnesium to 50.0 cm3 of 0.400 mol dm–3 AgNO3(aq).

Predict, with reasons, how this modification would affect the maximum temperature reached. Assume that there are no heat losses.

[1]
Reveal mark scheme Mark scheme for question 17(a)(ii)
(b)

Nitric acid is manufactured from ammonia in a multi-stage process. The equation for the first stage in this process is shown in Reaction 17.1.

Question 17, shown as an image

Some standard enthalpy changes of formation are shown in the table.

Compoundf H / kJ mol–1
NH3(g)–46
H2O(l)–286
(b)(i)

Explain the term enthalpy change of formation.

[1]
Reveal mark scheme Mark scheme for question 17(b)(i)
(b)(ii)

Calculate the standard enthalpy change of formation, maths symbolfHmaths symbol, of NO(g).

maths symbolfHmaths symbol of NO(g) =........ kJ mol–1

[2]
Reveal mark scheme Mark scheme for question 17(b)(ii)
(c)

Carbon disulfide, CS2, reacts with dinitrogen oxide, N2O, as shown in Reaction 17.2.

Question 17, shown as an image

Standard entropies, Smaths symbol, are shown in the table.

SubstanceCS2(l)N2O(g)S8(s)CO2(g)N2(g)
S/ J K–1 mol–1151220256214192
(c)(i)

Explain the term entropy.

[1]
Reveal mark scheme Mark scheme for question 17(c)(i)
(c)(ii)

The free energy change, maths symbolG, of Reaction 17.2 is –2672 kJ mol–1 at 25 Β°C.

Calculate the enthalpy change, maths symbolH, of Reaction 17.2, in kJ mol–1.

maths symbolH =........ kJ mol–1

[3]
Reveal mark scheme Mark scheme for question 17(c)(ii)
(c)(iii)

A student concludes that Reaction 17.2 is feasible at all temperatures.

Explain whether the student is correct or not.

[2]
Reveal mark scheme Mark scheme for question 17(c)(iii)

Periodic table, elements and physical chemistry November 2021, Question 22 16 marks

This question is about acids, bases and buffers.

(a)

Sodium hydroxide, NaOH, is a strong base.

Calculate the pH of 0.140 mol dm–3 NaOH(aq) at 298 K.

Give your answer to 2 decimal places.

[2]
Reveal mark scheme Mark scheme for question 22(a)
(b)

Sulfuric acid reacts with sodium hydroxide as shown in Equation 22.1. Equation 22.1 H2SO4(aq) + 2NaOH(aq) β†’ Na2SO4(aq) + 2H2O(l) maths symbolH1 This is a neutralisation reaction.

A student carries out an experiment to determine the enthalpy change maths symbolH1 and uses this value to deduce the enthalpy change of neutralisation, maths symbolneutH.

The student measures out two solutions:

β€’ 25.0 cm3 of 1.60 mol dm–3 H2SO4(aq)

β€’ 55.0 cm3 of 1.50 mol dm–3 NaOH(aq) (an excess).

The temperature of each solution is the same.

The student mixes the two solutions. The temperature increases by 13.0 Β°C.

(b)(i)

Show that NaOH is in excess.

[2]
Reveal mark scheme Mark scheme for question 22(b)(i)
(b)(ii)

Calculate the enthalpy change, maths symbolH1, for Equation 22.1, and deduce the value for the enthalpy change of neutralisation, βˆ†neutH, in kJ mol–1.

Assume that the densities of all solutions and the specific heat capacity, c, of the reaction mixture are the same as for water.

enthalpy change, maths symbolH1 =........ kJ mol–1

enthalpy change of neutralisation, maths symbolneutH =........ kJ mol–1

[4]
Reveal mark scheme Mark scheme for question 22(b)(ii)
(b)(iii)

The student repeats the experiment using 50.0 cm3 of 1.60 mol dm–3 H2SO4 and 110.0 cm3 of 1.50 mol dm–3 NaOH.

Predict the increase in temperature.

Explain your reasoning.

[2]
Reveal mark scheme Mark scheme for question 22(b)(iii)
(c)

* Nitrous acid, HNO2, is a weak BrΓΈnsted–Lowry acid with a pKa value of 3.34 at room temperature.

HNO2 can be prepared by reacting N2O3 with water. HNO2 is the only product.

A chemist makes up a buffer solution by the following method.

Step 1 The chemist weighs a sample of N2O3. Water is then added to form 100 cm3 of 0.500 mol dm–3 HNO2(aq).

Step 2 The chemist adds 100 cm3 of 0.150 mol dm–3 NaOH(aq) to the 100 cm3 solution of 0.500 mol dm–3 solution of HNO2(aq). The resulting solution is made up to 1.00 dm3.

Explain why a buffer solution forms in Step 2. Determine the pH of this buffer solution and the mass of N2O3 that was used in Step 1.

[6]
Reveal mark scheme Mark scheme for question 22(c)

Periodic table, elements and physical chemistry June 2019, Question 10 1 mark

Radical reactions are responsible for the catalysed breakdown of the ozone layer.

The overall equation is shown below.

Question 10, shown as an image

The molar gas volume in the ozone layer is approximately 2.5 m3 molβˆ’1.

What is the energy released, in kJ, during the breakdown of 1.0 m3 of ozone in the ozone layer?

A 56.8

B 113.6

C 355

D 710

Your answer

[1]
Reveal mark scheme Mark scheme for question 10

Periodic table, elements and physical chemistry June 2018, Question 13 1 mark

Which enthalpy change(s) is/are endothermic?

1 The bond enthalpy of the C–H bond

2 The second electron affinity of oxygen

3 The standard enthalpy change of formation of magnesium

A 1, 2 and 3

B Only 1 and 2

C Only 2 and 3

D Only 1

Your answer

[1]
Reveal mark scheme Mark scheme for question 13

Unified chemistry June 2018, Question 5 11 marks

A student carries out two experiments in the laboratory based on succinic acid (butanedioic acid), (CH2COOH)2.

(a)
Figure for question 5 Figure for question 5
[4]
Reveal mark scheme Mark scheme for question 5(a)
(b)

Succinic acid is esterified by ethanol, C2H5OH, in the presence of an acid catalyst to form an equilibrium mixture.

The equilibrium constant, Kc, for this equilibrium can be calculated using the amounts, in moles, of the components in the equilibrium mixture, using expression 5.1.

Question 5, shown as an image

A student carries out an experiment to determine the value of Kc for this equilibrium.

β€’ The student mixes together 0.0500 mol of succinic acid and 0.150 mol of ethanol, with a small amount of an acid catalyst.

β€’ The mixture is allowed to reach equilibrium.

β€’ The student determines that 0.0200 mol of succinic acid are present in the equilibrium mixture.

(b)(i)

Which technique could be used to determine the equilibrium amount of succinic acid?

[1]
Reveal mark scheme Mark scheme for question 5(b)(i)
(b)(ii)

Write the equation for the equilibrium reaction that takes place.

[1]
Reveal mark scheme Mark scheme for question 5(b)(ii)
(b)(iii)

Draw the skeletal formula of the ester present in the equilibrium mixture.

[1]
Reveal mark scheme Mark scheme for question 5(b)(iii)
(b)(iv)

Kc is the equilibrium constant in terms of equilibrium concentrations.

Why can expression 5.1 be used to calculate Kc for this equilibrium?

[1]
Reveal mark scheme Mark scheme for question 5(b)(iv)
(b)(v)

Calculate the value of Kc for this reaction.

Show your working.

[3]
Reveal mark scheme Mark scheme for question 5(b)(v)

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