Energetics I (Edexcel AS Chemistry)

Exam Questions

1 hour9 questions
1a
Sme Calculator
5 marks

Phosphorus(V) chloride, PCl5 , can be thermally decomposed to phosphorus(III) chloride, PCl3 , and chlorine, Cl2 . The equation for this reaction is

PCl5 (g)  →  PCl3 (g) + Cl2 (g)

The enthalpy change for this reaction cannot be measured directly.

i)
Complete the Hess’s Law cycle to include the enthalpy change of formation of both phosphorus chlorides.

Include the labels of the missing enthalpy changes.

ΔH is the enthalpy change for the vaporisation of the substance from the state shown to the gaseous state.
(3)
q2ai-8cho-as-2-oct-2021-edexcel-a-level-chem

ii)
Calculate the enthalpy change for the thermal decomposition of PCl5 (g) to PCl3 (g) and Cl2 (g), using the data given in the table.
Include a sign and units in your answer.
(2)

  Enthalpy change / kJ mol–1
ΔfH [PCl5 (s)] –443.5
ΔfH [PCl3 (l)] –319.7
ΔvH [PCl5 (s)] +64.9
ΔvH [PCl3 (l)] +30.5
1b2 marks

Another source gave a different value for the enthalpy change of this reaction.

PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)                 ΔrH = +87.9 kJ mol–1


Explain the effect, if any, of increasing the temperature on the position of the equilibrium at constant volume.

Did this page help you?

2a6 marks

This question is about enthalpy changes.

i)
State what is meant by the term ‘standard enthalpy change of combustion’.

(2)

ii)
Write the equation, including state symbols, for the reaction that occurs when the standard enthalpy change of combustion of octane, C8H18 (l), is measured.

(2)

iii)

The standard enthalpy change of combustion of octane is –5 470 kJ mol–1.

Complete the reaction profile diagram for the combustion of octane.
Include labels showing the standard enthalpy change of combustion, ΔcH to the power of ⦵ and the activation energy, Ea.

(2)

q3aiii-8cho-as-2-june-2019-qp-edexcel-a-level-chem

2b
Sme Calculator
4 marks

Enthalpy changes of reactions which cannot be measured directly can be calculated using standard enthalpy changes of combustion.

The table shows some of these values.

Substance ΔcH to the power of ⦵/ kJ mol−1
C(s) −394
H2(g) −286
CH4(g) −890

Complete the Hess cycle and use it to calculate the standard enthalpy change for the following reaction.

q3b-8cho-as-2-june-2019-qp-edexcel-a-level-chem

2c
Sme Calculator
3 marks

The equations for the combination of gaseous atoms of carbon and hydrogen to form methane, CH4, and propane, C3H8, are

C(g) + 4H(g) → CH4(g) ΔH = −1652 kJ mol−1
3C(g) + 8H(g) → C3H8(g) ΔH = −3998 kJ mol−1

Calculate:

i)
the mean bond enthalpy of a C—H bond.

(1)

ii)
the mean bond enthalpy of a C—C bond.

(2)

Did this page help you?

3a
Sme Calculator
4 marks

The standard molar enthalpy change of neutralisation is the enthalpy change when an acid and an alkali react under standard conditions to form one mole of water.

An experiment was carried out to determine the enthalpy change of neutralisation for the reaction between propanoic acid and sodium hydroxide.

The equation for this reaction is

CH3CH2COOH (aq) + NaOH (aq) → CH3CH2COO-Na+ (aq) + H2O (l)


50.0 cm3 of sodium hydroxide solution, of concentration 1.00 mol dm-3, was placed in a polystyrene cup. The initial temperature was measured.

i)
Which piece of equipment has the smallest measurement uncertainty for the measurement of 50.0 cm3 of sodium hydroxide solution?
(1)
      Equipment Measurement uncertainty for each reading
  A burette ±0.05 cm3
  B

50 cm3  measuring cylinder

±1 cm3
  C

25 cm3 pipette

±0.06 cm3
  D

50 cm3 pipette

±0.08 cm3


ii)
50.0 cm3 of propanoic acid solution, of concentration 1.00 mol dm-3, was added and thoroughly mixed with the sodium hydroxide solution in the polystyrene cup.

The maximum temperature rise was 6.5 °C.

Calculate the enthalpy change of neutralisation for propanoic acid, in kJ mol -1, giving your answer to the nearest whole number. 

[Assume density of the mixture = 1.00 g cm-3, specific heat capacity of the mixture = 4.18 J g –1 °C –1]
(3)
3b3 marks

Another experiment was carried out with a solution of ethanoic acid and sodium hydroxide solution of the same concentration.

i)
Which graph shows the correct way that the maximum temperature rise should be determined?

(1)

q3b-9cho-al-1-june-2019-qp-edexcel-a-level-chem

ii)
Explain why the data book value for the standard enthalpy change of neutralisation of ethanoic acid with sodium hydroxide is –55.2 kJ mol –1 but the value for hydrochloric acid is –57.1 kJ mol –1.

(2)

Did this page help you?

4a1 mark

Nitric acid reacts with sodium hydroxide solution in a neutralisation reaction.

HNO3 (aq) + NaOH (aq) → NaNO3 (aq) + H2O (l)

In an experiment to determine the enthalpy change of neutralisation, the following results were obtained.

Volume of 1.00 mol dm−3 HNO3 = 25.0 cm3

Volume of 1.05 mol dm−3 NaOH = 25.0 cm3

Temperature rise = 6.8°C

Give a reason why excess sodium hydroxide was used.

4b
Sme Calculator
4 marks

Calculate the enthalpy change of neutralisation for the reaction between nitric acid and sodium hydroxide solution, using the results of the experiment.

Give your answer to an appropriate number of significant figures.

open square brackets Assume colon space density space of space the space reaction space mixture space space space space space space space space space space equals space 1.0 space straight g space cm to the power of negative 3 end exponent
space specific space heat space capacity space of space the space reaction space mixture space equals space 4.18 space straight J space straight g to the power of negative 1 end exponent degree straight C to the power of negative 1 end exponent close square brackets

Did this page help you?

5a1 mark

When solid calcium sulfate dihydrate, CaSO4·2H2O, is heated in a crucible, it forms solid calcium sulfate hemihydrate, CaSO4·1⁄2H2O.

Write an equation, including state symbols, for this reaction.

5b1 mark

Which two terms could be used to describe this reaction?

      Enthalpy change Type of process
  A endothermic hydration
  B exothermic hydration
  C exothermic dehydration
  D endothermic dehydration
5c
Sme Calculator
8 marks

When water is added to calcium sulfate hemihydrate, there is a rise in temperature.

A student decided to investigate this reaction using the following procedure:

Step 1 10 cm3 of distilled water is measured using a measuring cylinder having an uncertainty of ±0.5 cm3 , and is placed in an insulated cup with a lid.
Step 2 A thermometer with an uncertainty of ±0.5 °C is placed in the water.
Step

Exactly 10.00 g of calcium sulfate hemihydrate is weighed out using a balance with an uncertainty of ±0.005 g.

Step 4  The weighed quantity of calcium sulfate hemihydrate is added to the

water in the insulated cup.

Step 5 

The mixture in the insulated cup is stirred until no further temperature change is observed.

Results

Temperature of the water before adding the solid = 23.5 °C
Maximum temperature of the mixture after adding the solid = 26.3 °C


Other data

Molar mass of calcium sulfate hemihydrate, CaSO4·1⁄2H2O = 145.2 g mol–1
Density of water = 1.00 g cm-3

i)
Calculate the minimum volume of water needed to convert 10.00g of CaSO4·1⁄2H2O into CaSO4·2H2O.

(2)

ii)
Calculate the enthalpy change, in kJmol–1, for this reaction.
Include a sign in your answer and give your answer to an appropriate number of significant figures.


Assume that the liquid has a mass of 10.00 g and a specific heat capacity of 4.18 J g–1 °C–1.

(4)

iii)
Deduce which measurement has the greatest uncertainty in this experiment.
Justify your answer by calculating the percentage uncertainty of this piece of apparatus.

(2)

Did this page help you?

6a
Sme Calculator
4 marks

Methanol, CH3OH, is a liquid fuel.
An experiment was carried out to determine the enthalpy change of combustion of liquid methanol.

q4-8cho-as-2-nov-2020-qp-edexcel-a-level-chem

The energy obtained from burning 2.08 g of methanol was used to heat 75.0 g of water.

The temperature of the water rose from 25.0 °C to 91.0 °C.

[Specific heat capacity of water = 4.18 J g−1 °C−1]

Use the data to calculate a value for the enthalpy change of combustion of one mole of methanol.

Give your answer to an appropriate number of significant figures and include a sign and units.

6b7 marks

Methanol can be synthesised from methane and steam by a process that occurs in two steps.

Step 1          CH4 (g) + H2O (g)  3H2 (g) + CO (g)             ΔH = +206 kJ mol−1

Step 2           CO (g) + 2H2 (g)  CH3OH (g)                        ΔH = −91 kJ mol−1

i)
Explain the effects of increasing the pressure on the yield of the products and on the rate of the reaction in Step 1.

(4)

ii)
Step 2 is carried out at a compromise temperature of 500 K.

Explain why 500 K is considered to be a compromise for Step 2 by considering what would happen at higher and lower temperatures.

(3)

6c
Sme Calculator
3 marks

Calculate a value for the standard enthalpy change of combustion of gaseous methanol using the enthalpy change for Step 2 and the standard enthalpy change of combustion of gaseous carbon monoxide and of hydrogen.

Substance Standard enthalpy change
of combustion/ kJ mol−1
CO -283
H2 -286

Did this page help you?

7a1 mark

This question is about 2-methylpropan-2-ol.

Draw the fully displayed formula of 2-methylpropan-2-ol.

7b2 marks

The mass spectrum of 2-methylpropan-2-ol is shown.

 
screenshot-2022-11-22-233750
 
i)
The relative molecular mass of 2-methylpropan-2-ol is 74. Give a possible reason why there is no molecular ion peak in the mass spectrum of 2-methylpropan-2-ol.
(1)
 
ii)
Write the formula for a species that could be responsible for the peak at m / z = 59.
(1)
7c
Sme Calculator
7 marks

The equation for the complete combustion of 2-methylpropan-2-ol is

 C4H10O (l) + 6O2 (g) → 4CO2 (g) + 5H2O (l)
 
i)
Using the bond enthalpies shown in the table, calculate a value for the enthalpy change, in kJ mol−1, for the complete combustion of 2-methylpropan-2-ol.
(4)
 
Bond Mean bond enthalpy / kJ mol−1
C–C 347
C–H 413
C–O 358
O–H 464
O=O 498
C=O 805
 
ii)
2-methylpropan-2-ol burns in air with a smoky flame. Explain how burning with a smoky flame affects the value of the experimentally determined enthalpy change of combustion.
(2)
 
iii)
A Data Book value for the enthalpy change of combustion of 2-methylpropan-2-ol is −2643.8 kJ mol−1. Give the main reason for the difference between this value and your answer to part 8(c)(i).
(1)
7d1 mark

Which observation would be expected when 2-methylpropan-2-ol is heated with potassium dichromate(VI) and dilute sulfuric acid?

  A orange to green
  B green to orange
  C purple to colourless
  D no change

Did this page help you?

8a
Sme Calculator
7 marks

An equation for the formation of ammonia using the Haber process is shown.

N2 (g) + 3H2 (g)  ⇌  2NH3 (g)

i)
Calculate the enthalpy change for the forward reaction shown in the equation, selecting from the bond enthalpies in the table.
Include a sign in your answer.

(3)

Bond Mean bond enthalpy / kJ mol–1
begin mathsize 14px style N bond N end style 158
begin mathsize 14px style N double bond N end style 410
straight N identical to straight N 945
begin mathsize 14px style N bond H end style 391
begin mathsize 14px style H bond H end style 436

ii)
A data book gives the standard enthalpy change of formation of ammonia as –46.1 kJ mol–1.

Give two reasons for the difference between this value and the value that you calculated in (a)(i).

(2)

iii)
What is the percentage atom economy, by mass, for ammonia in the forward reaction?

N2 (g) + 3H2 (g)  ⇌  2NH3 (g)

(1)

  A 17.6 %
  B 50.0 %
  C 82.4 %
  D 100 %

iv)
What is the equilibrium expression for Kc?

(1)
  A begin mathsize 16px style K subscript C equals fraction numerator open square brackets straight N subscript 2 close square brackets open square brackets 3 straight H subscript 2 close square brackets over denominator open square brackets 2 NH subscript 3 close square brackets end fraction end style
  B begin mathsize 16px style K subscript C equals fraction numerator open square brackets 2 NH subscript 3 close square brackets over denominator open square brackets straight N subscript 2 close square brackets open square brackets 3 straight H subscript 2 close square brackets end fraction end style
  C K subscript C equals fraction numerator open square brackets NH subscript 3 close square brackets squared over denominator open square brackets straight N subscript 2 close square brackets open square brackets straight H subscript 2 close square brackets cubed end fraction
  D begin mathsize 16px style K subscript C equals fraction numerator open square brackets straight N subscript 2 close square brackets open square brackets straight H subscript 2 close square brackets cubed over denominator open square brackets NH subscript 3 close square brackets squared end fraction end style

8b3 marks

In the chemical industry, many processes involve reversible reactions. The product is often removed before equilibrium is attained.

Give three reasons why the product may be removed before its maximum concentration is achieved.

8c2 marks

Ammonia is stable in air but can be oxidised on the surface of a copper catalyst.

An equation for this reaction is

4NH3 (g) + 5O2 (g) → 6H2O (g) + 4NO (g)                   ΔrH = –905.2 kJmol–1

The catalyst is usually warmed to approximately 300 °C to start the reaction, but after a short reaction time the copper catalyst often melts.

i)
Give a reason why the catalyst is warmed and a reason why the catalyst may melt.

(2)

ii)
Complete the reaction profile for this catalysed oxidation of ammonia, showing the enthalpy change, ΔrH.
(2)
q9cii-8cho-as-2-oct-2021-edexcel-a-level-chem
iii)
Describe the processes that occur on the surface of a heterogeneous catalyst during the oxidation of ammonia in air.
(3)

Did this page help you?

1a
Sme Calculator
2 marks

Hess’s law can be used to determine enthalpy changes for reactions which cannot be obtained directly.
An example is the reaction of anhydrous copper(II) sulfate with water to form hydrated copper(II) sulfate, CuSO4.5H2O.

The following outline procedure was carried out.

Step 1

42.75 g of deionised water was weighed out in a polystyrene cup and the temperature measured.

Step 2 0.0250 mol of hydrated copper(II) sulfate was added to the water in the polystyrene cup with stirring, making a total of 45.00 g of water.
Step 3 The temperature change was recorded.
Step 4

Steps 1 to 3 were repeated using 45.00 g of deionised water and 0.0250 mol of anhydrous copper(II) sulfate.

Calculate the mass of 0.0250 mol of hydrated copper(II) sulfate, CuSO4.5H2O.

1b
Sme Calculator
3 marks

The reaction of hydrated copper(II) sulfate with water is shown.

CuSO4.5H2O (s) + aq   →  CuSO4 (aq)           ∆H1 = +18.2 kJ mol−1

Calculate the temperature change that would have given this enthalpy change for the stated experimental procedure.
Give your answer to a measurable number of significant figures and state whether the temperature increases or decreases.

[Specific heat capacity of the solution = 4.18 J g−1 oC−1]

1c
Sme Calculator
4 marks

The reaction of anhydrous copper(II) sulfate with water is shown.

CuSO4 (s) + aq → CuSO4 (aq)      H2 = −84.5 kJ mol−1

i)
Draw to scale, on the graph paper, a labelled enthalpy level diagram which shows the enthalpy changes for the reactions of water with hydrated copper(II) sulfate (∆H1) and anhydrous copper(II) sulfate (∆H2).
(3)

q4ci-9cho-al-3-nov-2020-qp-edexcel-a-level-chem

ii)
Use your enthalpy level diagram in (c)(i) to determine the enthalpy change, ∆rH, for the reaction


CuSO4 (s) + 5H2O (l) → CuSO4.5H2O (s)

You must show your working on the diagram.

(1)

1d1 mark

State why the enthalpy change for the reaction of one mole of anhydrous copper(II) sulfate with five moles of water to form hydrated copper(II) sulfate, CuSO4.5H2O, cannot be measured directly.

Did this page help you?