DP IB Chemistry: SL

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IBChemistry: SLDPTopic Questions5. Energetics / Thermochemistry5.2 Hess's Law

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First teaching 2014

Last exams 2024

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5.2 Hess's Law

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11 mark

Enthalpy changes that are difficult to measure directly can often be determined using Hess’ Law to construct an enthalpy cycle. 

Which enthalpy change is indicated by X in the enthalpy cycle shown?

q1_5-2_ib_sl_easy_mcq

  • + 1 x Enthalpy of formation of water

  • - 1 x Enthalpy of formation of water

  • + 3 x Enthalpy of formation of water

  • - 3 x Enthalpy of formation of water

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21 mark

A student drew a Hess cycle to calculate the enthalpy of reaction to produce ethane from ethene and hydrogen.

q2_5-2_ib_sl_easy_mcq

 The student used the following enthalpy of combustion data 

 

C2H4 (g)

H2 (g)

C2H6 (g)

ΔHϴc / kJ mol-1

- 1411

- 286

- 1560

 

What are the correct labels for the arrows for the student’s Hess cycle?

 

Arrow 1

Arrow 2

Arrow 3

Arrow 4

A

ΔHϴr

- 1411

- 286

- 1560

B

ΔHϴc

- 1411

+ 286

- 1560

C

ΔHϴc

+ 1411

- 286

+ 1560

D

ΔHϴr

- 1411

- 286

+ 1560

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31 mark

Which is the correct equation to calculate the enthalpy of reaction using enthalpy of formation data? 

  • ΔHr = ΣΔHf products + ΣΔHf reactants

  • ΔHr = ΣΔHf products - ΣΔHf reactants

  • ΔHr = ΣΔHf reactants - ΣΔHf products 

  • ΔHr = ΣΔHf reactants + ΣΔHf products 

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41 mark

Enthalpy of formation data is often used to calculate the enthalpy of reaction. Which of the following does not have a standard enthalpy of formation of 0 kJ mol-1?

  • C (s)

  • N2 (l)

  • O2 (g)

  • F2 (g)

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51 mark

Iron(II,III) oxide, Fe3O4, is reduced by graphite according to the following equation: 

Fe3O4 (s) + 2C (s) → 3Fe (s) + 2CO2 (g) 

The standard enthalpy of formation values for iron(II,III) oxide and carbon dioxide are -1118 kJ mol-1 and -394 kJ mol-1 respectively. 

The standard enthalpy of reaction can be determined using which of the following calculations?

  • -1118 - (2 x -394)

  • -394 + 1118

  • (2 x -394) - 1118

  • (2 x -394) + 1118

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61 mark

Hydrogen peroxide slowly decomposes to form water and oxygen: 

2H2O2 (l) → 2H2O (l)  + O2 (g)

 

ΔHӨf, kJ mol-1

H2O2 (l)

-188

H2O (l)

-286

 

Which calculation gives the correct enthalpy of reaction for the decomposition of hydrogen peroxide?

  • -286 + 188

  • (2 x -286) - (2 x -188)

  • -286 + (2 x -188)

  • (2 x -188) - (2 x -286)

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71 mark

The diagram shows two possible reaction pathways for the reaction of A → D:
q7_5-2_ib_sl_easy_mcq

 Which of the following statements are correct? 

  1. A → D            ΔH = +45 kJ
  2. C → D            ΔH = -25 kJ
  3. D → C            ΔH = -65 kJ

  • I and II only

  • I and III only

  • II and III only 

  • I, II and III

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81 mark

The standard enthalpy of formation of sulfur dioxide cannot be measured directly. 

Which equation(s) provide a suitable alternative pathway to determine the standard enthalpy of formation of sulfur dioxide?

  • S (s) + O2 (g) → SO2 (g)

  • S (s) + O (g) → SO (g)

    SO (g) + O (g) → SO2 (g)

  • S (s) + 1.5O2 (g) → SO3 (g)

    SO3 (g) → SO2 (g) + 0.5O2 (g)

  • S (s) + 0.5O2 (g) → SO (g)

    SO (g) + O2 (g) → SO3 (g)

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91 mark

The thermal decomposition of calcium carbonate is very slow and requires a high temperature to go to completion. This makes it impractical to measure the enthalpy change for the direct reaction. 

The enthalpy change for the thermal decomposition of calcium carbonate can be determined by two chemical reactions with dilute hydrochloric acid.      
   q9_5-2_ib_sl_easy_mcq

 Which set of chemicals correctly completes the Hess cycle diagram?

  • CaCO3 + CaO + HCl

  • CaCl + H2O + CO2 

  • CaCl2 + H2O

  • CaCl2 + H2O + CO2

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101 mark

Hess’s Law can be used to calculate the enthalpy change for reactions that are difficult to measure experimentally, such as the conversion of graphite to diamond.q10_5-2_ib_sl_easy_mcq

 Which equation shows the correct application of Hess’s law to calculate the enthalpy change for the conversion of graphite to diamond?

  • ΔHr = ΔH1 + ΔH2

  • ΔHr = ΔH1 - ΔH2

  • ΔHr = ΔH2 - ΔH1

  • ΔHr = ΔH1 x ΔH2

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11 mark

The equations below show the formation of sulfur oxides from sulfur and oxygen. 

S(s) + O2(g) → SO2(g)                   begin mathsize 14px style increment H subscript f superscript ⦵ end style = –297 kJ mol–1 

S(s) + 1½O2(g) →SO3(g)               begin mathsize 14px style increment H subscript f superscript ⦵ end style = –395 kJ mol–1 

What is the enthalpy change of reaction,begin mathsize 14px style increment straight H to the power of ⦵ end style,of 2SO2(g) + O2(g) → 2SO3(g) in kJ mol–1?

  • (794 − 594)

  • (296 + 395)

  • (− 395 + 297)

  • (−790 + 594)

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21 mark

Some bond energy values are listed below.

bond

bond energy / kJ mol-1

Br–Br 

Cl–Cl

C–H

C–Cl

193

242

414

324

These bond energy values relate to the following four reactions. 

W

Br2 → 2Br

X

2Cl → Cl 2

Y

CH3 + Cl → CH3Cl
Z

CH4 → CH3 + H 

What is the correct order of enthalpy changes of the above reactions from most negative to most positive? 

  • Y → Z → W → X

  • Z → W → X → Y

  • Y → X → W → Z

  • X → Y → Z → W

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31 mark

A student calculated the standard enthalpy change of formation of propane, C3H8, using a method based on standard enthalpy changes of combustion. 

He used correct values for the standard enthalpy change of combustion of propane

 (–2220 kJ mol–1) and hydrogen (–286 kJ mol–1) but he used an incorrect value for the standard enthalpy change of combustion of carbon. He then performed his calculation correctly. His final answer was –158 kJ mol–1. 

What did he use for the standard enthalpy change of combustion of carbon?

  • - 2220 + (286 x 4) + 158

  • begin mathsize 14px style fraction numerator negative sign space 2220 space plus space left square bracket 286 space cross times 4 right square bracket space space plus space 158 over denominator 3 end fraction end style

  • begin mathsize 14px style fraction numerator plus space 2220 space minus space left square bracket 286 space cross times 4 right square bracket space space minus space 158 over denominator 3 end fraction end style

  • begin mathsize 14px style fraction numerator 3 space over denominator negative space 2220 space plus space left square bracket 286 space cross times 4 right square bracket space space plus space 158 end fraction end style

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41 mark

Given the following enthalpy changes, 

                                    I2(s) → I2(g)                                     ∆HƟ = +38 kJ mol–1 

                                            I2(g) + 3Cl2(g) → 2ICl3(s)                ∆HƟ = –214 kJ mol–1                  

What is the correct value for ∆HfƟ of iodine trichloride, ICl3(s)? 

  • begin mathsize 14px style 2 space left parenthesis 38 space minus space 214 right parenthesis end style

  • begin mathsize 14px style 2 space left parenthesis 214 space minus space 38 right parenthesis end style

  • begin mathsize 14px style ½ space left parenthesis 38 space minus space 214 right parenthesis end style

  • begin mathsize 14px style ½ space left parenthesis 214 minus 38 right parenthesis end style

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51 mark

 Using the following information: 

CO(g) + ½O2(g)   → CO2 (g)           ∆HƟ  = –283 kJ mol–1 

H2(g) + ½O2(g)    → H2O (I)            ∆HƟ  = –286 kJ mol–1 

H2O(g)    →   H2O (I)                        ∆HƟ = –44 kJ mol–1 

What is the enthalpy change, ∆HƟ, for the following reaction? 

                                    CO2(g) + H2(g) → CO(g) + H2O(g)

  • begin mathsize 14px style negative 286 space minus space 44 space minus space 283 end style

  • begin mathsize 14px style negative 286 space plus space 44 space plus space 283 end style

  • begin mathsize 14px style negative 286 space minus space 44 space plus space 283 end style

  • begin mathsize 14px style negative 286 space plus space 44 space minus space 283 end style

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61 mark

Iodine trichloride, ICl3, is made by reacting iodine with chlorine. 

                                    I2(s) + Cl2(g) → 2ICl(s)        ∆Ho = +14 kJ mol–1

                                                ICl(s) + Cl2(g) → ICl3(s)      ∆Ho= –88 kJ mol–1

By using the data above, what is the enthalpy change of the formation for solid iodine trichloride?

  • –162 kJ mol–1

  • –81 kJ mol–1

  • –74 kJ mol–1

  • –60 kJ mol–1

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71 mark

Shown below are three enthalpy changes: 

CH4(g)     +    O2(g)   →  HCHO(l)   +   H2O(l)                 ΔH = x 

HCHO(l)   +  ½O2(g)  → HCOOH(l)                                 ΔH = y 

2HCOOH(l)  +  ½O2(g)  → (COOH)2(l) + H2O(l)             ΔH = z 

Use the information given to deduce the correct expression for the enthalpy change of the following reaction: 

2CH4(g)   +   3½ O2(g)    →   (COOH)2(l)  + 3H2O(l)

  • x + y + z

  • 2x + y + z

  • 2x + 2y + z

  • 2x + 2y + 2z

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81 mark

The hydration enthalpy of anhydrous copper(II) sulfate, labelled as ΔHexp, cannot be measured directly. It can be found indirectly by determining the solution enthalpies of anhydrous and hydrated copper(II) sulfate.q8_5-2_ib_sl_hard_mcq

Which of the following statements correctly explains why the value for ΔHexp for this reaction cannot be measured directly? 

  1. Hydrated copper(II) sulfate is not produced in a controlled manner
  2. Dissolving of the solid is difficult to avoid
  3. Heat energy is trapped inside the solid copper(II) sulfate

  • I and II only  

  • I and III only  

  • II and III only 

  • I, II and III

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91 mark

Lithium iodide solution can be produced by two different reaction paths, according to the following diagram:q9_5-2_ib_sl_hard_mcq

Which labels could be added to complete the diagram

 

ΔH1

ΔH2

ΔH3

A

+364 kJ mol-1

ΔHhyd

+82 kJ mol-1

B

ΔHhyd

ΔHsol

+82 kJ mol-1

C

ΔHhyd

-307 kJ mol-1

ΔHsol

D

-364 kJ mol-1

ΔHsol

ΔHhyd

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101 mark

Bond energy calculations show the enthalpy of combustion for propene to be -1572.0 kJ mol-1.

Compound

C3H6 (g)

CO2 (g)

H2O (l)

H2O (g)

ΔHӨf / kJ mol-1

+20.0

-393.5

-285.8

-241.8

  

Using the enthalpy of formation data, which calculation correctly shows the percentage error between propene’s enthalpy of combustion values obtained from bond energy calculations and Hess’s Law calculations, assuming the bond energy calculation value is correct?

  • begin mathsize 14px style fraction numerator negative 1572.0 over denominator left parenthesis left parenthesis 3 space cross times space minus 393.5 right parenthesis space plus space left parenthesis 3 space cross times space minus 241.8 right parenthesis space minus space left parenthesis 20 right parenthesis right parenthesis space minus space 1572.0 end fraction x space 100 end style

  • begin mathsize 14px style fraction numerator left parenthesis 3 space cross times space minus 393.5 right parenthesis space plus space left parenthesis 3 space cross times space minus 241.8 right parenthesis space minus space left parenthesis 20 right parenthesis over denominator negative 1572.0 end fraction x space 100 end style

  • begin mathsize 14px style fraction numerator left parenthesis 3 space cross times space minus 393.5 right parenthesis space plus space left parenthesis 3 space cross times space minus 241.8 right parenthesis space plus space left parenthesis 20 right parenthesis over denominator negative 1572.0 end fraction cross times 100 end style

  • begin mathsize 14px style fraction numerator left parenthesis left parenthesis 3 space cross times space minus 393.5 right parenthesis space plus space left parenthesis 3 space cross times space minus 241.8 right parenthesis space minus space left parenthesis 20 right parenthesis right parenthesis space minus space left parenthesis negative 1572.0 right parenthesis over denominator negative 1572.0 end fraction cross times space 100 end style

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11 mark

The first stage in the industrial production of nitric acid from ammonia can be represented by the following equation. 

4NH3(g)  +  5O2(g)  ⇌  4NO(g)  +  6H2O(g) 

Using the following standard enthalpy change of formation data, what is the value of the standard enthalpy change begin mathsize 14px style italic increment H to the power of italic ⦵ end style for this reaction?

Compound

 begin mathsize 14px style increment H subscript f superscript ⦵ end style/kJ mol– 1

NH3 (g)

-46.1

NO (g)

+90.3

H2O (g)

-241.8

  • begin mathsize 14px style left parenthesis 4 space cross times space left parenthesis negative 46.1 right parenthesis right parenthesis space plus left parenthesis left parenthesis 4 space cross times space 90.3 right parenthesis plus left parenthesis 6 space cross times space left parenthesis negative 241.8 right parenthesis right parenthesis space end style

  • begin mathsize 14px style left parenthesis left parenthesis 4 space cross times space 90.3 right parenthesis plus left parenthesis 6 space cross times space left parenthesis negative 241.8 right parenthesis right parenthesis space minus left parenthesis 4 space cross times space left parenthesis negative 46.1 right parenthesis right parenthesis end style

  • begin mathsize 14px style left parenthesis 4 space cross times space left parenthesis negative 46.1 right parenthesis right parenthesis space minus left parenthesis left parenthesis 4 space cross times space 90.3 right parenthesis plus left parenthesis 6 space cross times space left parenthesis negative 241.8 right parenthesis right parenthesis right parenthesis end style

  • begin mathsize 14px style left parenthesis negative 46.1 right parenthesis space minus left parenthesis 90.3 right parenthesis plus left parenthesis negative 241.8 right parenthesis end style

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21 mark

Titanium occurs naturally as the mineral rutile, TiO2. One possible method of extraction of titanium is to reduce the rutile by heating with carbon. 

TiO2(s) + 2C(s)  →  Ti(s) + 2CO(g) 

The standard enthalpy changes of formation of TiO2(s) and CO(g) are –890 kJ mol-1 and –110.5 kJ mol-1 respectively. 

What is the standard enthalpy change of the extraction of titanium?

  • + 669 kJ mol–1

  • + 779.5 kJ mol–1

  • – 779.5 kJ mol–1

  • – 669 kJ mol–1

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31 mark

The combustion of ethanol (C2H5OH) is increasingly being used to fuel cars. The combustion reaction is shown below 

                         C2H5OH(l)  +   3O2 (g)   →  2CO2(g)    +  3H2O(l) 

The standard enthalpy change of formation of carbon dioxide is a kJ mol–1.

The standard enthalpy change of formation of water is b kJ mol–1.

The standard enthalpy change of formation of ethanol is c kJ mol–1. 

What is the standard enthalpy change of combustion for ethanol?

  • 2a + 3b + c

  • 2a + 3b − c

  • −2a − 3b + c

  • −2a − 3b −c

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41 mark

Propanone has the molecular formula C3H6O. Use the following information to calculate the enthalpy change for the formation of propanone? 

The enthalpy change of combustion of carbon is x kJ mol -1

The enthalpy change of combustion of hydrogen is y kJ mol -1

The enthalpy change of combustion of propanone is z kJ mol -1

  • x + y - z

  • 3x + 3y - z

  • z - 3x + 3y 

  • 3x + 3y + z

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51 mark

Combustion of ethene proceeds via the reaction shown in the Hess cycle diagram.q5_5-2_ib_sl_medium_mcq

 What are the correct elements missing from the Hess cycle?

  • 2C (g) + 2H2 (g) + 3O2 (g)

  • C (s) + H2 (g) + O2 (g)

  • 2C (s) + 2H2 (g) + 3O2 (g)

  • C2 (s) + 2H2 (g) + 3O2 (g)

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61 mark

Combustion of propene proceeds via the reaction shown in the Hess cycle diagram.q6_5-2_ib_sl_medium_mcq

Which are the correct labels for the enthalpies shown in the Hess cycle?

 

P

Q

R

S

A

ΔHӨc

ΔHӨr

ΔHӨc

ΔHӨc

B

ΔHӨc

ΔHӨf

ΔHӨr

ΔHӨr

C

ΔHӨc

ΔHӨf

ΔHӨc

ΔHӨf

D

ΔHӨr

ΔHӨr

ΔHӨf

ΔHӨf

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71 mark

Hess’s Law can be used to calculate the enthalpy change for reactions that are difficult to measure experimentally, such as the conversion of graphite to diamond. 

Which enthalpy data could be used to calculate the enthalpy change for the conversion of graphite to diamond? 

  1. Enthalpy of reaction
  2. Enthalpy of combustion
  3. Enthalpy of formation

  • I and II only

  • I and III only 

  • II and III only 

  • I, II and III

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81 mark

A basic definition of Hess’s Law states that the overall enthalpy change for a reaction is the same independent of the route taken. 

The addition of which of the following statements makes the definition of Hess’s Law more complete? 

  1. Providing that the reactants are the same
  2. Providing that the products are the same
  3. Providing that the conditions at the start and the end of the reaction are the same

  • I and II only 

  • I and III only 

  • II and III only 

  • I, II and III

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91 mark

The hydration enthalpy of anhydrous copper(II) sulfate cannot be measured directly. It can be found indirectly by determining the solution enthalpies of anhydrous and hydrated copper(II) sulfate. q9_5-2_mcq_medium_ib-chemistry-sl

Which are the correct labels for the enthalpies shown in the Hess cycle? 

 

1

2

3

A

ΔHӨsol

ΔHӨhyd

ΔHӨhyd

B

ΔHӨr

ΔHӨhyd

ΔHӨhyd

C

ΔHӨhyd

ΔHӨsol

ΔHӨhyd

D

ΔHӨr

ΔHӨsol

ΔHӨsol

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101 mark

The hydration enthalpy of anhydrous copper(II) sulfate, labelled as ΔHexp, cannot be measured directly. It can be found indirectly by determining the solution enthalpies of anhydrous and hydrated copper(II) sulfate.

q10_5-2_mcq_medium_ib-chemistry-sl

Which of the following statements correctly explains why the value for ΔHexp for this reaction cannot be measured directly?

  • Measuring the temperature change in a solid is difficult

  • The reaction is very slow

  • The reaction has high energy requirements

  • The reaction is endothermic

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