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DP IB Chemistry: SL

Topic Questions

Home / IB / Chemistry: SL / DP / Topic Questions / 5. Energetics / Thermochemistry / 5.2 Hess's Law / Multiple Choice: Paper 1


5.2 Hess's Law

Question 1

Marks: 1

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)

Choose your answer
  

Question 2

Marks: 1

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

Choose your answer
  

Question 3

Marks: 1

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

Choose your answer
  

Question 4

Marks: 1

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

Choose your answer
  

Question 5

Marks: 1

 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

Choose your answer
  

Question 6

Marks: 1

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

Choose your answer
  

Question 7

Marks: 1

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

Choose your answer
  

Question 8

Marks: 1

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

Choose your answer
  

Question 9

Marks: 1

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

    Choose your answer
      

    Question 10

    Marks: 1

    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

    Choose your answer