Lesson 8.2.1

8.2.1 Enthalpy cycles and Hess's Law Quiz: Pearson Edexcel Chemistry, Unit 8

20 questions

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Lesson 8.2.1, Enthalpy cycles and Hess's Law: 20 multiple choice questions for the Pearson Edexcel Chemistry (9CH0), Unit 8: Energetics I, written with Revision Ninja.

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The 20 questions

  1. What does Hess's Law state?

    • The enthalpy change is equal to the activation energy of the slowest step in the reaction mechanism
    • Enthalpy changes can only be added if they are for gases, so solids must be excluded from the sum
    • The total enthalpy change of a reaction is the same, regardless of the route taken
    • The enthalpy change depends on the number of steps taken, so each route gives a different overall result
  2. Given C(s) + O2(g) -> CO2(g), delta H = -393.5 kJ mol-1, and CO(g) + 1/2 O2(g) -> CO2(g), delta H = -283.0 kJ mol-1, what is delta H for C(s) + 1/2 O2(g) -> CO(g)?

    • -110.5 kJ mol-1
    • -283.0 kJ mol-1
    • +110.5 kJ mol-1
    • -676.5 kJ mol-1
  3. What is the correct way to construct an enthalpy cycle for a reaction whose enthalpy change cannot be measured directly?

    • Use only combustion data, since these are always measurable
    • Add the activation energies of each step
    • Use a route through known enthalpy changes that goes from the same reactants to the same products
    • Draw the cycle with the reactants at the top and the products at the bottom and ignore the signs
  4. When an equation is reversed in a Hess cycle calculation, what happens to the sign of its enthalpy change?

    • The value is doubled
    • The value becomes zero
    • The sign changes
    • The sign stays the same
  5. When an equation is multiplied by 2 in a Hess calculation, what happens to its enthalpy change?

    • It becomes negative
    • It is halved
    • It is doubled
    • It stays the same
  6. Using enthalpy of formation data, delta Hf of CO2 = -393.5, H2O(l) = -285.8, and C2H5OH(l) = -277.7 kJ mol-1, what is delta Hc of ethanol? C2H5OH + 3O2 -> 2CO2 + 3H2O

    • -1366.7 kJ mol-1
    • +1366.7 kJ mol-1
    • -956.9 kJ mol-1
    • -1088.0 kJ mol-1
  7. What is the standard enthalpy change of formation of an element in a Hess cycle calculation?

    • The same as its bond enthalpy
    • Equal to the combustion enthalpy
    • Equal to the enthalpy of neutralisation
    • Zero
  8. In an enthalpy cycle, what is the relationship between the direct route and the indirect routes?

    • They have the same overall enthalpy change
    • The direct route is always more exothermic
    • The direct route has a zero enthalpy change
    • The indirect routes have no enthalpy change
  9. Given delta H1 = -285.8 kJ mol-1 for H2 + 1/2 O2 -> H2O(l) and delta H2 = -241.8 kJ mol-1 for H2 + 1/2 O2 -> H2O(g), what is delta H for H2O(l) -> H2O(g)?

    • +241.8 kJ mol-1
    • -527.6 kJ mol-1
    • +44.0 kJ mol-1
    • -44.0 kJ mol-1
  10. What is the enthalpy change for the reaction C(s) + 2H2(g) -> CH4(g), given C(s) + O2 -> CO2 = -393.5, H2 + 1/2 O2 -> H2O(l) = -285.8 and CH4 + 2O2 -> CO2 + 2H2O(l) = -890.3 kJ mol-1? Use Hess's Law.

    • -74.8 kJ mol-1
    • -1569.6 kJ mol-1
    • -890.3 kJ mol-1
    • +74.8 kJ mol-1
  11. Why are indirect enthalpy calculations useful?

    • They give the rate of a reaction at any temperature by using the enthalpy values as a rate constant
    • They prove that the reaction is reversible by showing the enthalpy change changes sign on the reverse
    • They show the activation energy of the reaction from the difference between the reactant and product levels
    • They allow enthalpy changes that cannot be measured directly to be found from other measurable changes
  12. What must be true of the enthalpy cycle diagram before the sum is written?

    • All enthalpy changes must be positive so that the sum of the cycle can be taken without any sign errors
    • Only one route is needed, because a single route gives the full answer without any checks
    • The reactants and products must be placed at the same two states for each route
    • All reactants must be in the gas state so that the enthalpy changes can be combined in a single cycle
  13. Which enthalpy change is used in a Hess cycle calculation of a reaction's enthalpy of formation from combustion data?

    • Activation energies of each step in the mechanism, which are added to give the overall value
    • Bond enthalpies only, with no data for the compounds needed, so that the cycle can be closed
    • pH values of the reactants, which are converted into enthalpy changes using a standard formula
    • Enthalpies of combustion of reactants and products, with the sum determined by Hess's Law
  14. Given that the enthalpy of combustion of hydrogen is -285.8 kJ mol-1, how much heat is released when 4.00 g of hydrogen (Mr = 2.0) burns to form liquid water?

    • 285.8 kJ
    • 571.6 kJ
    • 142.9 kJ
    • 1143.2 kJ
  15. In a Hess cycle, a student gets +120 kJ mol-1 for a reaction that is known to release heat. What is the most likely cause?

    • The activation energy was included in the sum
    • The reaction is catalysed
    • A sign error when reversing one or more equations
    • The mass of water was measured too precisely
  16. Which statement about the enthalpy of formation of an element in its standard state is correct?

    • Its enthalpy of formation is zero, so it is excluded from the sum
    • It must be measured in each experiment
    • Its enthalpy of formation is equal to its ionisation energy
    • It is always positive
  17. What is the enthalpy change for C(s) + 1/2 O2(g) -> CO(g), given C(s) + O2 -> CO2 = -393.5 and CO + 1/2 O2 -> CO2 = -283.0 kJ mol-1, if the second equation is multiplied by -1?

    • -283.0 kJ mol-1
    • -110.5 kJ mol-1
    • -676.5 kJ mol-1
    • +110.5 kJ mol-1
  18. What is the sign convention used for a standard enthalpy change?

    • Negative for exothermic reactions and positive for endothermic reactions
    • Always negative regardless of the process
    • Positive for exothermic reactions and negative for endothermic reactions
    • Always positive for combustion
  19. A Hess cycle is set up with two routes. Route A gives -200 kJ mol-1 and route B has one step of -50 and one step of -160 kJ mol-1. Which statement is correct?

    • Route B is always more exothermic than route A
    • The routes do not agree, so one of the measured values is wrong
    • Route A is a catalysed route with lower enthalpy
    • Both routes are correct because they have different intermediates
  20. The standard enthalpy of combustion of carbon is -393.5 kJ mol-1. How much heat is released when 6.00 g of carbon (Ar = 12.0) burns completely?

    • 393.5 kJ
    • 196.8 kJ
    • 98.4 kJ
    • 787.0 kJ

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