Energy Cycles in Reactions (DP IB Chemistry)

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  • True or False?

    Bond breaking is an exothermic process.

    False.

    Bond breaking is an endothermic process.

  • Define the term bond dissociation enthalpy.

    Bond dissociation enthalpy is the energy required to break a particular chemical bond.

  • State the equation for calculating enthalpy change of reaction using bond energies.

    The equation for calculating enthalpy change of reaction using bond energies is:

    ΔH = Σ(bonds broken) - Σ(bonds formed)

  • Define average bond energy.

    Average bond energy is the energy needed to break one mole of bonds in a gaseous molecule averaged over similar compounds.

  • True or False?

    A reaction is exothermic if more energy is released when new bonds are formed than the energy required to break bonds.

    True.

    A reaction is exothermic if more energy is released when new bonds are formed than the energy required to break bonds.

  • In terms of bond making and breaking, why is a reaction endothermic?

    A reaction is endothermic because more energy is required to break bonds than the energy released when new bonds are formed.

  • Calculate the enthalpy change of reaction for this equation:

    N2 (g) + 3H2 (g) ⇌ 2NH3 (g)

    Bond

    Average Bond Energy (kJ mol-1)

    Nidentical toN

    945

    HminusH

    436

    NminusH

    391

    The enthalpy change of reaction is:

    • Bonds broken = (1 x 945) + (3 x 436) = +2253

    • Bonds formed= 6 x 391 = -2346

    • ΔHr = +2253 - 2346 = -93 kJ mol-1

  • Define Hess's Law.

    Hess's Law states that the total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place as long as the initial and final conditions are the same.

  • What does Hess's Law allow us to calculate?

    Hess's Law allows us to calculate the standard enthalpy change of a reaction from known standard enthalpy changes.

  • True or False?

    In Hess's Law energy cycles, the direct route always has a larger enthalpy change than the indirect route.

    False.

    In Hess's Law energy cycles, the enthalpy change of the direct route is equal to the enthalpy change of the indirect route.

  • Give the equation that would be used to calculate the enthalpy change for the conversion of graphite to diamond.

    Diagram showing the formation of CO2 gas from graphite and diamond. Arrows indicate transitions with enthalpy changes: graphite to CO2 (ΔH1) and diamond to CO2 (ΔH2).

    The equation that would be used to calculate the enthalpy change for the conversion of graphite to diamond is ΔHr = ΔH1 - ΔH2

    Diagram showing the formation of CO2 gas from graphite and diamond. Arrows indicate transitions with enthalpy changes: graphite to CO2 (ΔH1) and diamond to CO2 (ΔH2).
  • True or False?

    Hess's Law can be used to calculate enthalpy changes that can be found experimentally using calorimetry.

    False.

    Hess's Law is used to calculate enthalpy changes which cannot be found experimentally using calorimetry.

  • Give the two common methods used to solve Hess's law problems.

    The two common methods for solving Hess's Law problems are:

    • Using Hess's law cycles

    • Using equations

  • True or False?

    When using cycles to solve Hess's Law problems, if you follow the direction of the arrow you subtract the quantity.

    False.

    When using cycles to solve Hess's Law problems, if you follow the direction of the arrow you add the quantity.

  • True or False?

    In Hess's Law calculations, you always need to adjust for different molar amounts.

    True.

    In Hess's Law calculations, you always need to adjust for different molar amounts.

  • State the general equation for solving Hess's Law problems using equations.

    The general equation for solving Hess's Law problems using equations is:

    ΔH(reaction) = Σ(ΔH products) - Σ(ΔH reactants)

  • Use the Hess cycle to calculate the enthalpy change for the reaction.

    A chemical equation showing 4FeO + O2 forming 2Fe2O3 with an enthalpy change, and two alternative paths involving 4Fe and O2 with different enthalpy changes.

    The enthalpy change for the reaction is:

    A chemical equation showing 4FeO + O2 forming 2Fe2O3 with an enthalpy change, and two alternative paths involving 4Fe and O2 with different enthalpy changes.

    ΔHr = - ( - 544 x 2) + (- 1648) = - 560 kJ

  • Define standard enthalpy of formation.

    The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements under standard conditions.

  • True or False?

    In enthalpy of formation cycles, arrows always point downwards.

    False.

    In enthalpy of formation cycles, arrows always point upwards because the definition of enthalpy of formation must go from elements to compounds.

  • State the equation used to calculate enthalpy changes from enthalpy of formation data.

    The equation used to calculate enthalpy changes from enthalpy of formation data is:

    ΔH = ΣΔHf(products) - ΣΔHf(reactants)

  • Calculate the enthalpy of combustion using the data below.

    B2H6 (g) + 3O2 (g)  B2O3 (s) + 3H2O (g)

    B2H6 (g)

    B2O3 (g)

    H2O (g)

    Hf / kJmol-1

    +31.4

    -1270

    -242

    The enthalpy of combustion is:

    • ΔH = ΔHf products  - ΔHf reactants

    • ΔH = - 1996 - (+ 31.4) = -2027.4 kJ

  • Draw a Hess's law cycle for the reaction below that could be used to determine the overall enthalpy change using enthalpy of formation data.

    NH4NO3 (s)  +  ½C (s)  →  N2 (g)  +  2H2O (g)  +  ½CO2 (g)

    A Hess's law cycle that would allow the determination of overall enthalpy change for the following reaction using enthalpy of formation data is:

    Chemical equation: NH4NO3(s) + 1/2 C(s) -> N2(g) + 2H2O(g) + 1/2 CO2(g). Underneath: N2(g) + 1.5O2(g) + C(s) + 2H2(g).
  • Calculate the enthalpy change for this reaction using the Hess's law cycle below.

    Chemical equation of NH4NO3 and 1/2 C producing N2, water, and CO2 with annotations: -365, 2x - 242 = -484, 0.5x - 394 = -197. Box with N2 + 1.5 O2 + C + 2 H2.

    The enthalpy change for this reaction using the enthalpy of formation data given is:

    Chemical equation of NH4NO3 and 1/2 C producing N2, water, and CO2 with annotations: -365, 2x - 242 = -484, 0.5x - 394 = -197. Box with N2 + 1.5 O2 + C + 2 H2.

    ΔHr = +365 - 484 - 197 = -316 kJ mol-1

  • The enthalpy of formation of elements in their standard states is always ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ .

    The enthalpy of formation of elements in their standard states is always ‎zero.

  • True or False?

    The enthalpy of formation is always a positive value.

    False.

    The enthalpy of formation can be a positive or negative value.

  • Define standard enthalpy of combustion.

    Standard enthalpy of combustion is the enthalpy change that occurs when one mole of a substance burns completely under standard conditions.

  • True or False?

    In enthalpy of combustion cycles, arrows always point upwards.

    False.

    In enthalpy of combustion cycles, arrows should be pointing downwards.

  • What is the general equation for calculating ΔH using enthalpy of combustion data?

    The general equation for calculating ΔH using enthalpy of combustion data is:

    ΔH = ΣΔHc(reactants) - ΣΔHc(products)

  • True or False?

    The enthalpy of combustion is always an exothermic process.

    True.

    The enthalpy of combustion is always an exothermic process.

  • Draw a Hess's law cycle for the reaction below that could be used to determine the overall enthalpy change using enthalpy of combustion data.

    3C (s) + 3H2 (g) + ½ O2 (g)    CH3COCH3 (l)

    A Hess's law cycle that could be used to determine the overall enthalpy change using enthalpy of combustion data is:

    Chemical reaction showing 3C + 3H2 + 0.5O2 yielding CH3COCH3, and arrows pointing from the left, below to 3CO2 + 3H2O.
  • Calculate the enthalpy change of formation of propanone using the Hess's law cycle below.

    Chemical reaction diagram: 3C(s) + 3H₂(g) + ½O₂(g) → CH₃COCH₃(l). Arrows show enthalpy values (-1182, -858, -1821) leading to 3CO₂ + 3H₂O.

    The enthalpy change of formation of propanone using the Hess's law cycle is:

    Chemical reaction diagram: 3C(s) + 3H₂(g) + ½O₂(g) → CH₃COCH₃(l). Arrows show enthalpy values (-1182, -858, -1821) leading to 3CO₂ + 3H₂O.

    ΔHf = -1182 - 858 + 1821 = -219 kJ mol-1

  • The sign for enthalpy change of combustion values will always be ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ .

    The The sign for enthalpy change of combustion values will always be negative.

  • True or False?

    Enthalpy of combustion can be used to calculate enthalpy of formation.

    True.

    Enthalpy of combustion can be used to calculate enthalpy of formation using Hess's Law.

  • Calculate the enthalpy of formation of propane using the following information.

    3C (s) + 4H2 (g) C3H8 (g)

    C3H8(g)

    C (s)

    H2 (g)

    Hc / kJmol-1

    -2220

    -393

    -286

    The enthalpy of formation of propane is:

    • ΔH = ΣΔHc(reactants) - ΣΔHc(products)

    • ΔH = (3x -393)+ (4 x -286) - (-2220)

    • ΔH = -103 kJ mol-1

  • What is the significance of enthalpy of combustion in practical applications?

    Enthalpy of combustion is significant in practical applications as it helps determine the energy content of fuels and food.

  • The symbol to represent the enthalpy change of combustion under standard conditions is ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ .

    The symbol to represent the enthalpy change of combustion under standard conditions is ‎ ΔHc.

  • What is a Born-Haber cycle?

    A Born-Haber cycle is a specific application of Hess's Law for ionic compounds that enables the calculation of lattice enthalpy.

  • True or False?

    The first ionisation energy is always an exothermic process.

    False.

    The first ionisation energy is always an endothermic process.

  • Define the term lattice enthalpy.

    Lattice enthalpy is the energy required to separate one mole of an ionic compound into its constituent gaseous ions.

  • What is the enthalpy of atomisation?

    The enthalpy of atomisation is the energy required to convert one mole of an element in its standard state to gaseous atoms.

  • The energy change when one mole of gaseous atoms gains one mole of electrons to form one mole of gaseous anions is known as ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ .

    The energy change when one mole of gaseous atoms gains one mole of electrons to form one mole of gaseous anions is known as electron affinity.

  • True or False?

    Electron affinity is always an exothermic process.

    False.

    While electron affinity is often exothermic, it can be endothermic for some elements.

  • In a Born-Haber cycle for NaCl, what does an arrow from Na (s) to Na (g) represent?

    In a Born-Haber cycle for NaCl, the arrow from Na (s) to Na (g) represents the enthalpy of atomisation of sodium.

  • True or False?

    Arrows pointing upwards in a Born-Haber cycles represent endothermic reactions.

    True.

    Arrows pointing upwards in a Born-Haber cycles represent endothermic reactions.

  • Write an equation to represent the first electron affinity of chlorine.

    An equation to represent the first electron affinity of chlorine is:

    Cl (g) + e- → Cl- (g)

  • Write an equation to represent the second ionisation energy of magnesium.

    An equation to represent the second ionisation energy of magnesium:

    Mg+ (g) → Mg2+ (g) + e-

  • Which enthalpy change is shown in Step 1?

    Energy level diagram with 6 steps showing reactions from KF (s) at the bottom to K+(g) + e- + F(g) at the top, including intermediates like K(s) and F2(g).

    The enthalpy change is shown in Step 1 is the enthalpy change of formation.

  • Which enthalpy change is shown in Step 2?

    Energy level diagram with 6 steps showing reactions from KF (s) at the bottom to K+(g) + e- + F(g) at the top, including intermediates like K(s) and F2(g).

    The enthalpy change is shown in Step 2 is the enthalpy of atomisation of potassium.

  • Which enthalpy change is shown in Step 3?

    Energy level diagram with 6 steps showing reactions from KF (s) at the bottom to K+(g) + e- + F(g) at the top, including intermediates like K(s) and F2(g).

    The enthalpy change is shown in Step 3 is the enthalpy of atomisation of fluorine.

  • Which enthalpy change is shown in Step 4?

    Energy level diagram with 6 steps showing reactions from KF (s) at the bottom to K+(g) + e- + F(g) at the top, including intermediates like K(s) and F2(g).

    The enthalpy change is shown in Step 4 is the first ionisation energy of potassium.

  • Which enthalpy change is shown in Step 5?

    Energy level diagram with 6 steps showing reactions from KF (s) at the bottom to K+(g) + e- + F(g) at the top, including intermediates like K(s) and F2(g).

    The enthalpy change is shown in Step 5 is the first electron affinity of fluorine.

  • Which enthalpy change is shown in Step 6?

    Energy level diagram with 6 steps showing reactions from KF (s) at the bottom to K+(g) + e- + F(g) at the top, including intermediates like K(s) and F2(g).

    The enthalpy change is shown in Step 6 is the lattice enthalpy of formation.

  • Write the equations for the three missing steps in the Born-Haber cycle.

    Energy level diagram of the formation of Li2O from elemental Li and O2. Diagram is labeled with steps 1, 2, and 3, showing transitions and electron gain.

    Th equations for the three missing steps in the Born-Haber cycle are:

    Diagram showing a series of reactions with lithium and oxygen, moving from solid and gaseous phases to ionic states and culminating in the formation of lithium oxide (Li₂O).
  • Using the information below calculate the lattice enthalpy of KF, ΔHθlattice KF.

    A diagram to show the born haber cycle for KF

    To calculate lattice enthalpy for KF, ΔHθlattice KF:

    A diagram to show the born haber cycle for KF

    ΔHθlattice KF = -563-90-79-419+328

    ΔHθlattice KF = -823 (kJ mol−1)

  • Using the information below calculate the enthalpy of formation LiF. ΔHθf LiF.

    A Born-Haber cycle for LiF

    To calculate the enthalpy of formation LiF. ΔHθf LiF:

    A Born-Haber cycle for LiF

    ΔHθf LiF = 216 + 79 + 520 + (-348) + (-1061)

    ΔHθf LiF = -594 (kJ mol-1)

  • What is the enthalpy of atomisation of chlorine?

    ΔHθBE Cl2 = 242 kJ mol-1

    The enthalpy of atomisation of chlorine is 121 kJ mol-1:

    ΔHθat Cl2 = 242 / 2 = 121 kJ mol-1

  • Calculate the enthalpy change for the following conversion:

    Ca (s) → Ca2+ (g) + 2e-

    • ΔHθIE1 Ca = 590 kJ mol-1   

    • ΔHθIE2 Ca = 1145 kJ mol-1                       

    • ΔHθat Ca = 178 kJ mol-1

    To calculate the enthalpy change for the following conversions:

    Ca (s) → Ca2+ (g) + 2e-

    ΔHθIE1 Ca = 590 kJ mol-1   

    ΔHθIE2 Ca = 1145 kJ mol-1                       

    ΔHθat Ca = 178 kJ mol-1

    • ΔHθat Ca + ΔHθIE1 Ca + ΔHθIE2 Ca = 178 + 590 + 1145 = (+)1913 (kJ mol-1)

  • True or False?

    When performing a Born-Haber cycle calculation for MgCl2, the value for the first electron affinity of chlorine would need to be doubled.

    True.

    When performing a Born-Haber cycle calculation for MgCl2, the value for the first electron affinity of chlorine would need to be doubled.

  • Calculate the enthalpy change for the following conversion:

    Cl2 (g) + 2e- → 2Cl- (g)

    • ΔHθBE Cl2 = 242 kJ mol-1

    • ΔHθEA Cl = -349 kJ mol-1

    To calculate the enthalpy change for the following conversion:

    Cl2 (g) + 2e- → 2Cl- (g)

    ΔHθBE Cl2 = 242 kJ mol-1

    ΔHθEA Cl = -349 kJ mol-1

    • ΔHθBE Cl2 + (2 x ΔHθEA Cl) = 242 + (2 x -349) = -456 (kJ mol-1)

  • Rearrange the following equation to calculate lattice enthalpy,  ΔHlat.

    ΔHf = ΔHat + ΔHat + ΔHIE + ΔHEA - ΔHlat

    To calculate lattice enthalpy,  ΔHlat:

    ΔHf = ΔHat + ΔHat + ΔHIE + ΔHEA - ΔHlat

    ΔHlat = - ΔHf + ΔHat + ΔHat + ΔHIE + ΔHEA

  • Using the information below calculate the lattice enthalpy of CaCl2, ΔHθlattice CaCl2.

    CaCl2 Born Haber cycle

    To calculate the lattice enthalpy of CaCl2, ΔHθlattice CaCl2.

    CaCl2 Born Haber cycle

    ΔHθlattice CaCl2 = -814 - 178 - 590 - 1145 - 242 + (349 x 2)

    ΔHθlattice CaCl2 = -2271 (kJ mol−1)