IB Chemistry Quiz: Understand Energy Cycles In Reactions
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Understand Energy Cycles In ReactionsQuestion 1 of 20

The enthalpy change for the complete combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), is estimated using average bond enthalpies. Why does this calculated value differ from the experimentally determined value for this specific reaction?

Average bond enthalpies are mean values from various molecules and may not match the actual bond energy in CH₄.
The calculation assumes the reaction goes to completion, which is not always true experimentally.
Hess's Law cannot be applied when using bond enthalpies, only for standard enthalpies of formation.
The experimental value is measured under standard conditions, while bond enthalpies apply only at 0 K.
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IB Chemistry Quiz

IB Chemistry Quiz: Understand Energy Cycles In Reactions

Practice Understand Energy Cycles In Reactions in IB Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Understand Energy Cycles In Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

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Question 1

The enthalpy change for the complete combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), is estimated using average bond enthalpies. Why does this calculated value differ from the experimentally determined value for this specific reaction?

  1. Average bond enthalpies are mean values from various molecules and may not match the actual bond energy in CH₄. (correct answer)
  2. The calculation assumes the reaction goes to completion, which is not always true experimentally.
  3. Hess's Law cannot be applied when using bond enthalpies, only for standard enthalpies of formation.
  4. The experimental value is measured under standard conditions, while bond enthalpies apply only at 0 K.
Explanation: The primary limitation of using average bond enthalpies is that they are averaged over a wide range of different chemical compounds. The actual energy of a specific bond (e.g., a C−H bond in methane) depends on its molecular environment. Therefore, the calculated enthalpy change is only an estimate and will differ from the precise experimental value. B is incorrect because while incomplete combustion can occur, the experimental value is determined under conditions ensuring complete combustion. C is incorrect as bond enthalpy calculations are a direct application of Hess's Law. D is incorrect as both values are typically determined or corrected to standard conditions (298 K), not 0 K.

Question 2

Use the data below to calculate the first electron affinity of bromine, EA₁, in kJ mol⁻¹. Enthalpy of atomization of Br₂(l) to form Br(g): +112 First ionization energy of K(g): +419 Enthalpy of atomization of K(s): +89 Enthalpy of formation of KBr(s): −394 Lattice enthalpy of formation of KBr(s): −682

  1. −332 kJ mol⁻¹ (correct answer)
  2. +332 kJ mol⁻¹
  3. −1696 kJ mol⁻¹
  4. +456 kJ mol⁻¹
Explanation: The Born-Haber cycle relates the enthalpy of formation to the component enthalpy changes via Hess's Law: ΔH_f^⦵ = ΔH_at^⦵(K) + IE₁(K) + ΔH_at^⦵(Br) + EA₁(Br) + LE(KBr).
Substituting the given values:
−394 = (+89) + (+419) + (+112) + EA₁(Br) + (−682).
First, sum the known positive and negative terms on the right side:
−394 = (89 + 419 + 112) + EA₁(Br) - 682
−394 = 620 + EA₁(Br) - 682
−394 = −62 + EA₁(Br).
Rearranging to solve for EA₁(Br):
EA₁(Br) = −394 + 62 = −332 kJ mol⁻¹.

Question 3

Using the following data, what is the enthalpy change for the incomplete combustion of two moles of propane? 2C₃H₈(g) + 7O₂(g) → 6CO(g) + 8H₂O(l) ΔH_c^⦵(C₃H₈(g)) = −2220 kJ mol⁻¹ ΔH_c^⦵(CO(g)) = −283 kJ mol⁻¹

  1. −6138 kJ
  2. −1371 kJ
  3. −2742 kJ (correct answer)
  4. −1937 kJ
Explanation: We can construct a Hess's cycle where the reactants and products of the target reaction are both completely combusted to CO₂(g) and H₂O(l).
Route 1 (combustion of reactants): Complete combustion of 2 moles of C₃H₈. ΔH₁ = 2 × ΔH_c^⦵(C₃H₈) = 2 × (−2220) = −4440 kJ.
Route 2 (target reaction + combustion of products): Let the unknown enthalpy be ΔH_x. The products are 6CO and 8H₂O. Only CO can be further combusted. ΔH for combustion of 6CO = 6 × ΔH_c^⦵(CO) = 6 × (−283) = −1698 kJ. So, ΔH₂ = ΔH_x + (−1698).
By Hess's Law, ΔH₁ = ΔH₂.
−4440 = ΔH_x − 1698.
ΔH_x = −4440 + 1698 = −2742 kJ.

Question 4

The conversion of graphite to diamond is an endothermic process: C(graphite) → C(diamond). The standard enthalpies of combustion are ΔH_c^⦵(graphite) = −393.5 kJ mol⁻¹ and ΔH_c^⦵(diamond) = −395.4 kJ mol⁻¹. What is the enthalpy change for the graphite → diamond transition?

  1. +788.9 kJ mol⁻¹
  2. −1.9 kJ mol⁻¹
  3. −788.9 kJ mol⁻¹
  4. +1.9 kJ mol⁻¹ (correct answer)
Explanation: We can use Hess's Law. The target reaction is C(gr) → C(dia).
The combustion reactions are:
  1. C(gr) + O₂(g) → CO₂(g) ΔH = −393.5 kJ
  2. C(dia) + O₂(g) → CO₂(g) ΔH = −395.4 kJ
    To get the target equation, we keep equation (1) as is and reverse equation (2).
  3. C(gr) + O₂(g) → CO₂(g) ΔH = −393.5 kJ
    2(rev). CO₂(g) → C(dia) + O₂(g) ΔH = +395.4 kJ
    Adding these two equations, CO₂(g) and O₂(g) cancel out, leaving C(gr) → C(dia).
    The enthalpy change is the sum of the ΔH values: ΔH = −393.5 + 395.4 = +1.9 kJ mol⁻¹.

Question 5

In the Born-Haber cycle for calcium oxide (CaO), the overall process for the second electron affinity of oxygen, O(g) + 2e⁻ → O²⁻(g), is strongly endothermic. Which statement provides the best explanation for this?

  1. Oxygen is a highly electronegative element, which means it requires energy to be forced to gain electrons.
  2. The second electron is being added to a negatively charged ion (O⁻), and energy is required to overcome electrostatic repulsion. (correct answer)
  3. The formation of CaO from its elements is an overall endothermic process, requiring energy input.
  4. The O²⁻ ion is smaller than the O atom, causing electron-electron repulsion that requires energy input.
Explanation: The first electron affinity of oxygen (O + e⁻ → O⁻) is exothermic, as the neutral atom attracts an electron. However, the second electron affinity (O⁻ + e⁻ → O²⁻) is endothermic. This is because the second electron is being added to an ion that is already negatively charged (O⁻). There is a strong electrostatic repulsion between the negative ion and the electron being added, so a significant amount of energy must be supplied to force the electron onto the ion. B is incorrect; high electronegativity corresponds to a favorable (exothermic) first electron affinity. C is incorrect as the formation of CaO is highly exothermic. D is incorrect as the O²⁻ ion is larger than the O atom due to increased electron-electron repulsion and the same nuclear charge.

Question 6

When 0.050 mol of solid X is dissolved in 100 cm³ of water, the temperature rises by 5.0 °C. When 0.050 mol of solid Y is dissolved in 100 cm³ of water, the temperature falls by 3.0 °C. The standard enthalpy change for the solid-state conversion X(s) → Y(s) is +10.0 kJ mol⁻¹. What is the enthalpy of solution for Y, ΔH_sol(Y), in kJ mol⁻¹? (Assume solution specific heat capacity = 4.18 J g⁻¹ °C⁻¹ and density = 1.0 g cm⁻³).

  1. −42 kJ mol⁻¹
  2. +25 kJ mol⁻¹ (correct answer)
  3. −52 kJ mol⁻¹
  4. −25 kJ mol⁻¹
Explanation: For the dissolution of Y, calculate the heat absorbed from the water: q = mcΔT. The mass of water is 100 cm³ × 1.0 g cm⁻³ = 100 g. q = (100 g) × (4.18 J g⁻¹ °C⁻¹) × (3.0 °C) = 1254 J. Since the temperature fell, the dissolution absorbed heat from the water, making it endothermic (ΔH positive). ΔH_sol(Y) = +q/n = +1254 J / 0.050 mol = +25,080 J mol⁻¹ = +25 kJ mol⁻¹.

Question 7

Determine the enthalpy change for the reaction: P₄(s) + 6Cl₂(g) → 4PCl₃(l), using the following thermochemical equations:

  1. P₄(s) + 10Cl₂(g) → 4PCl₅(s) ΔH = −1774 kJ

  2. PCl₃(l) + Cl₂(g) → PCl₅(s) ΔH = −124 kJ

  1. +1278 kJ
  2. −1650 kJ
  3. −2270 kJ
  4. −1278 kJ (correct answer)
Explanation: We use Hess's Law to combine the given equations to form the target equation.
We need P₄(s) on the reactant side, so we use equation 1 as written:
P₄(s) + 10Cl₂(g) → 4PCl₅(s) ΔH = −1774 kJ.
We need 4PCl₃(l) on the product side. Equation 2 has PCl₃(l) as a reactant, so we must reverse it and multiply by 4 to get 4 moles.
4 × [PCl₅(s) → PCl₃(l) + Cl₂(g)] ΔH = 4 × (+124 kJ) = +496 kJ.
Now, add the two manipulated equations:
(P₄(s) + 10Cl₂(g)) + (4PCl₅(s)) → (4PCl₅(s)) + (4PCl₃(l) + 4Cl₂(g))
Cancel species that appear on both sides (4PCl₅(s) and 4Cl₂(g) from the 10Cl₂(g)):
P₄(s) + 6Cl₂(g) → 4PCl₃(l).
Sum the enthalpy changes: ΔH = −1774 kJ + 496 kJ = −1278 kJ.

Question 8

Given the following standard enthalpies of combustion, ΔH_c^⦵, in kJ mol⁻¹: C(graphite): −394 H₂(g): −286 C₂H₅OH(l): −1367 What is the standard enthalpy of formation, ΔH_f^⦵, of liquid ethanol (C₂H₅OH)?

  1. −279 kJ mol⁻¹ (correct answer)
  2. +279 kJ mol⁻¹
  3. −2047 kJ mol⁻¹
  4. −689 kJ mol⁻¹
Explanation: The target equation for the formation of ethanol is: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l). We can construct a Hess's cycle using the combustion data.
  1. Combustion of C: 2C(s) + 2O₂(g) → 2CO₂(g); ΔH = 2 × (−394) = −788 kJ.
  2. Combustion of H₂: 3H₂(g) + 1.5O₂(g) → 3H₂O(l); ΔH = 3 × (−286) = −858 kJ.
  3. Combustion of C₂H₅OH: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l); ΔH = −1367 kJ.
    To get the target equation, we do (1) + (2) - (3).
    ΔH_f^⦵ = (−788) + (−858) − (−1367) = −1646 + 1367 = −279 kJ mol⁻¹.

Question 9

The hydrogenation of ethyne to ethane is exothermic: C₂H₂(g) + 2H₂(g) → C₂H₆(g), with ΔH^⦵ = −314 kJ mol⁻¹. Use the following average bond enthalpies to calculate the C−C bond enthalpy in ethane.
C≡C: 839 kJ mol⁻¹
C−H: 414 kJ mol⁻¹
H−H: 436 kJ mol⁻¹

  1. 369 kJ mol⁻¹ (correct answer)
  2. 346 kJ mol⁻¹
  3. 259 kJ mol⁻¹
  4. 424 kJ mol⁻¹
Explanation: The enthalpy change of reaction is calculated as the sum of energies of bonds broken minus the sum of energies of bonds formed.
Bonds broken in reactants:
1 × C≡C in C₂H₂ = 839 kJ
2 × C−H in C₂H₂ = 2 × 414 = 828 kJ
2 × H−H in 2H₂ = 2 × 436 = 872 kJ
Total energy for bonds broken = 839 + 828 + 872 = 2539 kJ.
Bonds formed in products:
1 × C−C in C₂H₆ = x
6 × C−H in C₂H₆ = 6 × 414 = 2484 kJ
Total energy from bonds formed = x + 2484 kJ.
ΔH = (Bonds Broken) - (Bonds Formed)
−314 = 2539 - (x + 2484)
−314 = 2539 - x - 2484
−314 = 55 - x
x = 55 + 314 = 369 kJ mol⁻¹.

Question 10

The combustion of hydrazine, N₂H₄(l), is used in rocket propulsion: N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(l). Given the standard enthalpies of formation: ΔH_f^⦵(N₂H₄(l)) = +50.6 kJ mol⁻¹ and ΔH_f^⦵(H₂O(l)) = −285.8 kJ mol⁻¹, what is the heat energy released when 16.0 g of liquid hydrazine is completely combusted? (M_r(N₂H₄) = 32.0)

  1. 311 kJ (correct answer)
  2. 622 kJ
  3. 261 kJ
  4. 521 kJ
Explanation: First, calculate the standard enthalpy change of the reaction using ΔH_rxn^⦵ = ΣΔH_f^⦵(products) - ΣΔH_f^⦵(reactants). Remember that the standard enthalpy of formation for elements in their standard state (N₂(g) and O₂(g)) is zero.
ΔH_rxn^⦵ = [ΔH_f^⦵(N₂) + 2 × ΔH_f^⦵(H₂O)] - [ΔH_f^⦵(N₂H₄) + ΔH_f^⦵(O₂)]
ΔH_rxn^⦵ = [0 + 2(−285.8)] - [+50.6 + 0] = −571.6 − 50.6 = −622.2 kJ mol⁻¹.
Next, calculate the moles of hydrazine combusted: n = m/M_r = 16.0 g / 32.0 g mol⁻¹ = 0.500 mol.
Finally, calculate the heat released: Heat = n × |ΔH_rxn^⦵| = 0.500 mol × 622.2 kJ mol⁻¹ = 311.1 kJ. The closest answer is 311 kJ.

Question 11

The dissociation of ozone is exothermic: 2O₃(g) → 3O₂(g). The structure of ozone, O₃, has two equivalent O−O bonds with a bond order of 1.5. The bond enthalpy of the O=O double bond in O₂ is 498 kJ mol⁻¹. What can be deduced about the average bond enthalpy of the O−O bond in an ozone molecule?

  1. It is equal to 498 kJ mol⁻¹.
  2. It is less than 374 kJ mol⁻¹. (correct answer)
  3. It is greater than 498 kJ mol⁻¹.
  4. It is exactly 249 kJ mol⁻¹.
Explanation: For the reaction 2O₃(g) → 3O₂(g), bonds are broken in the reactants (ozone) and formed in the products (oxygen). Let x be the average bond enthalpy in ozone. Each O₃ molecule has two such bonds.
Energy for bonds broken = 2 mol O₃ × (2 bonds/molecule) × x = 4x.
Energy from bonds formed = 3 mol O₂ × (1 bond/molecule) × 498 kJ mol⁻¹ = 1494 kJ.
ΔH = (Bonds Broken) - (Bonds Formed) = 4x - 1494.
Since the reaction is exothermic, ΔH < 0.
4x - 1494 < 0
4x < 1494
x < 1494 / 4
x < 373.5 kJ mol⁻¹.
Therefore, the bond enthalpy in ozone must be less than 374 kJ mol⁻¹.

Question 12

For a certain gas-phase reaction, the total energy required to break all bonds in the reactants is 1500 kJ mol⁻¹, and the total energy released upon forming all bonds in the products is 1800 kJ mol⁻¹. Which statement about this reaction is correct?

  1. The reaction is endothermic, and the products are more stable than the reactants.
  2. The reaction is endothermic, and the products are less stable than the reactants.
  3. The reaction is exothermic, but the reactants are more stable than the products.
  4. The reaction is exothermic, and the products are more stable than the reactants. (correct answer)
Explanation: The enthalpy change of a reaction is calculated as ΔH = Σ(Energy for bonds broken) - Σ(Energy from bonds formed).
ΔH = (+1500 kJ mol⁻¹) - (+1800 kJ mol⁻¹) = -300 kJ mol⁻¹.
Since ΔH is negative, the reaction is exothermic, meaning it releases energy to the surroundings. In an exothermic reaction, the products have lower enthalpy (are more stable) than the reactants. Therefore, the reaction is exothermic and the products are more stable.

Question 13

Calculate the standard enthalpy change for the Ostwald process reaction: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g).
ΔH_f^⦵(NH₃(g)) = −46 kJ mol⁻¹
ΔH_f^⦵(NO(g)) = +90 kJ mol⁻¹
ΔH_f^⦵(H₂O(g)) = −242 kJ mol⁻¹

  1. +908 kJ
  2. −1276 kJ
  3. −908 kJ (correct answer)
  4. −106 kJ
Explanation: The standard enthalpy change of reaction is calculated as ΔH_rxn^⦵ = ΣΔH_f^⦵(products) - ΣΔH_f^⦵(reactants).
Sum of enthalpies of formation for products:
ΣΔH_f^⦵(products) = [4 × ΔH_f^⦵(NO)] + [6 × ΔH_f^⦵(H₂O)] = [4 × (+90)] + [6 × (−242)] = 360 − 1452 = −1092 kJ.
Sum of enthalpies of formation for reactants:
ΣΔH_f^⦵(reactants) = [4 × ΔH_f^⦵(NH₃)] + [5 × ΔH_f^⦵(O₂)] = [4 × (−46)] + [5 × 0] = −184 kJ.
ΔH_rxn^⦵ = (−1092) − (−184) = −1092 + 184 = −908 kJ.

Question 14

The theoretical lattice enthalpy for silver iodide (AgI) calculated using a perfect ionic model is −770 kJ mol⁻¹. The experimental value determined from a Born-Haber cycle is −889 kJ mol⁻¹. What is the most likely reason for this discrepancy?

  1. Experimental errors in measuring atomization enthalpies cause large inaccuracies in the cycle.
  2. The Born-Haber cycle does not account for intermolecular forces between AgI units.
  3. The perfect ionic model overestimates the repulsive forces between the electron clouds.
  4. Silver iodide has significant covalent character, which strengthens the overall bonding. (correct answer)
Explanation: When the experimental lattice enthalpy (from Born-Haber cycle) is more exothermic than the theoretical value (from a pure ionic model), it indicates that the bonding is stronger than predicted by ionic interactions alone. This additional stability is attributed to a degree of covalent character in the bond. The Ag⁺ ion polarizes the large I⁻ ion, leading to electron sharing and a stronger, partially covalent bond. B is incorrect as lattice enthalpy concerns the ionic lattice, not intermolecular forces. C is incorrect as the model underestimates attraction, not overestimates repulsion. D is less likely to be the primary reason for such a large, systematic discrepancy.

Question 15

For the reaction A + B → C, the forward activation energy is 85 kJ mol−1^{-1} and the reverse activation energy is 125 kJ mol−1^{-1}. If a catalyst is introduced that lowers the forward activation energy to 45 kJ mol−1^{-1}, what happens to the equilibrium constant and the reverse activation energy?

  1. KeqK_{eq} decreases; reverse EaE_a becomes 45 kJ mol−1^{-1}
  2. KeqK_{eq} increases; reverse EaE_a becomes 85 kJ mol−1^{-1}
  3. KeqK_{eq} remains unchanged; reverse EaE_a becomes 165 kJ mol−1^{-1}
  4. KeqK_{eq} remains unchanged; reverse EaE_a becomes 85 kJ mol−1^{-1} (correct answer)
Explanation: When you encounter activation energy problems involving catalysts, remember two fundamental principles: catalysts don't change equilibrium positions, and they lower activation energies equally in both directions. First, let's find the enthalpy change for this reaction. Since ΔH=Ea(forward)−Ea(reverse)=85−125=−40\Delta H = E_{a(forward)} - E_{a(reverse)} = 85 - 125 = -40 kJ mol−1^{-1}, this is an exothermic reaction. The equilibrium constant depends only on the enthalpy change and temperature through the relationship ln⁡K=−ΔH/RT\ln K = -\Delta H/RT. Since catalysts don't change ΔH\Delta H, they cannot change KeqK_{eq}. When the catalyst lowers the forward activation energy from 85 to 45 kJ mol−1^{-1} (a decrease of 40 kJ mol−1^{-1}), it must lower the reverse activation energy by the same amount. Therefore: reverse Ea=125−40=85E_a = 125 - 40 = 85 kJ mol−1^{-1}. Option A is wrong because KeqK_{eq} cannot decrease when only a catalyst is added, and the reverse activation energy wouldn't equal the forward value unless ΔH=0\Delta H = 0. Option B incorrectly suggests KeqK_{eq} increases, which again violates the principle that catalysts don't affect equilibrium positions. Option C correctly identifies that KeqK_{eq} remains unchanged but incorrectly adds the activation energy changes instead of maintaining the same ΔH\Delta H. Option D correctly recognizes both principles: KeqK_{eq} stays constant, and the reverse activation energy becomes 85 kJ mol−1^{-1}. Remember: catalysts are "shortcuts" that lower both activation energies equally, leaving the overall energy change—and therefore the equilibrium—completely unchanged.

Question 16

In comparing the lattice enthalpies of NaF, NaCl, MgO, and CaO using Born-Haber cycles, which factor analysis correctly explains the observed trend: MgO > CaO > NaF > NaCl?

  1. Polarizability effects: smaller, more highly charged ions create more covalent character, strengthening the lattice beyond purely electrostatic predictions
  2. Size dominates over charge: smaller ions (Mg2+^{2+}, Na+^+, F−^-, O2−^{2-}) create stronger lattices regardless of charge magnitude
  3. Electronegativity differences determine lattice strength: the greater the difference, the stronger the ionic bonding
  4. Charge dominates over size: MgO and CaO have 2+/2- charges vs 1+/1- for sodium halides, while size differences account for the variations within each pair (correct answer)
Explanation: When analyzing lattice enthalpies using Born-Haber cycles, you need to understand how ionic charge and size affect the electrostatic attraction between ions in a crystal lattice. Lattice enthalpy follows Coulomb's Law: it's proportional to the product of the charges and inversely proportional to the distance between ion centers. The trend MgO > CaO > NaF > NaCl demonstrates that charge effects dominate over size effects. MgO and CaO both involve 2+ and 2- ions, while NaF and NaCl involve 1+ and 1- ions. Since lattice enthalpy depends on the product of charges (q1×q2q_1 \times q_2), the 2+/2- compounds have four times the charge interaction compared to 1+/1- compounds. This massive charge advantage explains why both MgO and CaO exceed both sodium compounds. Within each charge type, size matters: Mg²⁺ is smaller than Ca²⁺, and F⁻ is smaller than Cl⁻, making MgO > CaO and NaF > NaCl respectively. Option A incorrectly emphasizes polarizability and covalent character, which aren't the primary factors here. Option B wrongly suggests size dominates over charge - the data clearly shows charge is more important since MgO vastly exceeds NaF despite similar ion sizes. Option C focuses on electronegativity differences, but lattice enthalpy primarily depends on electrostatic interactions in the formed ionic solid, not the bond formation process. Remember: for lattice enthalpy trends, always check charge first (higher charges win), then consider size effects within the same charge combinations. Charge effects typically outweigh size effects dramatically.

Question 17

Consider two reactions with identical ΔH\Delta H values but different activation energies. Reaction 1 has Ea=60E_a = 60 kJ mol−1^{-1} and Reaction 2 has Ea=90E_a = 90 kJ mol−1^{-1}. At 25°C, if both reactions are subject to the same temperature increase to 75°C, which statement best describes the relative change in their rate constants?

  1. Both reactions will show identical percentage increases in rate constant since ΔH\Delta H values are the same
  2. Reaction 2 will show a larger absolute increase in rate constant due to its higher initial activation energy
  3. Reaction 2 will show a larger percentage increase in rate constant because higher EaE_a reactions are more temperature-sensitive (correct answer)
  4. Reaction 1 will show a larger percentage increase because lower EaE_a reactions respond more dramatically to temperature changes
Explanation: According to the Arrhenius equation, k=Ae−Ea/RTk = Ae^{-E_a/RT}, the percentage change in rate constant with temperature is more dramatic for reactions with higher activation energies. This is because the exponential term e−Ea/RTe^{-E_a/RT} is more sensitive to temperature changes when EaE_a is larger. The ratio k2k1=e(Ea/R)(1/T1−1/T2)\frac{k_2}{k_1} = e^{(E_a/R)(1/T_1 - 1/T_2)} shows that higher EaE_a values lead to larger ratios when temperature increases. Option A is wrong because ΔH\Delta H doesn't directly affect temperature sensitivity. Option B confuses absolute vs. percentage change. Option D reverses the correct relationship.

Question 18

Consider the energy profile diagram for a catalyzed versus uncatalyzed reaction. If the uncatalyzed reaction has ΔH=−50\Delta H = -50 kJ mol−1^{-1} and Ea=120E_a = 120 kJ mol−1^{-1}, and the catalyst lowers the activation energy by 40%, what is the ratio of the rate constant for the catalyzed reaction to the uncatalyzed reaction at 298 K? (Use the approximation that a 10 kJ mol−1^{-1} decrease in EaE_a roughly triples the rate constant at room temperature.)

  1. Approximately 34.8^{4.8} ≈ 140 (correct answer)
  2. Approximately 34.0^{4.0} ≈ 81
  3. Approximately 37.2^{7.2} ≈ 800
  4. Approximately 312^{12} ≈ 531,000
Explanation: The catalyst lowers EaE_a by 40%, so the new activation energy is 120×0.6=72120 \times 0.6 = 72 kJ mol−1^{-1}. The reduction in activation energy is 120−72=48120 - 72 = 48 kJ mol−1^{-1}. Using the given approximation that each 10 kJ mol−1^{-1} decrease triples the rate constant, a 48 kJ mol−1^{-1} decrease corresponds to 48/10=4.848/10 = 4.8 factors of 3. Therefore, the ratio is approximately 34.8≈1403^{4.8} ≈ 140. Option B uses 40 kJ mol−1^{-1} instead of 48. Option C incorrectly uses 72 kJ mol−1^{-1} as the decrease. Option D uses the full 120 kJ mol−1^{-1} as if the catalyst completely eliminated the activation energy.

Question 19

In a Born-Haber cycle for the formation of calcium fluoride (CaF2_2), which energy term contributes most significantly to making the overall lattice formation process energetically favorable despite the high second ionization energy of calcium?

  1. The first ionization energy of calcium being relatively low compared to other alkaline earth metals
  2. The lattice enthalpy of CaF2_2 being exceptionally large due to the 2+ and 1- charge combination (correct answer)
  3. The sublimation enthalpy of calcium being compensated by favorable bond dissociation of F2_2
  4. The electron affinity of fluorine being the most negative of all elements in the periodic table
Explanation: In the Born-Haber cycle for CaF2_2, the lattice enthalpy (energy released when gaseous ions form the solid lattice) is proportional to q1×q2r\frac{q_1 \times q_2}{r} where q1q_1 and q2q_2 are the charges and rr is the distance. The Ca2+^{2+} and F−^- combination gives a charge product of 2, and the small size of F−^- makes the lattice enthalpy very large (around -2630 kJ mol−1^{-1}), which more than compensates for the large second ionization energy of Ca (1145 kJ mol−1^{-1}). Option A is incorrect because Ca's first ionization energy doesn't offset the second. Option C is wrong because F2_2 bond dissociation is actually unfavorable (+158 kJ mol−1^{-1}). Option D is incorrect because while F has the most negative electron affinity, this alone doesn't compensate for all the unfavorable terms.

Question 20

In the Born-Haber cycle for the formation of magnesium chloride (MgCl₂), which of the following statements is correct?

  1. The second ionization energy of magnesium is larger than the first, and both are endothermic processes. (correct answer)
  2. The lattice enthalpy is a positive value representing energy released when gaseous ions form a solid lattice.
  3. The enthalpy of atomization for chlorine is equal to the Cl−Cl bond enthalpy.
  4. The electron affinity of chlorine involves forming a Cl²⁻ ion from a chlorine atom in the gaseous state.
Explanation: A is correct because removing an electron from a neutral atom (first ionization energy) requires energy (endothermic). Removing a second electron from an already positive ion (Mg⁺) requires significantly more energy due to increased electrostatic attraction, and is also endothermic. B is incorrect because energy released corresponds to an exothermic process, which has a negative enthalpy value; lattice enthalpy of formation is negative. C is incorrect; the enthalpy of atomization for chlorine is for the process ½Cl₂(g) → Cl(g), so it is half the Cl−Cl bond enthalpy. D is incorrect as the formation of MgCl₂ involves forming Cl⁻ ions, not Cl²⁻ ions.