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Understand Energy Cycles in Reactions - IB Chemistry

  1. 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?
    Answer: Silver iodide has significant covalent character, which strengthens the overall bonding. 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.
  2. In the Born-Haber cycle for the formation of magnesium chloride (MgCl₂), which of the following statements is correct?
    Answer: The second ionization energy of magnesium is larger than the first, and both are endothermic processes. 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.
  3. Consider the following generalized reactions and their enthalpy changes: 1. A + 2B → C ΔH₁ = x 2. C + B → D ΔH₂ = y What is the enthalpy change for the reaction 2A + 6B → 2D?
    Answer: 2x + 2y To obtain the target equation, 2A + 6B → 2D, we must manipulate the given equations according to Hess's Law. We need 2A, so we multiply the first equation by 2: 2(A + 2B → C), which gives 2A + 4B → 2C with ΔH = 2x. We need 2D, so we multiply the second equation by 2: 2(C + B → D), which gives 2C + 2B → 2D with ΔH = 2y. Now, we add the two new equations together: (2A + 4B) + (2C + 2B) → 2C + 2D. The 2C on both sides cancels out, and the B terms combine: 2A + 6B → 2D. The total enthalpy change is the sum of the enthalpy changes of the manipulated equations: ΔH_total = 2x + 2y.
  4. 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
    Answer: −332 kJ mol⁻¹ 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.
  5. 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)?
    Answer: −279 kJ mol⁻¹ 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.
  6. 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?
    Answer: Reaction 2 will show a larger percentage increase in rate constant because higher EaE_a reactions are more temperature-sensitive 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.
  7. 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?
    Answer: KeqK_{eq} remains unchanged; reverse EaE_a becomes 85 kJ mol−1^{-1} 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.
  8. Hess's Law states that the total enthalpy change for a reaction is independent of the route taken. This law is a direct consequence of which fundamental principle?
    Answer: The Law of Conservation of Energy, because enthalpy is a state function. Hess's Law is a manifestation of the Law of Conservation of Energy (also known as the First Law of Thermodynamics). Enthalpy is a state function, meaning its value depends only on the current state of the system (e.g., temperature, pressure, composition), not on how that state was reached. Therefore, the change in enthalpy (ΔH) between an initial and final state is always the same, regardless of the intermediate steps or pathway taken. B, C, and D are important chemical principles but are not the basis for Hess's Law.
  9. 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).
    Answer: +25 kJ mol⁻¹ 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.
  10. Which process corresponds to the lattice dissociation enthalpy of sodium chloride and which corresponds to the bond dissociation enthalpy of chlorine?
    Answer: NaCl(s) → Na⁺(g) + Cl⁻(g); Cl₂(g) → 2Cl(g) Lattice dissociation enthalpy is the enthalpy change when one mole of a solid ionic compound is broken down into its constituent gaseous ions. For sodium chloride, this is NaCl(s) → Na⁺(g) + Cl⁻(g). Bond dissociation enthalpy is the energy required to break one mole of a specific covalent bond in the gaseous state. For chlorine, this corresponds to breaking the Cl−Cl bond in one mole of Cl₂ molecules: Cl₂(g) → 2Cl(g). Option B shows the enthalpy of atomization for chlorine, not bond dissociation. Option D shows the reverse processes (lattice formation and bond formation).
  11. 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?
    Answer: The reaction is exothermic, and the products are more stable than the reactants. 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.
  12. 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.)
    Answer: Approximately 34.8^{4.8} ≈ 140 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.
  13. 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?
    Answer: Average bond enthalpies are mean values from various molecules and may not match the actual bond energy in CH₄. 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.
  14. 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
    Answer: −908 kJ 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.
  15. 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?
    Answer: +1.9 kJ mol⁻¹ 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). 1. 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.
  16. 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? (MrM_r(N₂H₄) = 32.0)
    Answer: 311 kJ 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/MrM_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.
  17. 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
    Answer: −1278 kJ 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.
  18. 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?
    Answer: The lattice enthalpy of CaF2_2 being exceptionally large due to the 2+ and 1- charge combination 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.
  19. 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
    Answer: −2742 kJ 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 ΔHxH_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 = ΔHxH_x + (−1698). By Hess's Law, ΔH₁ = ΔH₂. −4440 = ΔHxH_x − 1698. ΔHxH_x = −4440 + 1698 = −2742 kJ.
  20. Which pair of ionic compounds is expected to have the most endothermic (most positive) and the least endothermic (least positive) lattice dissociation enthalpies, respectively?
    Answer: Most: MgO; Least: CsI Lattice enthalpy is governed by Coulomb's law, depending on ionic charge and ionic radius. It is most exothermic (or has the most endothermic dissociation enthalpy) for ions with higher charges and smaller radii. MgO consists of Mg²⁺ and O²⁻ ions, which have high charges (+2, -2) and are relatively small. This leads to very strong electrostatic attraction and the largest lattice enthalpy. CsI consists of Cs⁺ and I⁻ ions, which have low charges (+1, -1) and are very large (Cs in Period 6, I in Period 5). This leads to weak electrostatic attraction and the smallest lattice enthalpy. Therefore, MgO has the 'most endothermic' dissociation enthalpy and CsI has the 'least endothermic' dissociation enthalpy.

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