All questions
Question 1
The combustion of hydrazine, N₂H₄(g), is a highly exothermic reaction used in rocket fuel: N₂H₄(g) + O₂(g) → N₂(g) + 2H₂O(g), ΔH = -579 kJ mol⁻¹. Given the average bond enthalpies N-N (160 kJ mol⁻¹), N-H (391 kJ mol⁻¹), O=O (498 kJ mol⁻¹), and O-H (464 kJ mol⁻¹), what is the bond enthalpy of the N≡N bond in gaseous nitrogen?
- 945 kJ mol⁻¹ (correct answer)
- 785 kJ mol⁻¹
- 447 kJ mol⁻¹
- 1873 kJ mol⁻¹
Explanation: The enthalpy change of a reaction can be estimated as the energy required to break bonds minus the energy released when forming bonds. ΔH = Σ(Bonds Broken) - Σ(Bonds Formed). Bonds broken are one N-N, four N-H, and one O=O: (1 × 160) + (4 × 391) + (1 × 498) = 160 + 1564 + 498 = 2222 kJ. Bonds formed are one N≡N and four O-H (from 2 H₂O molecules): (1 × N≡N) + (4 × 464) = N≡N + 1856 kJ. So, -579 = 2222 - (N≡N + 1856). Rearranging gives -579 = 366 - N≡N. Therefore, N≡N = 366 + 579 = 945 kJ mol⁻¹.
Question 2
Given the following thermochemical equations:
C(s) + 2H₂(g) → CH₄(g) ΔH₁
C(s) + O₂(g) → CO₂(g) ΔH₂
H₂(g) + ½O₂(g) → H₂O(l) ΔH₃
What is the correct expression for the standard enthalpy of combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)?
- ΔH₂ + 2ΔH₃ - ΔH₁ (correct answer)
- ΔH₁ + ΔH₂ + 2ΔH₃
- ΔH₁ - ΔH₂ - 2ΔH₃
- ΔH₂ + ΔH₃ - ΔH₁
Explanation: To obtain the target equation, we must manipulate the given equations. The target equation has CH₄ as a reactant, so we must reverse the first equation (changing the sign of ΔH₁ to -ΔH₁). The target has CO₂ as a product, so we use the second equation as is (+ΔH₂). The target has 2H₂O(l) as a product, so we must multiply the third equation by 2 (+2ΔH₃). Summing these gives: (-ΔH₁) + (ΔH₂) + (2ΔH₃), which is ΔH₂ + 2ΔH₃ - ΔH₁.
Question 3
Using the following standard enthalpies of formation, ΔH_f⦵, what is the standard enthalpy change for the hydrogenation of ethyne?
C₂H₂(g) + 2H₂(g) → C₂H₆(g)
ΔH_f⦵(C₂H₂(g)) = +227 kJ mol⁻¹
ΔH_f⦵(C₂H₆(g)) = -85 kJ mol⁻¹
- -312 kJ mol⁻¹ (correct answer)
- +312 kJ mol⁻¹
- -142 kJ mol⁻¹
- +142 kJ mol⁻¹
Explanation: The standard enthalpy change of a reaction is calculated as the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants. ΔH_reaction⦵ = ΣΔH_f⦵(products) - ΣΔH_f⦵(reactants). For this reaction, ΔH_reaction⦵ = [ΔH_f⦵(C₂H₆(g))] - [ΔH_f⦵(C₂H₂(g)) + 2ΔH_f⦵(H₂(g))]. The enthalpy of formation of an element in its standard state, H₂(g), is zero. So, ΔH_reaction⦵ = (-85 kJ mol⁻¹) - (+227 kJ mol⁻¹) = -312 kJ mol⁻¹.
Question 4
Given the following standard enthalpies of combustion, ΔH_c⦵, what is the standard enthalpy of formation for propane, C₃H₈(g), according to the equation 3C(s) + 4H₂(g) → C₃H₈(g)?
ΔH_c⦵(C(graphite)) = -394 kJ mol⁻¹
ΔH_c⦵(H₂(g)) = -286 kJ mol⁻¹
ΔH_c⦵(C₃H₈(g)) = -2220 kJ mol⁻¹
- -106 kJ mol⁻¹ (correct answer)
- +106 kJ mol⁻¹
- -4546 kJ mol⁻¹
- +1540 kJ mol⁻¹
Explanation: When using enthalpies of combustion, the enthalpy change of a reaction is calculated as the sum of the enthalpies of combustion of the reactants minus the sum for the products. ΔH_reaction⦵ = ΣΔH_c⦵(reactants) - ΣΔH_c⦵(products). Here, the reaction is a formation reaction, so ΔH_f⦵(C₃H₈) = [3 × ΔH_c⦵(C) + 4 × ΔH_c⦵(H₂)] - [ΔH_c⦵(C₃H₈)]. ΔH_f⦵(C₃H₈) = [3(-394) + 4(-286)] - [-2220] = [-1182 - 1144] - [-2220] = -2326 + 2220 = -106 kJ mol⁻¹.
Question 5
Ethene, C₂H₄(g), reacts with bromine, Br₂(g), to form 1,2-dibromoethane, C₂H₄Br₂(g). Using the average bond enthalpies below, estimate the enthalpy change for this reaction.
Bond Enthalpies (kJ mol⁻¹): C=C (614), Br-Br (193), C-C (346), C-Br (285).
- -645 kJ
- +109 kJ
- +176 kJ
- -109 kJ (correct answer)
Explanation: In this addition reaction, one C=C bond and one Br-Br bond are broken, while one C-C bond and two C-Br bonds are formed. The four C-H bonds remain unchanged. ΔH = Σ(Bonds Broken) - Σ(Bonds Formed). Bonds Broken = (C=C) + (Br-Br) = 614 + 193 = 807 kJ. Bonds Formed = (C-C) + 2(C-Br) = 346 + 2(285) = 346 + 570 = 916 kJ. ΔH = 807 - 916 = -109 kJ.
Question 6
The incomplete combustion of carbon produces carbon monoxide: C(s) + ½O₂(g) → CO(g). Determine the standard enthalpy change for this reaction using the following data:
Standard enthalpy of combustion of C(s) = -393.5 kJ mol⁻¹
Standard enthalpy of combustion of CO(g) = -283.0 kJ mol⁻¹
- +676.5 kJ
- +110.5 kJ
- -676.5 kJ
- -110.5 kJ (correct answer)
Explanation: We can construct a Hess's cycle. We want ΔH for reaction (T): C(s) + ½O₂(g) → CO(g). We are given (1): C(s) + O₂(g) → CO₂(g), ΔH = -393.5 kJ, and (2): CO(g) + ½O₂(g) → CO₂(g), ΔH = -283.0 kJ. To get the target equation (T), we can use equation (1) and the reverse of equation (2). ΔH(T) = ΔH(1) - ΔH(2) = (-393.5) - (-283.0) = -393.5 + 283.0 = -110.5 kJ.
Question 7
An energy cycle relates the enthalpy of formation of methane (ΔH₁), the enthalpy of atomization of solid carbon (ΔH₂), and the bond enthalpy of H-H (ΔH₃). The cycle shows that C(s) and 2H₂(g) can form C(g) and 4H(g), which then form CH₄(g). Which expression represents the total energy required to break all bonds in one mole of methane, CH₄(g) → C(g) + 4H(g)?
- ΔH₁ - ΔH₂ - 2ΔH₃
- ΔH₁ + ΔH₂ + 2ΔH₃
- ΔH₂ + 2ΔH₃ - ΔH₁ (correct answer)
- ΔH₂ + ΔH₃ - ΔH₁
Explanation: According to Hess's Law, the enthalpy change is independent of the path. Path 1: C(s) + 2H₂(g) → CH₄(g) is ΔH₁. Path 2 involves atomization: C(s) → C(g) is ΔH₂, and 2H₂(g) → 4H(g) is 2ΔH₃. So C(s) + 2H₂(g) → C(g) + 4H(g) has an enthalpy change of ΔH₂ + 2ΔH₃. To complete the cycle, we consider the reverse of the desired reaction: C(g) + 4H(g) → CH₄(g), which is the negative of the total bond energy. So, ΔH₁ = (ΔH₂ + 2ΔH₃) - ΔH_atom(CH₄). Rearranging for the atomization energy of methane gives ΔH_atom(CH₄) = ΔH₂ + 2ΔH₃ - ΔH₁.
Question 8
A key step in the Ostwald process for producing nitric acid is the oxidation of ammonia:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g), ΔH⦵ = -905 kJ.
Given ΔH_f⦵(NH₃(g)) = -46 kJ mol⁻¹ and ΔH_f⦵(H₂O(g)) = -242 kJ mol⁻¹, what is the standard enthalpy of formation of nitrogen monoxide, NO(g), in kJ mol⁻¹?
- +363
- -90.8
- +90.8 (correct answer)
- +182.8
Explanation: Use the formula ΔH⦵_reaction = ΣΔH_f⦵(products) - ΣΔH_f⦵(reactants). Let x = ΔH_f⦵(NO). -905 = [4x + 6(-242)] - [4(-46) + 5(0)]. -905 = [4x - 1452] - [-184]. -905 = 4x - 1452 + 184. -905 = 4x - 1268. 4x = 1268 - 905 = 363. x = 363 / 4 = +90.75. The closest answer is +90.8.
Question 9
Using the data below, what is the lattice enthalpy (for the dissociation of the solid into gaseous ions) of potassium oxide, K₂O, in kJ mol⁻¹?
Enthalpy of formation of K₂O(s): -361
Enthalpy of atomization of K(s): +89
First ionization energy of K(g): +419
Enthalpy of atomization of O(g) from ½O₂(g): +249
First electron affinity of O(g): -141
Second electron affinity of O(g): +798
- +2283 (correct answer)
- -2283
- -1561
- -969
Explanation: According to Hess's Law, the enthalpy of formation is the sum of all other steps in the Born-Haber cycle. ΔH_f = 2(ΔH_atom K) + 2(IE₁ K) + ΔH_atom O + EA₁ O + EA₂ O + ΔH_Lattice_formation. Lattice enthalpy of dissociation is -ΔH_Lattice_formation. So, -ΔH_Lattice_dissociation = ΔH_f - [2(ΔH_atom K) + 2(IE₁ K) + ΔH_atom O + EA₁ O + EA₂ O]. -ΔH_Lattice_dissociation = -361 - [2(89) + 2(419) + 249 + (-141) + 798] = -361 - [178 + 838 + 249 - 141 + 798] = -361 - [1922] = -2283. Therefore, the lattice enthalpy of dissociation is +2283 kJ mol⁻¹.
Question 10
The standard enthalpy of combustion of graphite is -393.5 kJ mol⁻¹ and that of diamond is -395.4 kJ mol⁻¹. What is the enthalpy change for the allotropic conversion of one mole of diamond to graphite?
C(diamond, s) → C(graphite, s)
- -788.9 kJ
- +1.9 kJ
- -1.9 kJ (correct answer)
- +788.9 kJ
Explanation: We can use Hess's Law. We are given:
(1) C(graphite) + O₂(g) → CO₂(g) ΔH = -393.5 kJ
(2) C(diamond) + O₂(g) → CO₂(g) ΔH = -395.4 kJ
To find the enthalpy change for C(diamond) → C(graphite), we can take reaction (2) as it is and add the reverse of reaction (1).
C(diamond) + O₂(g) → CO₂(g) ΔH = -395.4 kJ
CO₂(g) → C(graphite) + O₂(g) ΔH = +393.5 kJ
Summing these two equations gives: C(diamond) → C(graphite), and the enthalpy change is -395.4 + 393.5 = -1.9 kJ.
Question 11
The enthalpy change for H₂(g) + Cl₂(g) → 2HCl(g) calculated using average bond enthalpies is -185 kJ. The experimentally determined value is -184 kJ. What is the most likely reason for this very close agreement?
- The stoichiometry of the reaction is simple, which reduces the potential for calculation errors that cause discrepancies in more complex reactions.
- The bond enthalpies of diatomic elements like H₂ and Cl₂ are defined as the standard for the average bond enthalpy table.
- Reactions involving only single bonds, like this one, are known to have theoretical values that are nearly identical to experimental results.
- The reaction occurs entirely in the gas phase with simple molecules, closely matching the conditions for which average bond enthalpies are defined and derived. (correct answer)
Explanation: Average bond enthalpy calculations are based on a model of breaking and forming bonds between gaseous species. The high level of agreement occurs because this specific reaction involves simple diatomic molecules entirely in the gas phase, a situation that aligns very well with the idealized model. Discrepancies are often larger for reactions involving liquids, solids, or complex molecules where bond energies are more influenced by neighboring atoms and intermolecular forces.
Question 12
The standard enthalpy of combustion for methane (CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)) is experimentally determined to be -890 kJ mol⁻¹. A calculation using average bond enthalpies yields a value of -818 kJ mol⁻¹. Which statement best explains this discrepancy?
- Average bond enthalpies are derived from a range of compounds and may not reflect the specific bond energies in methane and its combustion products. (correct answer)
- The calculation does not account for the energy required to change reactants from the standard state to the gaseous state before bonds are broken.
- Hess's Law applies to standard enthalpies of formation but cannot be accurately applied using average bond enthalpies.
- Experimental calorimetry has inherent inaccuracies, while theoretical calculations from bond enthalpies are precise.
Explanation: The main reason for the discrepancy is that average bond enthalpies are averaged values from many different molecules. The actual energy of a C-H bond in methane or an O-H bond in water will be slightly different from this average value. The calculation is an approximation based on a model. The other options are incorrect: reactants are already gaseous in this equation, Hess's Law is the principle underlying the bond enthalpy calculation, and the experimental value is the true value while the calculation is the approximation.
Question 13
The standard enthalpy change for the combustion of carbon monoxide, 2CO(g) + O₂(g) → 2CO₂(g), is -566 kJ. The standard enthalpy of formation of carbon dioxide, CO₂(g), is -394 kJ mol⁻¹. What is the standard enthalpy of formation, ΔH_f⦵, of carbon monoxide, CO(g)?
- +111 kJ mol⁻¹
- -111 kJ mol⁻¹ (correct answer)
- -222 kJ mol⁻¹
- -677 kJ mol⁻¹
Explanation: Using ΔH_reaction⦵ = ΣΔH_f⦵(products) - ΣΔH_f⦵(reactants): -566 kJ = [2 × ΔH_f⦵(CO₂)] - [2 × ΔH_f⦵(CO) + ΔH_f⦵(O₂)]. Let x = ΔH_f⦵(CO). The ΔH_f⦵ of O₂(g) is 0. So, -566 = [2 × (-394)] - [2x + 0]. This simplifies to -566 = -788 - 2x. Rearranging the equation: 2x = -788 + 566 = -222. Therefore, x = -111 kJ mol⁻¹.
Question 14
In a thermochemical cycle, XCl2(s) dissolves according to: XCl2(s)→X2+(aq)+2Cl−(aq) with ΔHsolution=+15 kJ mol−1. If the hydration enthalpy of X2+ is −2176 kJ mol−1 and of Cl− is −364 kJ mol−1, and the Born-Landé equation predicts a lattice energy of 2850 kJ mol−1, what can be concluded about the ionic character of XCl2?
- The experimental lattice energy is 2889 kJ mol−1, suggesting 98.6% ionic character with minimal covalent contribution
- The experimental lattice energy is 2919 kJ mol−1, indicating the compound is more ionic than predicted theoretically (correct answer)
- The experimental lattice energy is 2719 kJ mol−1, showing significant covalent character reduces the lattice energy by 4.6%
- The solution process is too endothermic to determine lattice energy reliably using the given hydration enthalpy values
Explanation: Using ΔHsolution=−ΔHlattice+ΔHhydration: +15=−ΔHlattice+(−2176+2(−364)). So +15=−ΔHlattice+(−2176−728)=−ΔHlattice−2904. Therefore ΔHlattice=2904+15=2919 kJ mol−1. This is higher than the theoretical prediction of 2850, suggesting the compound is more ionic than expected, possibly due to better crystal packing or stronger electrostatic interactions than modeled. Choice A has a calculation error. Choice C incorrectly suggests lower experimental value. Choice D incorrectly dismisses the validity of the thermochemical method. Question 15
Consider the following energy cycle for the formation of Al2O3: Step 1: 2Al(s)→2Al(g) ΔH=+652 kJ; Step 2: 23O2(g)→3O(g) ΔH=+747 kJ; Step 3: 2Al(g)→2Al3+(g)+6e− ΔH=+11340 kJ; Step 4: 3O(g)+6e−→3O2−(g) ΔH=+2196 kJ. If ΔHf[Al2O3]=−1676 kJ mol−1, why is Step 4 endothermic despite oxygen gaining electrons?
- The second electron affinity of oxygen is highly endothermic, outweighing the exothermic first electron affinity for each atom (correct answer)
- The electron-electron repulsion in O2− makes it thermodynamically unstable compared to O− ions in the gas phase
- The lattice energy calculation requires positive values for all steps to maintain mathematical consistency in Hess's law
- Oxygen atoms in the gas phase have insufficient kinetic energy to accommodate additional electrons without external energy input
Explanation: Step 4 involves 3×[O(g)+e−→O−(g)]+3×[O−(g)+e−→O2−(g)]. The first electron affinity is exothermic (-141 kJ/mol), but the second electron affinity is highly endothermic (+878 kJ/mol) due to electron-electron repulsion. Net: 3×(−141+878)=3×737=+2211 kJ, close to given +2196. Choice B describes instability but doesn't explain the thermodynamics. Choice C incorrectly describes Hess's law requirements. Choice D gives an incorrect physical explanation. Question 16
Which statement provides the fundamental reason why Hess's Law is a valid method for calculating enthalpy changes?
- The total energy and mass in a closed system remain constant during a chemical reaction.
- Enthalpy is a state function, so the change in enthalpy is independent of the pathway between initial and final states. (correct answer)
- Energy released during bond formation is always greater than the energy absorbed during bond breaking for spontaneous reactions.
- Every chemical reaction has an equal and opposite reverse reaction, allowing for algebraic manipulation of reaction enthalpies.
Explanation: Hess's Law is a direct consequence of the fact that enthalpy is a state function. A state function is a property of a system that depends only on its current state, not on the path taken to reach that state. Therefore, the total enthalpy change for a reaction is the same whether it occurs in one step or multiple steps.
Question 17
Which compound is expected to have the lattice enthalpy with the largest magnitude?
- LiF
- NaCl
- KBr
- MgO (correct answer)
Explanation: Lattice enthalpy is principally determined by the charges on the ions and the distance between them (ionic radii), according to Coulomb's Law. The magnitude is proportional to (Q₁Q₂)/r. The charge product for MgO (Mg²⁺ and O²⁻) is (+2) × (-2) = -4. For LiF, NaCl, and KBr, the charge product is (+1) × (-1) = -1. The much larger product of charges in MgO results in a significantly stronger electrostatic attraction and therefore a lattice enthalpy of much greater magnitude.
Question 18
In the Born-Haber cycle for the formation of solid magnesium chloride (MgCl₂), which process corresponds to the second ionization energy of magnesium?
- Mg(g) → Mg⁺(g) + e⁻
- Mg⁺(g) → Mg²⁺(g) + e⁻ (correct answer)
- Mg(s) → Mg(g)
- Cl₂(g) → 2Cl(g)
Explanation: The second ionization energy is the energy required to remove a second electron from a gaseous ion. Mg⁺(g) → Mg²⁺(g) + e⁻ represents the removal of an electron from a gaseous Mg⁺ ion to form a gaseous Mg²⁺ ion. Choice A is the first ionization energy. Choice C is the enthalpy of atomization of magnesium. Choice D is the bond dissociation enthalpy of chlorine.
Question 19
Consider the Born-Haber cycles for the formation of sodium fluoride (NaF) and magnesium oxide (MgO). Which statement correctly compares the energy terms involved?
- The magnitude of the lattice enthalpy of NaF is greater than that of MgO.
- The second ionization energy of magnesium is significantly larger than the first ionization energy of sodium. (correct answer)
- The overall electron affinity for forming O²⁻(g) from O(g) is a large negative value (exothermic).
- The enthalpy of atomization of sodium is greater than that of magnesium.
Explanation: Removing a second electron (from Mg⁺) is much harder than removing the first (from Mg) due to the increased net positive charge. It is also much larger than removing the first electron from a neutral Na atom. Therefore, IE₂(Mg) >> IE₁(Na). B is incorrect; MgO has a much larger lattice enthalpy due to the +2 and -2 charges on its ions. C is incorrect; the second electron affinity is highly endothermic. D is incorrect; magnesium has stronger metallic bonding and a higher enthalpy of atomization than sodium.
Question 20
The magnitude of the lattice enthalpy for the alkali metal chlorides, MCl, decreases progressively from LiCl to CsCl. What is the primary reason for this trend?
- The electronegativity of the alkali metal decreases, reducing the degree of ionic character in the M-Cl bond.
- The first ionization energy of the alkali metal decreases, making it easier to form the cation, which destabilizes the lattice.
- The ionic radius of the alkali metal cation increases, increasing the internuclear distance and weakening the electrostatic forces. (correct answer)
- The mass of the alkali metal cation increases, leading to weaker vibrational energies within the crystal lattice.
Explanation: Lattice enthalpy is governed by Coulomb's law, where the electrostatic energy is inversely proportional to the distance between the ion centers (r). Moving down Group 1, the ionic radius of the M⁺ cation increases (Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺). This increased distance between the M⁺ and Cl⁻ ions weakens the electrostatic attraction, thus decreasing the magnitude of the lattice enthalpy. While ionization energy (B) also shows a trend, it is a factor in the overall enthalpy of formation, not the lattice enthalpy itself.