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201. In chemistry, “standard conditions” for thermodynamic data typically refer to which set?
ⓐ. 273.15 K, 1 atm, 1 M
ⓑ. 310 K, 1 bar, 1 M
ⓒ. 298 K, 1 bar, 1 m
ⓓ. 298 K, 1 atm, 1 M
Correct Answer: 298 K, 1 atm, 1 M
Explanation: Many general-chemistry sources use 298 K (25°C), 1 atm pressure, and 1 M solute concentration as standard conditions for reporting thermodynamic quantities like ΔH°, ΔG°, and E°. This creates a consistent reference for comparing reactions. Some conventions now adopt 1 bar instead of 1 atm, but if a problem states 1 atm explicitly, that convention should be followed to avoid mixing standards.
202. What does the superscript “°” in ΔH° or ΔG° signify?
ⓐ. Measured at constant volume only
ⓑ. Measured at 0 K and 1 atm
ⓒ. Measured under the specified standard conditions
ⓓ. Measured for pure solids only
Correct Answer: Measured under the specified standard conditions
Explanation: The “°” indicates a quantity evaluated in the standard state. For this problem set, the standard state is defined as 298 K, 1 atm pressure, and 1 M concentration for solutes. Using a standard reference removes ambiguity when comparing thermodynamic data across different labs or textbooks. It does not restrict the phase to solids; it applies to gases, liquids, solutions, and solids.
203. Under standard conditions, what is the activity assumed for a 1 M ideal solute?
ⓐ. 0
ⓑ. 0.5
ⓒ. 1
ⓓ. Depends on ionic strength only
Correct Answer: 1
Explanation: Activities are dimensionless effective concentrations used in rigorous thermodynamics. By convention, the activity of a 1 M ideal solute at the standard state is set to 1, simplifying many equations such as equilibrium constants and Nernst relationships. Real solutions deviate from ideality; activity coefficients correct for non-ideal behavior but the standard reference remains activity = 1.
204. Which statement about standard state for a gas is correct in this context?
ⓐ. Any gas at 0.1 atm is standard
ⓑ. A gas at 1 atm behaves ideally by definition
ⓒ. A gas at its partial pressure equals its activity
ⓓ. The standard state is the pure gas at 1 atm and 298 K
Correct Answer: The standard state is the pure gas at 1 atm and 298 K
Explanation: The gas standard state is chosen as the hypothetical ideal gas at 1 atm and the specified temperature (here 298 K). This allows us to assign activities and derive thermodynamic relations consistently. While real gases may deviate from ideality, referencing the pure gas at 1 atm provides a clear baseline for tabulated ΔH°, ΔG°, and S° values.
205. What is the standard enthalpy of formation of an element in its most stable form at 298 K and 1 atm?
ⓐ. Always positive
ⓑ. Always negative
ⓒ. Equal to its molar enthalpy
ⓓ. Defined as zero
Correct Answer: Defined as zero
Explanation: By convention, elements in their thermodynamically most stable form at standard conditions have ΔH°f = 0. Examples include O2(g), N2(g), H2(g), graphite for carbon, and metallic forms for many metals. This choice sets a reference point so that compound formation enthalpies are measured relative to elemental baselines, simplifying Hess’s law calculations.
206. Which pair is correctly matched to standard conditions in this set?
ⓐ. STP: 298 K and 1 atm
ⓑ. Standard solutions: 1 M concentration at 298 K
ⓒ. Standard gas volume: 24.465 L at 273.15 K
ⓓ. Standard electrode potential: defined at 2 M concentration
Correct Answer: Standard solutions: 1 M concentration at 298 K
Explanation: For solution thermodynamics, the standard state uses a reference molarity, here 1 M at 298 K. STP is a separate convention typically involving 273.15 K and 1 atm (or 1 bar in some definitions). Standard electrode potentials are referenced to unit activity, which corresponds to 1 M for solutes, not 2 M. Gas molar volumes depend on the exact definition of STP used.
207. A standard Gibbs energy change ΔG° is most directly related to which equilibrium quantity at 298 K?
ⓐ. Reaction quotient Q
ⓑ. Standard enthalpy ΔH° only
ⓒ. Equilibrium constant K
ⓓ. Heat capacity Cp
Correct Answer: Equilibrium constant K
Explanation: The relationship is ΔG° = −RT ln K, evaluated at the standard temperature with standard-state activities. When ΔG° is negative, K > 1, favoring products at equilibrium; when positive, K < 1. This link is central to predicting spontaneity and composition under standard conditions. Q is the instantaneous ratio, while K is the equilibrium value at the specified T.
208. Why are standard conditions specified when tabulating thermodynamic data?
ⓐ. To force all reactions to be spontaneous
ⓑ. To eliminate temperature dependence from all properties
ⓒ. To provide a common reference for comparison among reactions
ⓓ. To guarantee ideal behavior of real systems
Correct Answer: To provide a common reference for comparison among reactions
Explanation: Standard conditions define a uniform baseline so data measured at different places and times can be compared meaningfully. They do not ensure ideality or spontaneity; they simply state the reference pressure, temperature, and concentration. From this baseline, corrections to nonstandard conditions can be made using equations such as ΔG = ΔG° + RT ln Q.
209. For a solute A in solution, the standard chemical potential μ°A corresponds to which condition?
ⓐ. A at any concentration and temperature
ⓑ. A at saturation concentration only
ⓒ. A at infinite dilution with γA = 0
ⓓ. A at 1 M and 298 K with activity defined as 1
Correct Answer: A at 1 M and 298 K with activity defined as 1
Explanation: The correct description is that μ°A corresponds to the chemical potential of A in its standard state: a hypothetical ideal 1 M solution at 298 K with activity set to 1.
210. Which statement correctly distinguishes standard conditions from STP in many textbooks?
ⓐ. Both use 298 K and 1 atm
ⓑ. STP requires 1 M concentration for solutions
ⓒ. Standard conditions use 273.15 K; STP uses 298 K
ⓓ. Standard conditions use 298 K and 1 atm; STP often uses 273.15 K and 1 atm
Correct Answer: Standard conditions use 298 K and 1 atm; STP often uses 273.15 K and 1 atm
Explanation: This option is incorrect as written; the accurate distinction is that standard conditions in this problem set are 298 K, 1 atm, and 1 M, whereas STP is commonly 273.15 K and 1 atm (or 1 bar under some conventions).
211. What is meant by the standard enthalpy of a reaction (ΔH°rxn)?
ⓐ. The heat absorbed or evolved when reactants and products are at any condition
ⓑ. The heat change measured at constant volume and temperature
ⓒ. The total internal energy of the products only
ⓓ. The heat change when one mole of reaction occurs under standard conditions
Correct Answer: The heat change when one mole of reaction occurs under standard conditions
Explanation: The standard enthalpy of a reaction, ΔH°rxn, is defined as the enthalpy change when all reactants and products are in their standard states (298 K, 1 atm, 1 M). It reflects the energy absorbed or released at constant pressure. This value allows chemists to compare reactions under a uniform set of reference conditions, independent of experimental setup.
212. The standard enthalpy of reaction can be expressed as:
ⓐ. \(\Delta H^\circ_{\text{rxn}} = \sum \Delta H^\circ_{\text{f (reactants)}} – \sum \Delta H^\circ_{\text{f (products)}}\)
ⓑ. \(\Delta H^\circ_{\text{rxn}} = \Delta S \times T\) )
ⓒ. \(\Delta H^\circ_{\text{rxn}} = \Delta U – P\Delta V\)
ⓓ. \(\Delta H^\circ_{\text{rxn}} = \sum \Delta H^\circ_{\text{f (products)}} – \sum \Delta H^\circ_{\text{f (reactants)}}\)
Correct Answer: \(\Delta H^\circ_{\text{rxn}} = \sum \Delta H^\circ_{\text{f (products)}} – \sum \Delta H^\circ_{\text{f (reactants)}}\)
Explanation: The enthalpy of a chemical reaction equals the difference between the total standard enthalpies of formation of products and reactants, each multiplied by their stoichiometric coefficients. This is derived from Hess’s law, which states that total enthalpy change depends only on initial and final states, not the reaction path.
213. Which condition must be true for ΔH°rxn to represent the standard enthalpy change?
ⓐ. The pressure is 1 atm and all substances are in their standard states
ⓑ. The temperature must be 0°C
ⓒ. The volume must be constant during the process
ⓓ. The concentration of all solutions must be 0.1 M
Correct Answer: The pressure is 1 atm and all substances are in their standard states
Explanation: Standard enthalpy changes are measured when reactants and products are in their standard states (pure solids or liquids, 1 atm gases, and 1 M solutions) at 298 K. These controlled conditions ensure consistency and comparability between tabulated values. The temperature need not be 0°C, and the process is assumed to occur at constant pressure, not volume.
214. What is the standard enthalpy of formation (ΔH°f) of an element in its most stable form?
ⓐ. +100 kJ mol⁻¹
ⓑ. –273 kJ mol⁻¹
ⓒ. Zero
ⓓ. Depends on the element
Correct Answer: Zero
Explanation: By convention, the standard enthalpy of formation for an element in its most stable form (like O₂(g), N₂(g), graphite for C, H₂(g)) is zero at 298 K and 1 atm. This provides a consistent reference so that reaction enthalpies can be determined relative to these baseline states using Hess’s law.
215. Consider the reaction: \(2H_2(g) + O_2(g) → 2H_2O(l\). The standard enthalpy of formation of water is –286 kJ mol⁻¹. What is ΔH°rxn?
ⓐ. –286 kJ
ⓑ. –572 kJ
ⓒ. +572 kJ
ⓓ. +286 kJ
Correct Answer: –572 kJ
Explanation: Since two moles of water are formed, total enthalpy change \(ΔH°_{\text{rxn}} = 2 × (–286) = –572 , \text{kJ}\). The negative sign indicates an exothermic reaction where energy is released as heat. Standard enthalpy of reaction always refers to the stoichiometric quantities specified in the balanced equation.
216. Which of the following reactions has ΔH° = 0 by definition?
ⓐ. \(H_2(g) + \tfrac{1}{2}O_2(g) → H_2O(l\)
ⓑ. \(O_2(g) → 2O(g\)
ⓒ. ( C(graphite) → C(diamond) )
ⓓ. \(O_2(g) → O_2(g\)
Correct Answer: \(O_2(g) → O_2(g\)
Explanation: A reaction that does not change the chemical state or composition of the system has ΔH° = 0. In this case, both reactants and products are identical (oxygen gas in its standard state), so no heat is absorbed or released. Enthalpy depends only on state, so if initial and final states are identical, the change is zero.
217. The standard enthalpy of combustion is defined as:
ⓐ. The enthalpy change when elements combine to form one mole of compound
ⓑ. The heat change when one mole of oxygen reacts with hydrogen
ⓒ. The enthalpy required to decompose one mole of compound into elements
ⓓ. The heat change when one mole of a substance reacts completely with oxygen at standard conditions
Correct Answer: The heat change when one mole of a substance reacts completely with oxygen at standard conditions
Explanation: The standard enthalpy of combustion represents the heat released when one mole of a substance is burned in excess oxygen under standard conditions (298 K, 1 atm). Combustion enthalpies are typically negative, indicating exothermic reactions, and are widely used to calculate fuel efficiency and calorific values.
218. Which of the following statements about standard enthalpy of reaction is correct?
ⓐ. It depends on the path of reaction
ⓑ. It is measured at constant volume
ⓒ. It is always positive
ⓓ. It depends only on the initial and final states of reactants and products
Correct Answer: It depends only on the initial and final states of reactants and products
Explanation: Enthalpy is a state function, meaning its change depends solely on the starting and ending states of the system. Whether a reaction proceeds directly or through intermediate steps, the overall ΔH°rxn remains the same. This property allows enthalpy values to be determined indirectly using Hess’s law even when a reaction cannot be measured directly.
219. If the standard enthalpy of formation of NH₃(g) is –46 kJ mol⁻¹, what is ΔH° for the reaction \(N_2(g) + 3H_2(g) → 2NH_3(g\)?
ⓐ. –46 kJ
ⓑ. –92 kJ
ⓒ. +46 kJ
ⓓ. +92 kJ
Correct Answer: –92 kJ
Explanation: Since two moles of NH₃ are formed, \(ΔH°_{\text{rxn}} = 2 × (–46) = –92 , \text{kJ}\). The negative value indicates the process is exothermic. Formation of ammonia releases energy, which is why high pressures are used industrially in the Haber process to favor product formation according to Le Chatelier’s principle.
220. Why are standard enthalpies of reaction important in chemistry?
ⓐ. They predict the color of compounds
ⓑ. They help compare reaction rates
ⓒ. They depend on catalyst presence
ⓓ. They determine the energy changes accompanying chemical reactions under fixed conditions
Correct Answer: They determine the energy changes accompanying chemical reactions under fixed conditions
Explanation: Standard enthalpies provide a quantitative measure of the heat absorbed or released during a reaction under well-defined conditions (298 K, 1 atm, 1 M). These data are essential for thermodynamic calculations, energy efficiency assessments, and determining reaction feasibility. Catalysts may alter reaction rate but not ΔH°, since enthalpy depends only on states, not pathway.
221. What is meant by the heat of combustion of a substance?
ⓐ. The heat absorbed when one mole of substance is formed from its elements
ⓑ. The heat evolved when one mole of a substance is completely burnt in oxygen under standard conditions
ⓒ. The heat absorbed when a substance decomposes in air
ⓓ. The heat required to evaporate a liquid fuel
Correct Answer: The heat evolved when one mole of a substance is completely burnt in oxygen under standard conditions
Explanation: The heat (enthalpy) of combustion is the energy released when one mole of a compound reacts completely with oxygen at 298 K and 1 atm. It is usually exothermic, producing CO₂ and H₂O (and sometimes other oxides). Examples include the combustion of methane, ethanol, or benzene. These values are vital in calculating fuel efficiency and energy content of materials.
222. Which of the following represents the correct equation for the standard heat of combustion of methane?
ⓐ. \(CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l\)
ⓑ. \(CH_4(g) \rightarrow C(s) + 2H_2(g\)
ⓒ. \(CH_4(l) + 2O_2(l) \rightarrow CO_2(g) + 2H_2O(g\)
ⓓ. \(CH_4(g) + H_2O(l) \rightarrow CO_2(g) + 3H_2(g\)
Correct Answer: \(CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l\)
Explanation: The combustion reaction for methane involves complete oxidation of one mole of CH₄ to CO₂ and H₂O. The standard enthalpy of combustion (ΔH°comb) for this reaction is –890 kJ mol⁻¹. The negative sign indicates that energy is released as heat to the surroundings.
223. The standard enthalpy of combustion of hydrogen is:
ⓐ. +286 kJ mol⁻¹
ⓑ. –286 kJ mol⁻¹
ⓒ. +572 kJ mol⁻¹
ⓓ. –572 kJ mol⁻¹
Correct Answer: –286 kJ mol⁻¹
Explanation: The combustion of hydrogen gas, \(2H_2(g) + O_2(g) → 2H_2O(l\), releases 572 kJ of heat for two moles of water, or –286 kJ per mole. The negative sign indicates an exothermic process. Hydrogen’s high combustion enthalpy makes it a potential clean fuel, producing only water as the combustion product.
224. The standard heat of combustion of ethane (C₂H₆) is –1560 kJ mol⁻¹. What does the negative sign indicate?
ⓐ. Heat is absorbed from surroundings
ⓑ. Reaction is endothermic
ⓒ. Reaction occurs only at high temperature
ⓓ. Heat is released to surroundings
Correct Answer: Heat is released to surroundings
Explanation: A negative enthalpy value indicates an exothermic process. During the combustion of ethane, chemical energy in C–H and C–C bonds converts to heat as CO₂ and H₂O form. The reaction releases –1560 kJ per mole of ethane burned, demonstrating energy liberation typical of hydrocarbon fuels.
225. Which of the following substances will have the highest heat of combustion per mole?
ⓐ. Hydrogen
ⓑ. Methane
ⓒ. Ethanol
ⓓ. Octane
Correct Answer: Octane
Explanation: The heat of combustion generally increases with molecular complexity and the number of C–H bonds available for oxidation. Octane (C₈H₁₈) has more combustible bonds, releasing more energy per mole. However, per gram, hydrogen releases the most energy, as its molar mass is much smaller than that of hydrocarbons.
226. What is meant by the heat of neutralization?
ⓐ. The heat absorbed when one mole of acid reacts with base at any condition
ⓑ. The heat evolved when one mole of water is formed by reaction between an acid and a base under standard conditions
ⓒ. The heat absorbed during neutralization of a salt
ⓓ. The total enthalpy change of any chemical reaction
Correct Answer: The heat evolved when one mole of water is formed by reaction between an acid and a base under standard conditions
Explanation: Heat of neutralization is defined as the enthalpy change accompanying the formation of one mole of water from the reaction of an acid and a base at 298 K, 1 atm, and in dilute aqueous solution. For strong acid–strong base reactions, it is almost constant (–57.1 kJ mol⁻¹), because both fully ionize and form water via \(H^+ + OH^- \rightarrow H_2O\).
227. For the reaction \(HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l\), the heat of neutralization is approximately:
ⓐ. +57.1 kJ mol⁻¹
ⓑ. –57.1 kJ mol⁻¹
ⓒ. –28.6 kJ mol⁻¹
ⓓ. +28.6 kJ mol⁻¹
Correct Answer: –57.1 kJ mol⁻¹
Explanation: The neutralization of a strong acid by a strong base releases 57.1 kJ per mole of water formed. The process is exothermic because the recombination of \(H^+\) and \(OH^-\) ions forms a strong covalent O–H bond in water, releasing significant energy.
228. For the neutralization between a weak acid and strong base, the heat of neutralization is less than –57.1 kJ mol⁻¹ because:
ⓐ. Weak acids react slower
ⓑ. Temperature affects the solubility
ⓒ. The base becomes less reactive in water
ⓓ. The weak acid does not ionize completely and requires energy for dissociation
Correct Answer: The weak acid does not ionize completely and requires energy for dissociation
Explanation: When a weak acid (like CH₃COOH) neutralizes a strong base (NaOH), part of the released energy is used to ionize the weak acid before it reacts. Hence, the net heat evolved is less than for strong acid–strong base systems. The same applies when a weak base reacts with a strong acid.
229. Which of the following equations represents a heat of neutralization reaction?
ⓐ. \(CH_4 + 2O_2 → CO_2 + 2H_2O\)
ⓑ. \(N_2 + 3H_2 → 2NH_3\)
ⓒ. \(HNO_3(aq) + KOH(aq) → KNO_3(aq) + H_2O(l\)
ⓓ. \(C(s) + O_2 → CO_2(g\)
Correct Answer: \(HNO_3(aq) + KOH(aq) → KNO_3(aq) + H_2O(l\)
Explanation: This is a typical acid–base neutralization reaction where nitric acid reacts with potassium hydroxide to produce potassium nitrate and water. The process releases approximately –57 kJ mol⁻¹ per mole of water formed, consistent with the standard enthalpy of neutralization for strong acid–strong base systems.
230. Why is the heat of neutralization nearly constant for all strong acid–strong base reactions?
ⓐ. All acids and bases have equal dissociation energies
ⓑ. The reaction always produces the same number of moles of salt
ⓒ. The reaction is essentially \(H^+ + OH^- \rightarrow H_2O\) in every case
ⓓ. The enthalpy depends on the type of salt formed
Correct Answer: The reaction is essentially \(H^+ + OH^- \rightarrow H_2O\) in every case
Explanation: Strong acids and strong bases are completely ionized in aqueous solution, so the neutralization reaction is identical for all: the combination of hydrogen and hydroxide ions to form liquid water. Since this process always releases the same energy per mole of water formed (–57.1 kJ mol⁻¹), the heat of neutralization remains nearly constant for all strong acid–base combinations.
231. What is meant by the standard heat of formation (ΔH°f)?
ⓐ. The heat evolved when one mole of an element forms from its atoms
ⓑ. The heat released during complete combustion of a compound
ⓒ. The heat absorbed when a compound decomposes into its ions
ⓓ. The heat change when one mole of a compound forms from its constituent elements in their standard states
Correct Answer: The heat change when one mole of a compound forms from its constituent elements in their standard states
Explanation: The standard enthalpy (heat) of formation is defined as the enthalpy change when one mole of a compound forms from its elements in their most stable (standard) forms at 298 K and 1 atm. For example,
\(C(graphite) + 2H_2(g) → CH_4(g\), \(ΔH°_f = –74.8,kJ/mol\).
It provides a reference point for calculating enthalpies of reactions via Hess’s law.
232. Which of the following represents the standard heat of formation of water?
ⓐ. \(H_2(g) + \tfrac{1}{2}O_2(g) → H_2O(l\)
ⓑ. \(2H_2(g) + O_2(g) → 2H_2O(l\)
ⓒ. \(H_2O(l) → H_2(g) + \tfrac{1}{2}O_2(g\)
ⓓ. \(H_2(g) + O_2(g) → H_2O_2(l\)
Correct Answer: \(H_2(g) + \tfrac{1}{2}O_2(g) → H_2O(l\)
Explanation: The standard enthalpy of formation is always written for the formation of one mole of the compound from its elements. Therefore, water’s formation enthalpy uses ½ O₂ to ensure that exactly one mole of H₂O is produced. The standard ΔH°f for this reaction is –286 kJ/mol, indicating an exothermic process.
233. The standard heat of formation of any element in its most stable form is:
ⓐ. +100 kJ/mol
ⓑ. 0 kJ/mol
ⓒ. –100 kJ/mol
ⓓ. Depends on temperature
Correct Answer: 0 kJ/mol
Explanation: Elements like O₂(g), H₂(g), N₂(g), Cl₂(g), and C(graphite) in their most stable states have zero enthalpy of formation by definition. This provides a consistent reference frame for comparing other compounds. Any compound’s ΔH°f value is measured relative to these elemental baselines.
234. Which of the following equations correctly represents the standard enthalpy of formation of ammonia?
ⓐ. \(2NH_3(g) → N_2(g) + 3H_2(g\)
ⓑ. \(N_2(g) + 3H_2(g) → 2NH_3(g\)
ⓒ. \(\tfrac{1}{2}N_2(g) + \tfrac{3}{2}H_2(g) → NH_3(g\)
ⓓ. \(NH_3(g) → \tfrac{1}{2}N_2(g) + \tfrac{3}{2}H_2(g\)
Correct Answer: \(\tfrac{1}{2}N_2(g) + \tfrac{3}{2}H_2(g) → NH_3(g\)
Explanation: The standard enthalpy of formation must describe the creation of exactly one mole of product from its elements in their standard states. The standard ΔH°f for ammonia is –46 kJ/mol. The coefficients are fractional to ensure only one mole of NH₃ forms.
235. If the ΔH°f for CO₂(g) is –393.5 kJ/mol and for H₂O(l) is –286 kJ/mol, what is ΔH°rxn for combustion of CH₄(g)?
ⓐ. –890 kJ/mol
ⓑ. –1079.5 kJ/mol
ⓒ. –286 kJ/mol
ⓓ. –607.5 kJ/mol
Correct Answer: –890 kJ/mol
Explanation: The combustion reaction:
\(CH_4(g) + 2O_2(g) → CO_2(g) + 2H_2O(l\)
\(ΔH° = [–393.5 + 2(–286)] – [–74.8 + 0] = –965.5 + 74.8 = –890.7,kJ/mol\).
Thus, ΔH°combustion = –890 kJ/mol, showing that methane combustion is strongly exothermic.
236. What is meant by the heat of atomization of an element or compound?
ⓐ. The heat change during formation of one mole of atoms in gaseous state from the element in its standard state
ⓑ. The heat change when one mole of gaseous atoms combines to form a molecule
ⓒ. The heat evolved during combustion of an atom
ⓓ. The heat absorbed during neutralization
Correct Answer: The heat change during formation of one mole of atoms in gaseous state from the element in its standard state
Explanation: The heat (enthalpy) of atomization is defined as the energy required to break all bonds in one mole of a substance to produce free gaseous atoms. It is always positive because bond breaking requires energy input. Example:
\(H_2(g) → 2H(g), ΔH°_{atom} = +435.9,kJ/mol\).
237. The heat of atomization of chlorine gas (Cl₂) is 243 kJ/mol. What is the bond dissociation energy of the Cl–Cl bond?
ⓐ. 243 kJ/mol
ⓑ. 121.5 kJ/mol
ⓒ. 486 kJ/mol
ⓓ. 0 kJ/mol
Correct Answer: 121.5 kJ/mol
Explanation: One mole of Cl₂ produces two moles of Cl atoms. The enthalpy change per mole of Cl–Cl bonds is therefore half of the total heat of atomization:
\(\text{Bond energy} = \frac{243}{2} = 121.5,kJ/mol\).
This energy corresponds to the strength of one Cl–Cl covalent bond.
238. Which of the following processes represents heat of atomization?
ⓐ. \(N_2(g) → 2NH_3(g\)
ⓑ. \(H_2(g) + \tfrac{1}{2}O_2(g) → H_2O(l\)
ⓒ. \(Na(s) + Cl_2(g) → NaCl(s\)
ⓓ. ( C(graphite) → C(g) )
Correct Answer: ( C(graphite) → C(g) )
Explanation: The process ( C(graphite) → C(g) ) shows conversion of solid carbon in its standard state to individual gaseous atoms, matching the definition of atomization. The heat absorbed equals the enthalpy of atomization and is always positive because bonds within the solid lattice are broken.
239. The heat of atomization of water is the energy required for which transformation?
ⓐ. \(H_2O(l) → H_2(g) + \tfrac{1}{2}O_2(g\)
ⓑ. \(H_2O(l) → 2H(g) + O(g\)
ⓒ. \(2H(g) + O(g) → H_2O(l\)
ⓓ. \(H_2O(g) → H_2O(l\)
Correct Answer: \(H_2O(l) → 2H(g) + O(g\)
Explanation: The heat of atomization refers to converting one mole of a compound entirely into its gaseous atoms. For water, this involves breaking two O–H bonds to yield two hydrogen atoms and one oxygen atom in the gas phase. This process is highly endothermic due to strong O–H bond energies (~463 kJ/mol each).
240. Why are heats of atomization always positive?
ⓐ. Because forming gaseous atoms releases energy
ⓑ. Because atoms in a compound are unstable
ⓒ. Because bond breaking requires absorption of energy
ⓓ. Because of loss of entropy
Correct Answer: Because bond breaking requires absorption of energy
Explanation: Atomization involves breaking chemical bonds, which needs energy input to overcome attractive forces between atoms. Hence, the enthalpy change is positive (endothermic). When the atoms later recombine to form molecules, energy of equal magnitude is released (exothermic bond formation). This balance forms the basis of bond energy and thermochemical cycle calculations.
241. What is meant by the heat of solution?
ⓐ. The heat absorbed or evolved when one mole of solute dissolves completely in a large amount of solvent at constant pressure
ⓑ. The heat change when two solids are mixed
ⓒ. The heat evolved during neutralization of acid and base
ⓓ. The heat absorbed during evaporation of solvent
Correct Answer: The heat absorbed or evolved when one mole of solute dissolves completely in a large amount of solvent at constant pressure
Explanation: The heat of solution (ΔHsol) represents the enthalpy change when one mole of solute dissolves completely in solvent to form a homogeneous solution under standard conditions (298 K, 1 atm). It can be endothermic (absorbs heat) or exothermic (releases heat) depending on whether solute–solvent attractions overcome or are weaker than solute–solute interactions.
242. The process of dissolving anhydrous CuSO₄ in water is accompanied by:
ⓐ. Absorption of heat (endothermic)
ⓑ. Evolution of heat (exothermic)
ⓒ. No heat change
ⓓ. Heat absorbed and released equally
Correct Answer: Evolution of heat (exothermic)
Explanation: When anhydrous CuSO₄ dissolves in water, strong ion–dipole interactions form between Cu²⁺, SO₄²⁻, and water molecules, releasing energy. Hence, the dissolution is exothermic. This contrasts with dissolving CuSO₄·5H₂O (blue crystals), where less energy is released due to partial hydration already existing.
243. The heat of solution of NH₄NO₃ in water is positive because:
ⓐ. Bonds are formed between solute and solvent
ⓑ. It occurs at high pressure
ⓒ. It releases heat during dissolution
ⓓ. The hydration energy is less than the lattice energy
Correct Answer: The hydration energy is less than the lattice energy
Explanation: In dissolving ammonium nitrate, energy is absorbed to break the crystal lattice (endothermic), but hydration releases less energy than this input. Hence, the net process absorbs heat from surroundings, making the beaker feel cold. Thus, \(\Delta H_{sol}\) for NH₄NO₃ is positive (endothermic dissolution).
244. What is meant by the term heat of hydration?
ⓐ. The heat evolved when ions are surrounded by solvent molecules
ⓑ. The heat required to evaporate water
ⓒ. The heat absorbed during electrolysis
ⓓ. The heat change when solute crystallizes
Correct Answer: The heat evolved when ions are surrounded by solvent molecules
Explanation: Heat of hydration is the energy released when gaseous ions or solid salts become surrounded by water molecules, forming hydrated ions. It is always exothermic, as ion–dipole attractions release energy. For example, the hydration of Na⁺ and Cl⁻ in water contributes to the overall heat of solution of NaCl.
245. The overall heat of solution (ΔHsol) can be expressed as:
ⓐ. \(\Delta H_{sol} = \Delta H_{hydration} – \Delta H_{lattice}\)
ⓑ. \(\Delta H_{sol} = \Delta H_{lattice} – \Delta H_{hydration}\)
ⓒ. \(\Delta H_{sol} = \Delta H_{hydration} \times \Delta H_{lattice}\)
ⓓ. \(\Delta H_{sol} = \Delta H_{hydration} / \Delta H_{lattice}\)
Correct Answer: \(\Delta H_{sol} = \Delta H_{hydration} – \Delta H_{lattice}\)
Explanation: Dissolution involves two steps — breaking the solute lattice (endothermic) and hydrating ions (exothermic). The net enthalpy of solution equals the difference between hydration energy and lattice energy. If hydration > lattice energy, the solution process is exothermic; otherwise, it’s endothermic.
246. When NaOH pellets dissolve in water, the beaker becomes hot because:
ⓐ. The process is endothermic
ⓑ. Water absorbs heat from surroundings
ⓒ. The NaOH does not dissociate
ⓓ. The process is exothermic with large hydration energy
Correct Answer: The process is exothermic with large hydration energy
Explanation: Dissolution of NaOH releases a large amount of heat due to strong ion–dipole interactions between Na⁺, OH⁻, and water molecules. The hydration energy far exceeds the lattice energy, leading to an exothermic dissolution. The high heat release can even cause boiling in concentrated solutions.
247. What is meant by the heat of dilution?
ⓐ. The heat change when one mole of solute is added to an infinite amount of solvent
ⓑ. The heat evolved during condensation
ⓒ. The heat absorbed during vaporization
ⓓ. The heat change when a solution is diluted with more solvent
Correct Answer: The heat change when a solution is diluted with more solvent
Explanation: Heat of dilution is the enthalpy change when a solution of definite concentration is diluted to another concentration by adding solvent. It results from changes in solute–solvent interactions. Generally, it is exothermic for ionic solutions because hydration energy increases with dilution.
248. What happens to the heat of dilution as a solution becomes very dilute?
ⓐ. It increases
ⓑ. It becomes zero
ⓒ. It changes sign
ⓓ. It oscillates between positive and negative values
Correct Answer: It becomes zero
Explanation: When a solution becomes infinitely dilute, further addition of solvent causes negligible changes in solute–solvent interactions. Therefore, no additional heat is absorbed or released, and the heat of dilution approaches zero. The system reaches a state where solute particles are completely hydrated or solvated.
249. Which of the following pairs of processes are always exothermic?
ⓐ. Dissolution of NH₄Cl and heat of dilution
ⓑ. Dissolution of NH₄NO₃ and vaporization of water
ⓒ. Dissolution of NaOH and hydration of ions
ⓓ. Dilution of acid and dissolution of KNO₃
Correct Answer: Dissolution of NaOH and hydration of ions
Explanation: Both dissolution of NaOH and hydration of ions release heat due to strong ion–dipole interactions. Hydration energy is highly exothermic, explaining why ionic solids like NaOH and CaCl₂ cause significant temperature rise upon dissolving in water.
250. When concentrated sulfuric acid is diluted with water, the process is highly exothermic. What safety precaution should be followed?
ⓐ. Add acid into water slowly
ⓑ. Add water into acid slowly
ⓒ. Cool the acid before mixing
ⓓ. Mix both rapidly to finish faster
Correct Answer: Add acid into water slowly
Explanation: Always add acid to water, never the reverse. Adding water to concentrated acid causes rapid exothermic reaction, producing heat that can boil or splatter acid dangerously. By adding acid into water gradually, the heat dissipates safely, preventing violent boiling and ensuring controlled dilution.
251. What does Hess’s law state?
ⓐ. The total heat change in a reaction depends on the rate of reaction
ⓑ. The total enthalpy change of a reaction is independent of the path between initial and final states
ⓒ. The total heat change in a reaction depends on pressure only
ⓓ. The total enthalpy change depends on the catalyst used
Correct Answer: The total enthalpy change of a reaction is independent of the path between initial and final states
Explanation: Hess’s law states that the enthalpy change of an overall reaction equals the sum of the enthalpy changes of individual steps, provided the initial and final conditions are the same. This is due to the fact that enthalpy is a state function and depends only on the initial and final states, not the process path.
252. Which of the following is the correct expression of Hess’s law?
ⓐ. ΔH = ΔU + PΔV
ⓑ. ΔHoverall = ΔH1 + ΔH2 + ΔH3 + …
ⓒ. ΔH = qv
ⓓ. ΔH = CpΔT
Correct Answer: ΔHoverall = ΔH1 + ΔH2 + ΔH3 + …
Explanation: According to Hess’s law, the total enthalpy change for a chemical reaction is the algebraic sum of the enthalpy changes for individual steps into which the reaction may be divided. This principle is used to calculate enthalpy changes for reactions that cannot be measured directly, such as combustion or formation enthalpies.
253. Hess’s law is based on which fundamental concept?
ⓐ. Conservation of energy
ⓑ. Law of definite proportions
ⓒ. Law of multiple proportions
ⓓ. Kinetic theory of gases
Correct Answer: Conservation of energy
Explanation: Hess’s law follows directly from the first law of thermodynamics, which states that energy can neither be created nor destroyed. The total energy change in a chemical reaction depends only on the initial and final energy states, irrespective of the reaction path or steps taken.
254. Which of the following reactions can be used to illustrate Hess’s law?
ⓐ. Direct and indirect formation of CO₂ from carbon and oxygen
ⓑ. Neutralization of acid and base
ⓒ. Combustion of hydrogen
ⓓ. Fusion of ice
Correct Answer: Direct and indirect formation of CO₂ from carbon and oxygen
Explanation: Carbon can form CO₂ directly by combustion or indirectly by forming CO first and then oxidizing CO to CO₂. The total enthalpy change in both routes is the same, illustrating Hess’s law. This demonstrates that the total energy change is path-independent.
255. Which of the following is true according to Hess’s law?
ⓐ. ΔH of a reaction depends on the temperature only
ⓑ. ΔH for a reaction remains the same whether the reaction takes place in one or several steps
ⓒ. ΔH increases with pressure
ⓓ. ΔH changes with the catalyst used
Correct Answer: ΔH for a reaction remains the same whether the reaction takes place in one or several steps
Explanation: Since enthalpy is a state function, its change depends only on the difference between final and initial states. Therefore, whether a reaction occurs in one step or multiple steps, the overall enthalpy change remains constant under identical conditions.
256. Which of the following is an application of Hess’s law?
ⓐ. Determining rate constants
ⓑ. Calculating enthalpy of reaction indirectly
ⓒ. Measuring entropy change experimentally
ⓓ. Finding equilibrium constants directly
Correct Answer: Calculating enthalpy of reaction indirectly
Explanation: Hess’s law allows chemists to find enthalpy changes for reactions that cannot be measured directly in the laboratory. For example, the enthalpy of formation of CO can be calculated from enthalpy data of combustion of carbon and carbon monoxide using Hess’s law.
257. If ΔH₁ = –393.5 kJ and ΔH₂ = –283 kJ, what is ΔH for the reaction C(s) + ½O₂(g) → CO(g)?
ⓐ. –110.5 kJ
ⓑ. +110.5 kJ
ⓒ. –676.5 kJ
ⓓ. +676.5 kJ
Correct Answer: –110.5 kJ
Explanation: According to Hess’s law:
C + O₂ → CO₂, ΔH₁ = –393.5 kJ
CO + ½O₂ → CO₂, ΔH₂ = –283 kJ
Reversing the second reaction gives CO₂ → CO + ½O₂, ΔH = +283 kJ.
Adding both gives: C + ½O₂ → CO, ΔH = –393.5 + 283 = –110.5 kJ.
258. In Hess’s law calculations, reversing a reaction will:
ⓐ. Double the enthalpy change
ⓑ. Make no change in enthalpy
ⓒ. Change the sign of enthalpy
ⓓ. Divide enthalpy by two
Correct Answer: Change the sign of enthalpy
Explanation: When a reaction is reversed, the direction of heat flow is reversed. Therefore, an exothermic reaction (negative ΔH) becomes endothermic (positive ΔH) and vice versa. The magnitude of enthalpy remains the same because the same amount of energy is involved, only the direction changes.
259. If a chemical equation is multiplied by a factor n, then according to Hess’s law, ΔH becomes:
ⓐ. Unchanged
ⓑ. Multiplied by n
ⓒ. Divided by n
ⓓ. Reversed in sign
Correct Answer: Multiplied by n
Explanation: Enthalpy change is an extensive property. When the stoichiometric coefficients in a balanced equation are multiplied by n, the enthalpy change also scales proportionally by n. For instance, if ΔH for H₂ + ½O₂ → H₂O is –286 kJ, then for 2H₂ + O₂ → 2H₂O, ΔH = –572 kJ.
260. Which statement best summarizes Hess’s law?
ⓐ. The total enthalpy of a system changes with the number of steps
ⓑ. Enthalpy change depends only on the total pressure of the reaction
ⓒ. Enthalpy change of a reaction is equal to the sum of enthalpy changes of its individual steps
ⓓ. Enthalpy is not a state function
Correct Answer: Enthalpy change of a reaction is equal to the sum of enthalpy changes of its individual steps
Explanation: Hess’s law emphasizes that the total energy change between two states remains constant irrespective of the reaction pathway. It is a direct consequence of energy conservation and the state function nature of enthalpy. This principle enables calculation of ΔH° for complex reactions using data from simpler, known reactions.
261. Which of the following is an important use of Hess’s law?
ⓐ. To calculate molecular weights
ⓑ. To determine enthalpy changes that cannot be measured directly
ⓒ. To measure bond angles
ⓓ. To identify ionic radii
Correct Answer: To determine enthalpy changes that cannot be measured directly
Explanation: Many reactions, such as the formation of CO or certain decomposition processes, cannot be studied experimentally due to side reactions or safety reasons. Hess’s law allows indirect calculation of enthalpy changes by combining other reactions with known ΔH values. Since enthalpy is a state function, this method gives accurate results independent of the path taken.
262. Hess’s law helps to determine which of the following?
ⓐ. Enthalpy of formation
ⓑ. Enthalpy of neutralization
ⓒ. Enthalpy of combustion
ⓓ. All of these
Correct Answer: All of these
Explanation: Hess’s law can be applied to determine various thermodynamic quantities—enthalpy of formation, combustion, neutralization, and reaction—using known enthalpy data for related reactions. The additivity of enthalpy changes enables calculation of unknown ΔH° values when direct experimental determination is difficult or impossible.
263. The enthalpy of formation of CO(g) can be calculated using Hess’s law from:
ⓐ. Direct combination of carbon and oxygen
ⓑ. Combustion of carbon and combustion of CO
ⓒ. Decomposition of CO₂
ⓓ. Photosynthesis reaction
Correct Answer: Combustion of carbon and combustion of CO
Explanation: By Hess’s law, ΔH°f for CO(g) can be determined from two known reactions:
C + O₂ → CO₂, ΔH₁ = –393.5 kJ;
CO + ½O₂ → CO₂, ΔH₂ = –283 kJ.
Reversing the second reaction and adding gives C + ½O₂ → CO, ΔH = –110.5 kJ. Thus, the law allows indirect calculation of CO formation enthalpy.
264. Hess’s law can be applied to find the enthalpy of formation of CH₄ from which data set?
ⓐ. Enthalpies of combustion of C, H₂, and CH₄
ⓑ. Ionization energies of C and H
ⓒ. Lattice energies of C and H₂
ⓓ. Enthalpy of sublimation only
Correct Answer: Enthalpies of combustion of C, H₂, and CH₄
Explanation: The formation of CH₄ from its elements can’t be directly measured, so Hess’s law uses combustion data:
C + 2H₂ + 2O₂ → CO₂ + 2H₂O,
and known ΔH values for CO₂ and H₂O formation. Combining these reactions via Hess’s law gives the enthalpy of CH₄ formation indirectly, a standard approach in thermochemistry.
265. The enthalpy of decomposition of CaCO₃(s) → CaO(s) + CO₂(g) can be calculated using Hess’s law by combining:
ⓐ. Enthalpy of formation of CaCO₃, CaO, and CO₂
ⓑ. Ionization enthalpy of Ca and electron affinity of O
ⓒ. Heat of hydration of Ca²⁺ and CO₃²⁻
ⓓ. Enthalpy of solution of CaCO₃
Correct Answer: Enthalpy of formation of CaCO₃, CaO, and CO₂
Explanation: Using Hess’s law,
ΔHreaction = [ΔH°f(CaO) + ΔH°f(CO₂)] – ΔH°f(CaCO₃).
This method works because enthalpy is a state function, so combining formation enthalpies yields the decomposition enthalpy directly without experimental decomposition, which is otherwise difficult to measure.
266. In an indirect reaction path, if the sum of all ΔH values of the steps equals the direct ΔH of the reaction, this confirms:
ⓐ. Dalton’s law of partial pressures
ⓑ. Hess’s law of constant heat summation
ⓒ. Boyle’s law
ⓓ. Charles’s law
Correct Answer: Hess’s law of constant heat summation
Explanation: Hess’s law asserts that total enthalpy change remains constant regardless of the reaction path. Therefore, if a multistep reaction yields the same ΔH as a direct one, it validates the law and confirms that enthalpy depends only on initial and final states.
267. The enthalpy of formation of CO₂ is –393.5 kJ/mol and that of CO is –110.5 kJ/mol. What is the enthalpy change for oxidation of CO to CO₂?
ⓐ. +283.0 kJ/mol
ⓑ. –283.0 kJ/mol
ⓒ. –110.5 kJ/mol
ⓓ. +110.5 kJ/mol
Correct Answer: –283.0 kJ/mol
Explanation: According to Hess’s law,
ΔHreaction = ΔHf(CO₂) – ΔHf(CO) = (–393.5) – (–110.5) = –283.0 kJ/mol.
This negative sign indicates that the oxidation of carbon monoxide to carbon dioxide is exothermic, releasing 283 kJ of heat per mole.
268. The standard enthalpy of combustion of graphite is –393.5 kJ/mol, and that of CO is –283 kJ/mol. Using Hess’s law, calculate ΔH°f for CO.
ⓐ. –110.5 kJ/mol
ⓑ. –676.5 kJ/mol
ⓒ. +110.5 kJ/mol
ⓓ. 0 kJ/mol
Correct Answer: –110.5 kJ/mol
Explanation:
C + O₂ → CO₂, ΔH = –393.5 kJ/mol
CO + ½O₂ → CO₂, ΔH = –283 kJ/mol
Reverse the second reaction: CO₂ → CO + ½O₂, ΔH = +283 kJ/mol.
Adding both: C + ½O₂ → CO, ΔH = –110.5 kJ/mol.
Thus, the enthalpy of CO formation is –110.5 kJ/mol.
269. Hess’s law can also be applied in determining which other thermodynamic function besides ΔH?
ⓐ. Entropy
ⓑ. Internal energy
ⓒ. Gibbs free energy
ⓓ. All of these
Correct Answer: All of these
Explanation: Since entropy (S), Gibbs free energy (G), and internal energy (U) are also state functions, their total changes between initial and final states are path-independent. Therefore, Hess’s law of summation applies to these properties as well, enabling indirect calculations using known data for multiple steps.
270. Why is Hess’s law particularly useful in thermochemistry?
ⓐ. It explains why reactions are spontaneous
ⓑ. It allows estimation of reaction enthalpies using known data
ⓒ. It defines heat capacities of compounds
ⓓ. It predicts bond lengths of molecules
Correct Answer: It allows estimation of reaction enthalpies using known data
Explanation: Hess’s law provides a practical way to compute enthalpy changes for complex or dangerous reactions using known enthalpy values of related reactions. This method ensures accurate thermochemical calculations without requiring direct experimentation, which may be challenging or unsafe for certain reactions.
271. What is meant by bond enthalpy?
ⓐ. The energy released when a bond forms between two atoms in the gas phase
ⓑ. The energy required to break one mole of bonds in gaseous molecules
ⓒ. The heat evolved when one mole of atoms is formed
ⓓ. The energy absorbed when a solid melts into a liquid
Correct Answer: The energy required to break one mole of bonds in gaseous molecules
Explanation: Bond enthalpy (or bond dissociation enthalpy) is defined as the amount of energy required to break one mole of a specific type of bond in gaseous molecules, producing gaseous atoms. It is always positive because energy must be supplied to overcome the attractive forces holding the atoms together in a bond.
272. Which of the following has the highest bond enthalpy?
ⓐ. H–H
ⓑ. Cl–Cl
ⓒ. O=O
ⓓ. N≡N
Correct Answer: N≡N
Explanation: The triple bond in nitrogen (N≡N) is very strong, requiring 941 kJ/mol to break — the highest among common diatomic molecules. Multiple bonds involve greater electron sharing, leading to stronger attraction between atoms and higher bond enthalpy compared to single or double bonds.
273. What does a high bond enthalpy value indicate about a molecule?
ⓐ. The bond is weak
ⓑ. The molecule is highly reactive
ⓒ. The bond is strong and stable
ⓓ. The molecule has low bond energy
Correct Answer: The bond is strong and stable
Explanation: A large bond enthalpy means more energy is needed to break the bond, implying a stronger and more stable bond. Such bonds make molecules less reactive and more resistant to decomposition. Conversely, low bond enthalpies correspond to weaker, more reactive bonds.
274. Which of the following correctly represents the bond dissociation enthalpy of hydrogen gas?
ⓐ. \(H_2(g) → 2H(g\)
ⓑ. \(2H(g) → H_2(g\)
ⓒ. \(H_2(l) → 2H(g\)
ⓓ. \(H(g) → H^+ + e^-\)
Correct Answer: \(H_2(g) → 2H(g\)
Explanation: The dissociation enthalpy of H₂ refers to the energy required to break one mole of H–H bonds in hydrogen gas to form two separate gaseous hydrogen atoms. It is an endothermic process with a standard bond enthalpy of +435.9 kJ/mol.
275. Which of the following statements is correct about average bond enthalpy?
ⓐ. It is the same for all bonds of the same type in a molecule
ⓑ. It is used when a molecule contains identical bonds only
ⓒ. It is the average energy required to break each bond in one mole of gaseous molecules
ⓓ. It depends on temperature only
Correct Answer: It is the average energy required to break each bond in one mole of gaseous molecules
Explanation: Average bond enthalpy is used for molecules containing bonds of the same type but slightly different energies, like O–H bonds in H₂O. It is calculated by dividing the total energy needed to break all identical bonds by the number of bonds broken. It gives an approximate value used in thermochemical calculations.
276. The average bond enthalpy of O–H in H₂O is 463 kJ/mol. What is the total energy required to dissociate one mole of H₂O(g) into atoms?
ⓐ. 463 kJ
ⓑ. 926 kJ
ⓒ. 231.5 kJ
ⓓ. 500 kJ
Correct Answer: 926 kJ
Explanation: Each H₂O molecule has two O–H bonds. Hence, total energy required = 2 × 463 = 926 kJ/mol. Since both bonds are broken to form H(g) and O(g), the total enthalpy is the sum of both dissociation energies.
277. Why do polyatomic molecules have average bond enthalpies instead of specific ones?
ⓐ. Because each bond in the molecule has exactly the same strength
ⓑ. Because bonds of the same type may have different strengths due to the molecular environment
ⓒ. Because measuring bond energy in polyatomic gases is impossible
ⓓ. Because bond energies are always constant
Correct Answer: Because bonds of the same type may have different strengths due to the molecular environment
Explanation: In molecules like H₂O or CH₄, the first bond broken slightly changes the electronic environment of the molecule, affecting the energy required to break subsequent bonds. Thus, average bond enthalpy provides a representative value useful for calculations, even though individual bond strengths vary slightly.
278. Which of the following correctly represents the average bond enthalpy of C–H bonds in methane?
ⓐ. Energy required to break one C–H bond in CH₄
ⓑ. Energy required to break all four C–H bonds in one mole of CH₄ divided by 4
ⓒ. Energy released when CH₄ is formed from atoms
ⓓ. Energy absorbed when one C–C bond is broken
Correct Answer: Energy required to break all four C–H bonds in one mole of CH₄ divided by 4
Explanation: Average bond enthalpy = (Total energy to break all identical bonds) ÷ (Number of bonds). For methane,
\(CH_4(g) → C(g) + 4H(g\),
ΔH = 1660 kJ/mol.
Average bond enthalpy = 1660 ÷ 4 = 415 kJ/mol.
279. Which factor affects the bond enthalpy of a molecule?
ⓐ. Atomic mass only
ⓑ. Bond length and bond polarity
ⓒ. Temperature alone
ⓓ. Number of isotopes present
Correct Answer: Bond length and bond polarity
Explanation: Bond enthalpy increases as bond length decreases and bond polarity increases. Shorter and more polar bonds (like O–H) are stronger and require more energy to break. Longer or less polar bonds (like C–C) have lower bond enthalpy, making them weaker and easier to break.
280. Why is the bond enthalpy of O=O lower than that of N≡N?
ⓐ. Oxygen atoms are smaller than nitrogen atoms
ⓑ. O=O has higher bond energy
ⓒ. O=O bond is more polar than N≡N
ⓓ. N≡N is a triple bond while O=O is a double bond
Correct Answer: N≡N is a triple bond while O=O is a double bond
Explanation: The N≡N triple bond involves six shared electrons, resulting in much stronger attraction and higher bond energy than the O=O double bond. Hence, N₂ has a much higher bond enthalpy (941 kJ/mol) than O₂ (498 kJ/mol). The more shared electron pairs, the greater the bond strength and enthalpy.
281. What does a higher bond enthalpy indicate about the stability of a molecule?
ⓐ. The molecule is less stable and more reactive
ⓑ. The molecule is more stable and less reactive
ⓒ. The molecule has low activation energy
ⓓ. The molecule decomposes easily
Correct Answer: The molecule is more stable and less reactive
Explanation: A large bond enthalpy means more energy is needed to break the bonds, implying stronger bonding and higher stability. Molecules with high bond enthalpy resist decomposition and are thermally more stable. Conversely, molecules with low bond enthalpy break easily and are more chemically reactive.
282. Which of the following pairs of molecules shows greater stability based on bond enthalpy?
ⓐ. O₂ (498 kJ/mol) and N₂ (941 kJ/mol)
ⓑ. Cl₂ (243 kJ/mol) and F₂ (159 kJ/mol)
ⓒ. H₂ (436 kJ/mol) and N₂ (941 kJ/mol)
ⓓ. N₂ (941 kJ/mol) and O₂ (498 kJ/mol)
Correct Answer: N₂ (941 kJ/mol) and O₂ (498 kJ/mol)
Explanation: Nitrogen has a very high bond enthalpy (941 kJ/mol) due to its triple bond, making it extremely stable and inert at room temperature. Oxygen, with a double bond and lower enthalpy (498 kJ/mol), is more reactive. Thus, higher bond enthalpy directly correlates with greater molecular stability.
283. Why is the N≡N bond in nitrogen gas so stable?
ⓐ. Because nitrogen atoms are light
ⓑ. Because the bond is highly polar
ⓒ. Because it has a strong triple bond requiring large energy to break
ⓓ. Because nitrogen forms hydrogen bonds
Correct Answer: Because it has a strong triple bond requiring large energy to break
Explanation: Nitrogen’s triple bond consists of one sigma and two pi bonds, resulting in a very high bond enthalpy (941 kJ/mol). This makes N₂ thermodynamically stable and chemically inert. High bond energy means a very large amount of energy is required to dissociate the molecule, ensuring its stability.
284. Which of the following molecules is the least stable based on bond enthalpy values?
ⓐ. H₂ (436 kJ/mol)
ⓑ. F₂ (159 kJ/mol)
ⓒ. O₂ (498 kJ/mol)
ⓓ. N₂ (941 kJ/mol)
Correct Answer: F₂ (159 kJ/mol)
Explanation: The F–F bond has very low bond enthalpy because of strong repulsion between lone pairs on the small fluorine atoms, which weakens the bond. Hence, F₂ molecules are less stable and more reactive compared to others with higher bond enthalpies.
285. What is meant by lattice enthalpy?
ⓐ. The energy required to break one mole of covalent bonds in a solid
ⓑ. The enthalpy change when one mole of an ionic solid is formed from its gaseous ions
ⓒ. The heat evolved when an ionic solid dissolves in water
ⓓ. The heat absorbed when a solute forms ions in solution
Correct Answer: The enthalpy change when one mole of an ionic solid is formed from its gaseous ions
Explanation: Lattice enthalpy (ΔHlattice) is defined as the energy released when one mole of an ionic compound forms from its gaseous ions, e.g.,
\(Na^+(g) + Cl^-(g) → NaCl(s\).
It is always exothermic (negative), indicating the strong electrostatic attraction that stabilizes the crystal lattice.
286. Which of the following statements about lattice enthalpy is correct?
ⓐ. It is always positive
ⓑ. It increases with larger ionic radii
ⓒ. It decreases as the distance between ions decreases
ⓓ. It increases with greater ionic charge and smaller ionic radius
Correct Answer: It increases with greater ionic charge and smaller ionic radius
Explanation: Lattice enthalpy is proportional to the electrostatic attraction between ions, which increases with ionic charge and decreases with distance. Thus, small, highly charged ions produce large lattice enthalpies, as in MgO or Al₂O₃, making such compounds more stable.
287. Which of the following compounds has the highest lattice enthalpy?
ⓐ. NaCl
ⓑ. MgCl₂
ⓒ. AlCl₃
ⓓ. KCl
Correct Answer: AlCl₃
Explanation: Lattice enthalpy increases with the product of ionic charges and decreases with ionic size. Aluminum ions (Al³⁺) have a high charge and small radius, producing strong electrostatic attraction with Cl⁻. Therefore, AlCl₃ has a much higher lattice enthalpy than NaCl, KCl, or MgCl₂.
288. Which statement best explains the relationship between lattice enthalpy and ionic stability?
ⓐ. Higher lattice enthalpy means the ionic compound is less stable
ⓑ. Lower lattice enthalpy leads to more stable ionic solids
ⓒ. Higher lattice enthalpy means stronger ionic bonding and greater stability
ⓓ. Lattice enthalpy has no relation to stability
Correct Answer: Higher lattice enthalpy means stronger ionic bonding and greater stability
Explanation: Large lattice enthalpy values indicate strong electrostatic forces between cations and anions. This strong attraction increases the stability and hardness of ionic solids and raises their melting points. Compounds with lower lattice enthalpies, like CsI, are comparatively less stable.
289. Which of the following has the lowest lattice enthalpy?
ⓐ. NaF
ⓑ. MgF₂
ⓒ. LiF
ⓓ. KF
Correct Answer: KF
Explanation: Lattice enthalpy decreases with increasing ionic size because the electrostatic attraction weakens as the distance between ions increases. Potassium has the largest ionic radius among these cations, so KF has the lowest lattice enthalpy. LiF, having the smallest cation, has the highest.
290. Why is lattice enthalpy important in thermodynamics?
ⓐ. It determines the reactivity of covalent molecules
ⓑ. It helps calculate the solubility and stability of ionic compounds
ⓒ. It measures the energy of vaporization
ⓓ. It determines the speed of ionic reactions
Correct Answer: It helps calculate the solubility and stability of ionic compounds
Explanation: Lattice enthalpy is a crucial factor in predicting the solubility and stability of ionic solids. When an ionic solid dissolves, its lattice must break apart (requiring energy), while hydration releases energy. The balance between these two determines whether dissolution is endothermic or exothermic, influencing solubility and crystal stability.
291. What is meant by a spontaneous process?
ⓐ. A process that requires continuous external energy to proceed
ⓑ. A process that can occur by itself under given conditions without external help
ⓒ. A process that never occurs naturally
ⓓ. A reversible process that proceeds infinitely slowly
Correct Answer: A process that can occur by itself under given conditions without external help
Explanation: A spontaneous process is one that occurs naturally without the need for external energy once initiated. Examples include the melting of ice at room temperature, diffusion of gases, and rusting of iron. It may be fast or slow but proceeds in the direction of thermodynamic stability without external assistance.
292. Which of the following is an example of a spontaneous process?
ⓐ. Water flowing uphill
ⓑ. Heat flowing from cold to hot body
ⓒ. Compression of gas without applying pressure
ⓓ. Ice melting at 298 K
Correct Answer: Ice melting at 298 K
Explanation: At 298 K (above 0°C), ice melts spontaneously because the Gibbs free energy change (ΔG) is negative, indicating that the process is thermodynamically favorable. Processes that decrease free energy occur spontaneously, while those increasing it are non-spontaneous under the same conditions.
293. Which of the following is a non-spontaneous process?
ⓐ. Heat flowing from a hot body to a cold body
ⓑ. Evaporation of water at 373 K
ⓒ. Decomposition of water into hydrogen and oxygen without external energy
ⓓ. Iron rusting in moist air
Correct Answer: Decomposition of water into hydrogen and oxygen without external energy
Explanation: The decomposition of water into its elements requires an external energy input, such as electricity (electrolysis), because it is non-spontaneous. A process that cannot occur naturally under given conditions is non-spontaneous, even if it becomes possible when external work is applied.
294. Which of the following statements is true about spontaneous processes?
ⓐ. They always occur rapidly
ⓑ. They always release energy
ⓒ. They proceed in the direction of decreasing Gibbs free energy (ΔG < 0)
ⓓ. They require continuous energy supply
Correct Answer: They proceed in the direction of decreasing Gibbs free energy (ΔG < 0)
Explanation: A process is spontaneous if ΔG = ΔH – TΔS is negative. This means the process either releases energy or increases entropy sufficiently to make ΔG negative. The spontaneity criterion applies universally, regardless of the reaction rate or energy form.
295. What is true for a system undergoing a spontaneous process at constant temperature and pressure?
ⓐ. ΔG = 0
ⓑ. ΔG > 0
ⓒ. ΔG < 0
ⓓ. ΔH = 0
Correct Answer: ΔG < 0
Explanation: For a process to occur spontaneously under constant temperature and pressure, the Gibbs free energy change (ΔG) must be negative. This ensures that the process moves toward equilibrium and that energy is released or entropy increases enough to drive it forward naturally.
296. Which condition represents equilibrium between spontaneous and non-spontaneous behavior?
ⓐ. ΔG > 0
ⓑ. ΔH < 0
ⓒ. ΔS = 0
ⓓ. ΔG = 0
Correct Answer: ΔG = 0
Explanation: When ΔG = 0, the system is at equilibrium, meaning there is no net change in the forward or reverse directions. The system has reached maximum stability under given conditions, and any spontaneous change stops at this point, balancing both processes.
297. Which of the following is a spontaneous process at room temperature?
ⓐ. Combustion of methane in air
ⓑ. Decomposition of calcium carbonate
ⓒ. Electrolysis of molten NaCl
ⓓ. Formation of ozone from oxygen
Correct Answer: Combustion of methane in air
Explanation: The combustion of methane is highly exothermic, with a large negative ΔG value, making it spontaneous at room temperature. Although it requires ignition energy to start, once initiated, it proceeds without further external energy input, releasing large amounts of heat and CO₂.
298. What is meant by a non-spontaneous process?
ⓐ. A process that occurs naturally and continuously
ⓑ. A process that can occur only when external energy is supplied
ⓒ. A process that always decreases entropy
ⓓ. A process that is always exothermic
Correct Answer: A process that can occur only when external energy is supplied
Explanation: Non-spontaneous processes require external work or energy input to proceed, such as electrolysis, refrigeration, and compressing gases. These processes have positive Gibbs free energy changes (ΔG > 0), meaning they are thermodynamically unfavorable under given conditions.
299. Which of the following processes is spontaneous at high temperatures but not at low temperatures?
ⓐ. Freezing of water
ⓑ. Combustion of carbon
ⓒ. Rusting of iron
ⓓ. Melting of ice
Correct Answer: Melting of ice
Explanation: The melting of ice is endothermic (ΔH > 0) but accompanied by a large increase in entropy (ΔS > 0). At low temperatures, TΔS < ΔH, so ΔG > 0 (non-spontaneous). At higher temperatures (above 0°C), TΔS > ΔH, making ΔG < 0 and the process spontaneous.
300. Which of the following correctly describes spontaneity?
ⓐ. A spontaneous process is always exothermic
ⓑ. A spontaneous process can be either exothermic or endothermic depending on entropy change
ⓒ. All endothermic processes are spontaneous
ⓓ. Spontaneous processes violate the first law of thermodynamics
Correct Answer: A spontaneous process can be either exothermic or endothermic depending on entropy change
Explanation: While many spontaneous processes are exothermic, some endothermic processes like evaporation and melting are spontaneous because they result in a large increase in entropy (ΔS > 0), making ΔG negative. Thus, spontaneity depends on both enthalpy (ΔH) and entropy (ΔS) contributions as expressed by ΔG = ΔH – TΔS.
Welcome to Class 11 Chemistry MCQs – Chapter 6: Thermodynamics (Part 3). This section keeps you exam-ready for Boards, JEE, NEET with focused practice and crisp explanations.
Focus of Part 3 (100 MCQs): Hess’s Law and energy cycles; standard enthalpies (formation, combustion, solution, neutralization); bond enthalpy concepts and quick estimates; lattice/solvation ideas; typical data-based numericals. Great for strengthening concept → calculation flow.
How to practice here
- Attempt steadily, then Favourite the tricky/high-yield MCQs and review via the Favourite toggle.
- Use the Workspace below each question to write your own shortcut or error-fix; notes auto-save for quick revisits.
- Shuffle with Random to challenge recall from different angles.
- 👉 Total in chapter: 395 MCQs (100 + 100 + 100 + 95)
- 👉 This page: Third set of 100 solved MCQs
- 👉 Next: Proceed to Part 4 for the final set