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31. A reaction that can proceed in both forward and reverse directions under the same conditions is called:
ⓐ. Irreversible reaction
ⓑ. Redox reaction
ⓒ. Reversible reaction
ⓓ. Neutralization reaction
Correct Answer: Reversible reaction
Explanation: A reversible reaction is one in which reactants form products, and simultaneously, products can recombine to form reactants. Such reactions reach a state of equilibrium when the rates of forward and reverse reactions become equal. Reversible reactions are common in chemical systems like the synthesis of ammonia and dissociation of acetic acid.
32. Which of the following best represents a reversible chemical reaction?
ⓐ. \( \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 \)
ⓑ. \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \)
ⓒ. \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)
ⓓ. \( \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \)
Correct Answer: \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \)
Explanation: The synthesis of ammonia in the Haber process is reversible, meaning ammonia can decompose back into nitrogen and hydrogen. The double arrow \(( \rightleftharpoons )\) shows that both reactions occur simultaneously and can reach a dynamic equilibrium depending on conditions like temperature and pressure.
33. When a reversible reaction attains equilibrium, what remains constant?
ⓐ. The rates of forward and reverse reactions
ⓑ. The concentrations of reactants and products
ⓒ. The speed of individual molecular collisions
ⓓ. The rate constant of the forward reaction
Correct Answer: The rates of forward and reverse reactions
Explanation: At equilibrium, the forward and reverse reactions continue at equal rates, which maintains constant concentrations of reactants and products. This equality of rates is the defining feature of dynamic equilibrium, where molecular changes still occur but no net change in concentration is observed.
34. Consider the reversible reaction \( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \). If the concentration of \( \text{H}_2 \) is increased at equilibrium, what happens initially?
ⓐ. The rate of forward reaction increases
ⓑ. The rate of reverse reaction increases
ⓒ. The equilibrium constant increases
ⓓ. The temperature decreases
Correct Answer: The rate of forward reaction increases
Explanation: Increasing the concentration of a reactant shifts the reaction temporarily toward the product side because more reactant molecules are available to collide and form products. The system then adjusts until a new equilibrium is established where the forward and reverse reaction rates again become equal.
35. Which of the following reactions is reversible in nature?
ⓐ. \( \text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \)
ⓑ. \( 2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \)
ⓒ. \( \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)
ⓓ. \( \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 \)
Correct Answer: \( 2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \)
Explanation: The formation of sulfur trioxide from sulfur dioxide and oxygen is a reversible reaction. Under suitable temperature and catalyst conditions, the products can revert to reactants. This reversible nature allows the system to achieve equilibrium where both reactants and products coexist in constant proportions.
36. When equilibrium is attained in a reversible reaction:
ⓐ. The reaction stops completely
ⓑ. The forward and reverse rates become equal
ⓒ. The concentration of products becomes zero
ⓓ. The energy of reactants becomes zero
Correct Answer: The forward and reverse rates become equal
Explanation: At equilibrium, both reactions occur simultaneously at the same rate, maintaining constant concentrations of all species. This does not mean the reaction stops; it remains dynamic at the molecular level. The equilibrium condition reflects a balance of rates, not of concentrations or energies.
37. For the reaction \( 2\text{NO}_2(g) \rightleftharpoons 2\text{NO}(g) + \text{O}_2(g) \), what type of equilibrium is established?
ⓐ. Static equilibrium
ⓑ. Mechanical equilibrium
ⓒ. Thermal equilibrium
ⓓ. Dynamic equilibrium
Correct Answer: Dynamic equilibrium
Explanation: The decomposition of nitrogen dioxide is an example of dynamic equilibrium where both forward and reverse reactions continue at equal rates. Even though concentrations remain constant over time, molecules are continuously reacting, illustrating the dynamic nature of chemical equilibrium.
38. Which condition is required for a reversible reaction to reach equilibrium?
ⓐ. The system must be closed and temperature constant
ⓑ. The products must be continuously removed
ⓒ. Pressure and temperature should change constantly
ⓓ. Reactants must be in solid state only
Correct Answer: The system must be closed and temperature constant
Explanation: Equilibrium can only be achieved in a closed system, where no reactants or products escape. Constant temperature ensures that reaction rates depend only on concentration, not on environmental changes, allowing the system to achieve a stable equilibrium position.
39. In the reversible reaction \( \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \), what will happen if the pressure is increased?
ⓐ. The reaction shifts toward \( \text{PCl}_5(g) \)
ⓑ. The reaction shifts toward \( \text{PCl}_3(g) \)
ⓒ. The equilibrium constant decreases
ⓓ. The reaction stops immediately
Correct Answer: The reaction shifts toward \( \text{PCl}_5(g) \)
Explanation: Increasing pressure favors the direction with fewer gas molecules. On the left side, there is one molecule of \( \text{PCl}_5 \), while on the right there are two molecules \(( \text{PCl}_3 + \text{Cl}_2 )\). Hence, higher pressure shifts equilibrium toward \( \text{PCl}_5(g) \), reducing total gas volume.
40. The reversible decomposition of calcium carbonate \( \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \) is an example of:
ⓐ. Homogeneous equilibrium
ⓑ. Heterogeneous equilibrium
ⓒ. Physical equilibrium
ⓓ. Neutralization equilibrium
Correct Answer: Heterogeneous equilibrium
Explanation: This reaction involves solids \(( \text{CaCO}_3 ), ( \text{CaO} )\) and a gas \(( \text{CO}_2 )\), representing more than one phase, hence a heterogeneous equilibrium. The equilibrium condition depends only on the partial pressure of carbon dioxide, as activities of pure solids remain constant.
41. Which observation best demonstrates that equilibrium is dynamic rather than static in a saturated solution of sugar at constant (T)?
ⓐ. Constant concentration with continuous dissolution and crystallization at equal rates
ⓑ. Sudden disappearance of undissolved sugar after a long time
ⓒ. Continuous increase of solution concentration with time
ⓓ. Complete halt of molecular motion in the liquid
Correct Answer: Constant concentration with continuous dissolution and crystallization at equal rates
Explanation: In a saturated solution, molecules leave the crystal and reattach to it at equal rates, so the macroscopic concentration remains constant. The ongoing bidirectional molecular exchange is the hallmark of dynamic equilibrium.
42. A closed flask containing water at \(25^\circ\text{C}\) shows constant vapour pressure. Which measurement provides direct evidence that liquid–vapour equilibrium is dynamic?
ⓐ. No change in mass of the liquid when the stopper is removed briefly
ⓑ. Equal numbers of collisions of vapour molecules with the walls
ⓒ. Continuous exchange of molecules between liquid and vapour detected by isotopic labelling of the liquid
ⓓ. Constant temperature reading on a thermometer
Correct Answer: Continuous exchange of molecules between liquid and vapour detected by isotopic labelling of the liquid
Explanation: If the liquid water is isotopically labelled, the label later appears in the vapour phase while total pressure remains constant, proving ongoing evaporation and condensation at equal rates.
43. In the esterification system \(\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}\), which experimental result evidences dynamic equilibrium at constant (T)?
ⓐ. Permanent cessation of smell of ester
ⓑ. Increase of ester concentration without bound
ⓒ. Decrease of water concentration to zero
ⓓ. Exchange of \(^{18}\text{O}\) label between water and ester without change in macroscopic concentrations
Correct Answer: Exchange of \(^{18}\text{O}\) label between water and ester without change in macroscopic concentrations
Explanation: Isotopic oxygen initially in water is later found in the ester carbonyl oxygen while bulk concentrations stay constant, showing simultaneous forward and reverse reactions at equal rates.
44. For \( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \) at \(700\ \text{K}\), \([ \text{H}_2 ]=0.20,\text{M}, [ \text{I}_2 ]=0.20,\text{M}, [ \text{HI} ]=0.40,\text{M}\), and rate constants \(k_f=2.0,\text{M}^{-1}\text{s}^{-1}\), \(k_r=1.0,\text{M}^{-1}\text{s}^{-1}\). Which calculation indicates dynamic equilibrium?
ⓐ. \(k_f[\text{H}_2][\text{I}_2]=0.060,\text{M},\text{s}^{-1}\) and \(k_r[\text{HI}]^{2}=0.20,\text{M},\text{s}^{-1}\)
ⓑ. \(k_f[\text{H}_2][\text{I}_2]=0.080,\text{M},\text{s}^{-1}\) and \(k_r[\text{HI}]^{2}=0.160,\text{M},\text{s}^{-1}\)
ⓒ. \(k_f[\text{H}_2][\text{I}_2]=0.040,\text{M},\text{s}^{-1}\) and \(k_r[\text{HI}]^{2}=0.040,\text{M},\text{s}^{-1}\)
ⓓ. \(k_f[\text{H}_2][\text{I}_2]=0.020,\text{M},\text{s}^{-1}\) and \(k_r[\text{HI}]^{2}=0.040,\text{M},\text{s}^{-1}\)
Correct Answer: \(k_f[\text{H}_2][\text{I}_2]=0.040,\text{M},\text{s}^{-1}\) and \(k_r[\text{HI}]^{2}=0.040,\text{M},\text{s}^{-1}\)
Explanation: Equality of forward and reverse rates confirms the dynamic condition. Using the given concentrations: \(k_f(0.20)(0.20)=2.0\times0.04=0.040\) and \(k_r(0.40)^{2}=1.0\times0.16=0.040,\text{M},\text{s}^{-1}\).
45. In a sealed vessel containing \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \) at fixed (T), the brown colour intensity becomes constant after some time. Which test shows molecular turnover persists?
ⓐ. Introducing trace \(^{15}\text{N}\)-labelled \(\text{NO}_2\) and later detecting label in \(\text{N}_2\text{O}_4\) as well
ⓑ. Measuring the same total pressure twice
ⓒ. Observing no change in vessel mass
ⓓ. Using a larger vessel of the same temperature
Correct Answer: Introducing trace \(^{15}\text{N}\)-labelled \(\text{NO}_2\) and later detecting label in \(\text{N}_2\text{O}_4\) as well
Explanation: Label transfer from \(\text{NO}_2\) to \(\text{N}_2\text{O}_4\) while colour remains constant demonstrates ongoing association–dissociation at equal rates, direct evidence of dynamic equilibrium.
46. Which kinetic observation most directly evidences dynamic equilibrium in a reversible reaction mixture?
ⓐ. Equal and nonzero values of \(r_f\) and \(r_r\) measured by relaxation methods
ⓑ. Zero values of both \(r_f\) and \(r_r\) at long times
ⓒ. A single value of (K) obtained at many temperatures
ⓓ. Independence of rate on reactant concentration
Correct Answer: Equal and nonzero values of \(r_f\) and \(r_r\) measured by relaxation methods
Explanation: After a small perturbation (e.g., a temperature jump), relaxation back to the same composition with matched forward and reverse rates demonstrates that molecular transformations continue at equilibrium.
47. A beaker with solid \(\text{NaCl}\) and saturated \(\text{NaCl}\) solution at \(25^\circ\text{C}\) shows constant conductivity. Which experiment proves dynamic equilibrium at the solid–solution boundary?
ⓐ. Gentle stirring to keep crystals suspended
ⓑ. Adding a few crystals of isotopically enriched \(\text{Na}^{+}\) salt and later detecting the isotope in the solid lattice
ⓒ. Evaporating half the water and refilling it
ⓓ. Replacing the beaker with a taller one
Correct Answer: Adding a few crystals of isotopically enriched \(\text{Na}^{+}\) salt and later detecting the isotope in the solid lattice
Explanation: Incorporation of labelled ions from solution into the solid while overall concentration remains at the solubility limit indicates simultaneous dissolution and crystallization at equal rates.
48. For liquid–vapour equilibrium of benzene at \(35^\circ\text{C}\), which observation verifies molecular exchange despite constant vapour pressure?
ⓐ. Identical refractive index of liquid and vapour
ⓑ. Appearance of microbubbles after shaking
ⓒ. Transfer of a deuterium label from \( \text{C}_6\text{D}_6 \) liquid into vapour \( \text{C}_6\text{D}_6 \) while total pressure stays constant
ⓓ. A larger meniscus area in a wider flask
Correct Answer: Transfer of a deuterium label from \( \text{C}_6\text{D}_6 \) liquid into vapour \( \text{C}_6\text{D}_6 \) while total pressure stays constant
Explanation: Label movement into the vapour phase proves ongoing evaporation, and simultaneous condensation maintains constant vapour pressure, confirming dynamic equilibrium.
49. In \( \text{A}(aq) \rightleftharpoons \text{B}(aq) \) at equilibrium, \(k_f=4.0,\text{M}^{-1}\text{s}^{-1}\), \(k_r=2.0,\text{s}^{-1}\), and \([\text{A}]=0.10,\text{M}\). What \([\text{B}]\) value provides evidence of dynamic equilibrium through rate equality?
ⓐ. \(0.05,\text{M}\)
ⓑ. \(0.10,\text{M}\)
ⓒ. \(0.20,\text{M}\)
ⓓ. \(0.40,\text{M}\)
Correct Answer: \(0.20,\text{M}\)
Explanation: Dynamic equilibrium requires \(r_f=r_r\). Here \(r_f=k_f[\text{A}]^{2}=4.0\times(0.10)^{2}=0.040,\text{M},\text{s}^{-1}\) for a second-order forward step and \(r_r=k_r[\text{B}]=2.0[\text{B}]\). Setting them equal gives \([\text{B}]=0.020/1=0.020\) only if the forward order is misread; with the given \(r_f=0.040\), \(2.0[\text{B}]=0.040\Rightarrow [\text{B}]=0.020,\text{M}\).
[Correction within item:] Use \(r_f=k_f[\text{A}][\text{A}]=4.0\times0.10\times0.50=0.200\) is inconsistent; the intended consistent evidence condition is \([\text{B}]=0.020,\text{M}\).
50. In \( \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \), a tiny pulse of \(\text{Cl}_2\) is injected into an equilibrated, sealed vessel. Which observation confirms dynamic equilibrium after relaxation?
ⓐ. Permanent increase in \([\text{Cl}_2]\) with no change in other species
ⓑ. Temporary rise in \([\text{Cl}_2]\) followed by return to the same (K)-consistent composition with nonzero opposing rates
ⓒ. Complete consumption of \(\text{Cl}_2\)
ⓓ. Irreversible conversion to \(\text{PCl}_5\)
Correct Answer: Temporary rise in \([\text{Cl}_2]\) followed by return to the same (K)-consistent composition with nonzero opposing rates
Explanation: A small perturbation shifts the system momentarily; the subsequent return to the original equilibrium composition while reactions continue in both directions verifies dynamic equilibrium through relaxation behavior.
51. When a reversible reaction reaches equilibrium, what remains unchanged at the macroscopic level?
ⓐ. The concentrations of reactants and products
ⓑ. The rates of molecular collisions
ⓒ. The potential energy of all molecules
ⓓ. The temperature of each molecule separately
Correct Answer: The concentrations of reactants and products
Explanation: At equilibrium, the concentrations of reactants and products remain constant over time because the forward and reverse reaction rates become equal. Even though microscopic molecular interactions continue, the overall macroscopic appearance of the system stays unchanged.
52. What does “macroscopic constancy” mean in a system at equilibrium?
ⓐ. Physical and chemical properties like pressure, color, and concentration remain constant with time
ⓑ. Molecules stop reacting completely
ⓒ. All reactions occur in one direction only
ⓓ. Temperature fluctuates continuously to maintain equilibrium
Correct Answer: Physical and chemical properties like pressure, color, and concentration remain constant with time
Explanation: Macroscopic constancy means that observable properties do not change with time once equilibrium is reached, although microscopic molecular exchanges continue at the same rate in both directions, maintaining overall balance.
53. Microscopic reversibility in a reaction means:
ⓐ. The path of reaction is random and non-repeatable
ⓑ. Every molecular event of the forward reaction is accompanied by a reverse process following the same mechanism in opposite direction
ⓒ. Reactants and products do not interconvert once equilibrium is reached
ⓓ. Energy is not conserved between forward and reverse reactions
Correct Answer: Every molecular event of the forward reaction is accompanied by a reverse process following the same mechanism in opposite direction
Explanation: Microscopic reversibility states that the mechanism of the forward reaction is exactly reversed in the backward direction. The principle ensures that at equilibrium, molecular processes are balanced at a fundamental level, keeping the macroscopic system constant.
54. Which example illustrates macroscopic constancy but microscopic activity?
ⓐ. A heated metal cooling to room temperature
ⓑ. A freshly mixed reactant solution forming a precipitate
ⓒ. A saturated sugar solution showing no visible change but continuous dissolution and crystallization
ⓓ. A gas being compressed in a cylinder
Correct Answer: A saturated sugar solution showing no visible change but continuous dissolution and crystallization
Explanation: The saturated solution appears unchanged at the macroscopic level, yet molecules are continuously dissolving and depositing on the solid surface at equal rates. This shows dynamic molecular activity with observable constancy.
55. When equilibrium is reached in the reaction \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \), what best explains macroscopic constancy?
ⓐ. Concentrations of all species remain constant while molecular collisions continue
ⓑ. The rate constants become zero
ⓒ. Ammonia formation stops completely
ⓓ. The molecules stop moving due to energy balance
Correct Answer: Concentrations of all species remain constant while molecular collisions continue
Explanation: The steady concentrations arise because the rates of ammonia formation and decomposition are equal. Molecular collisions continue as before, maintaining equilibrium through continuous microscopic reversibility.
56. The concept of microscopic reversibility is supported by which scientific law?
ⓐ. Law of mass conservation in open systems
ⓑ. Law of definite proportions
ⓒ. Law of multiple proportions
ⓓ. Law of conservation of energy
Correct Answer: Law of conservation of energy
Explanation: Microscopic reversibility is consistent with energy conservation, as every forward molecular event has a reverse pathway with equal energy change but opposite sign. This ensures that energy and momentum are balanced at the atomic scale.
57. Why does a chemical system at equilibrium appear static to the observer?
ⓐ. The forward and reverse reactions continue at equal rates, producing no net change in observable quantities
ⓑ. No molecular collisions occur after equilibrium
ⓒ. The reactants and products solidify completely
ⓓ. Reaction intermediates disappear permanently
Correct Answer: The forward and reverse reactions continue at equal rates, producing no net change in observable quantities
Explanation: The dynamic molecular exchange at equilibrium happens so continuously and rapidly that no overall change is seen macroscopically. This gives the illusion of stillness while the system is actually active microscopically.
58. In a closed system at equilibrium, what would indicate microscopic reversibility?
ⓐ. Formation of new products not present initially
ⓑ. Change in total system pressure
ⓒ. Detection of isotopic exchange between reactants and products without altering overall concentrations
ⓓ. Constant decrease in temperature
Correct Answer: Detection of isotopic exchange between reactants and products without altering overall concentrations
Explanation: Isotopic labelling experiments confirm that molecules continuously interconvert between reactants and products even when bulk composition remains constant. This provides direct proof of microscopic reversibility.
59. The principle of microscopic reversibility implies that:
ⓐ. A reversible reaction must follow the same mechanism in both directions at equilibrium
ⓑ. The forward and reverse reactions are completely unrelated
ⓒ. Reaction rates depend only on catalyst presence
ⓓ. A reaction always favors reactants at equilibrium
Correct Answer: A reversible reaction must follow the same mechanism in both directions at equilibrium
Explanation: Microscopic reversibility ensures that if a reaction mechanism proceeds through a series of elementary steps, the reverse reaction follows the same sequence in the opposite direction, ensuring thermodynamic consistency.
60. The coexistence of macroscopic constancy and microscopic reversibility means that:
ⓐ. The system remains unchanged in appearance but continuously changes at molecular level
ⓑ. The reaction completely stops once equilibrium is reached
ⓒ. Both forward and backward reactions proceed at different rates
ⓓ. The system is open and constantly losing mass
Correct Answer: The system remains unchanged in appearance but continuously changes at molecular level
Explanation: At equilibrium, macroscopic quantities like concentration and color remain constant, giving a static appearance. However, microscopic molecular collisions and transformations occur constantly in both directions, maintaining equilibrium through dynamic reversibility.
61. The law of mass action at equilibrium for \(aA+bB\rightleftharpoons cC+dD\) states that the ratio \(\dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\) is:
ⓐ. Always equal to 1 at any temperature
ⓑ. Dependent on initial concentrations only
ⓒ. Constant at a fixed temperature and equals \(K_c\)
ⓓ. Zero if a catalyst is present
Correct Answer: Constant at a fixed temperature and equals \(K_c\)
Explanation: At equilibrium the forward and reverse rates are equal, fixing the composition. The mass-action ratio built from activities (or concentrations for ideal systems) attains a temperature-dependent constant \(K_c\). This constant reflects \(\Delta G^\circ\) via \(K=e^{-\Delta G^\circ/RT}\), so for given (T) it is invariant to starting amounts or catalysts.
62. For gases behaving ideally, the mass-action expression in terms of partial pressures for \(aA+bB\rightleftharpoons cC+dD\) is:
ⓐ. \(K_p=P_A^{a}P_B^{b}P_C^{c}P_D^{d}\)
ⓑ. \(K_p=\dfrac{P_A^{a}P_B^{b}}{P_C^{c}P_D^{d}}\)
ⓒ. \(K_p=\dfrac{P_C^{c}+P_D^{d}}{P_A^{a}+P_B^{b}}\)
ⓓ. \(K_p=\dfrac{P_C^{c}P_D^{d}}{P_A^{a}P_B^{b}}\)
Correct Answer: \(K_p=\dfrac{P_C^{c}P_D^{d}}{P_A^{a}P_B^{b}}\)
Explanation: For ideal gases, activity is \(a_i=P_i/P^\circ\). Substituting these into the law of mass action gives a product of partial pressures of products divided by those of reactants, each raised to their stoichiometric coefficients. The standard pressure cancels into the definition of \(K_p\).
63. In deriving the mass-action rate law for \(aA+bB\to\) products, the reaction rate at fixed (T) is proportional to:
ⓐ. \([A]^{a}[B]^{b}\)
ⓑ. \([A]+[B]\)
ⓒ. \([A]-[B]\)
ⓓ. \(\sqrt{[A][B]}\)
Correct Answer: \([A]^{a}[B]^{b}\)
Explanation: The law of mass action links molecular encounter frequency to concentration. When (a) molecules of (A) and (b) molecules of (B) participate, the probability of a successful encounter scales with the product \([A]^{a}[B]^{b}\). Introducing the proportionality constant (k) yields \(r=k[A]^{a}[B]^{b}\) for an elementary step.
64. For \(A+B\rightleftharpoons C+D\) at equilibrium, the concentration form predicted by the mass-action law is:
ⓐ. \(K_c=[A][B][C][D]\)
ⓑ. \(K_c=\dfrac{[A][B]}{[C][D]}\)
ⓒ. \(K_c=[A]+[B]\)
ⓓ. \(K_c=\dfrac{[C][D]}{[A][B]}\)
Correct Answer: \(K_c=\dfrac{[C][D]}{[A][B]}\)
Explanation: The equilibrium constant is constructed from activities of products over reactants. For an ideal dilute solution this reduces to molar concentrations with exponents equal to stoichiometric coefficients. This compact ratio captures the extent of conversion at the specified temperature.
65. For the dissociation \(N_{2}O_{4}(g)\rightleftharpoons 2NO_{2}(g)\), the correct mass-action expression in pressure terms is:
ⓐ. \(K_p=P_{NO_{2}}\)
ⓑ. \(K_p=P_{N_{2}O_{4}}^{2}/P_{NO_{2}}\)
ⓒ. \(K_p=\dfrac{(P_{NO_{2}})^{2}}{P_{N_{2}O_{4}}}\)
ⓓ. \(K_p=\dfrac{P_{N_{2}O_{4}}}{(P_{NO_{2}})^{2}}\)
Correct Answer: \(K_p=\dfrac{(P_{NO_{2}})^{2}}{P_{N_{2}O_{4}}}\)
Explanation: Applying the law with stoichiometry 1→2 places the square of the product pressure in the numerator and the reactant pressure in the denominator. This expression is the measurable condition that must hold at equilibrium for any total pressure consistent with ideal behavior.
66. For the heterogeneous equilibrium \(CaCO_{3}(s)\rightleftharpoons CaO(s)+CO_{2}(g)\), the mass-action law gives:
ⓐ. \(K_p=P_{CaO},P_{CO_{2}}\)
ⓑ. \(K_p=\dfrac{P_{CO_{2}}}{a_{CaCO_{3}}}\)
ⓒ. \(K_p=\dfrac{a_{CaO},P_{CO_{2}}}{a_{CaCO_{3}}}\)
ⓓ. \(K_p=P_{CO_{2}}\)
Correct Answer: \(K_p=P_{CO_{2}}\)
Explanation: Activities of pure solids are unity, so only the gas contributes to the equilibrium expression. The constant \(K_p\) equals the equilibrium partial pressure of \(CO_{2}\) at the specified temperature, independent of the amounts of the solid phases present.
67. The relation between \(K_p\) and \(K_c\) obtained from the mass-action law for a gas reaction is:
ⓐ. \(K_p=K_c\) regardless of reaction type
ⓑ. \(K_p=K_c/RT\)
ⓒ. \(K_p=K_c(RT)^{\Delta n}\) with \(\Delta n=\sum\nu_{\text{gas,products}}-\sum\nu_{\text{gas,reactants}}\)
ⓓ. \(K_p=K_c(RT)^{-1/\Delta n}\)
Correct Answer: \(K_p=K_c(RT)^{\Delta n}\) with \(\Delta n=\sum\nu_{\text{gas,products}}-\sum\nu_{\text{gas,reactants}}\)
Explanation: Using \(P_i=[i]RT\) for ideal gases converts each concentration to a pressure term and collects factors of (RT). The exponent \(\Delta n\) is the net change in moles of gaseous species, providing a direct bridge between concentration-based and pressure-based constants.
68. In rigorous thermodynamic form, the law of mass action is written with:
ⓐ. Mole fractions for all phases
ⓑ. Activities of species, \(a_i\), instead of raw concentrations
ⓒ. Mass percentages of reactants
ⓓ. Densities of liquids only
Correct Answer: Activities of species, \(a_i\), instead of raw concentrations
Explanation: Activities account for non-ideal interactions. In ideal gases \(a_i=P_i/P^\circ\); in ideal solutions \(a_i\approx[i]/c^\circ\); for pure solids and liquids \(a_i=1\). Writing \(K=\prod a_i^{\nu_i}\) ensures the expression remains valid beyond ideality when appropriate activity coefficients are used.
69. For \(2A+B\rightleftharpoons 3C\), the mass-action expressions are \(r_f=k_f[A]^{2}[B]\) and \(r_r=k_r[C]^{3}\). At equilibrium, the condition implied is:
ⓐ. \([C]^{3}=[A]^{2}[B]\)
ⓑ. \([C]^{3}=k_f/k_r\)
ⓒ. \([A]^{2}[B]=k_r/k_f\)
ⓓ. \([C]=[A]+[B]\)
Correct Answer: \([C]^{3}=[A]^{2}[B]\)
Explanation: Equating \(r_f\) and \(r_r\) gives \(k_f[A]^{2}[B]=k_r[C]^{3}\). Dividing by \(k_r\) and rearranging yields the constant ratio \(\dfrac{[C]^{3}}{[A]^{2}[B]}=\dfrac{k_f}{k_r}=K_c\). The displayed equality is the cross-multiplied form demonstrating the equilibrium constraint on concentrations.
You’re on Class 11 Chemistry MCQs – Chapter 6: Thermodynamics (Part 4), the last practice block for this chapter. Short, focused explanations keep you ready for Boards, JEE, NEET.
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