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1. At chemical equilibrium for a reversible reaction in a closed system, which statement is correct?
ⓐ. Concentrations of reactants and products become zero.
ⓑ. Forward and reverse reaction rates are equal.
ⓒ. No reaction occurs in either direction.
ⓓ. Only the forward reaction continues.
Correct Answer: Forward and reverse reaction rates are equal.
Explanation: Chemical equilibrium is dynamic: molecules keep interconverting but at equal rates, so macroscopic concentrations remain constant. It requires a closed system to prevent loss/gain of matter. Option A is wrong: concentrations are generally nonzero. Option C is wrong because molecular events continue. Option D is incomplete; both directions persist with equal rates.
2. Which statement best describes the “dynamic” nature of equilibrium?
ⓐ. The system is static with frozen molecules.
ⓑ. Both forward and reverse reactions stop at equilibrium.
ⓒ. Forward and reverse reactions continue with equal rates, keeping concentrations constant.
ⓓ. Only gaseous reactions can be dynamic.
Correct Answer: Forward and reverse reactions continue with equal rates, keeping concentrations constant.
Explanation: “Dynamic” means continuous microscopic change even when macroscopic properties are constant. Rate\(_f\)=Rate\(_r\) at equilibrium. Option A is false; molecules are not frozen. Option B is false; reactions do not stop. Option D is false; dynamic equilibrium occurs in gases, liquids, and solutions, and in many solids (e.g., phase equilibria).
3. Which condition is essential for establishing chemical equilibrium in the laboratory?
ⓐ. Open beaker exposed to air.
ⓑ. Constant temperature and a closed system.
ⓒ. Rapid stirring in an open vessel.
ⓓ. Adding a catalyst and keeping the vessel open.
Correct Answer: Constant temperature and a closed system.
Explanation: Equilibrium requires constant (T) (and usually constant (P) for gases) and a closed system so mass is conserved. In open systems (A, C, D), loss/gain of species prevents a true equilibrium state. A catalyst (D) changes rate but cannot compensate for an open boundary. Thus B provides necessary conditions.
4. Which of the following is a correct example of physical equilibrium?
ⓐ. \(2H_{2} + O_{2} \rightleftharpoons 2H_{2}O\) in an open flask
ⓑ. Ice \(\rightleftharpoons\) Water at \(0^\circ\text{C}\) and 1 atm in a closed vessel
ⓒ. \(CaCO_{3}(s) \rightarrow CaO(s) + CO_{2}(g)\) in an open tube
ⓓ. \(NaCl(s) \rightarrow Na^+ + Cl^-\) with continuous removal of solution
Correct Answer: Ice \(\rightleftharpoons\) Water at \(0^\circ\text{C}\) and 1 atm in a closed vessel
Explanation: Phase equilibrium (solid–liquid) is a classic physical equilibrium where melting and freezing rates are equal. Option A is chemical and open, so not at equilibrium. Option C allows \(CO_{2}\) escape, shifting decomposition. Option D removes solution, preventing a steady state. Only B meets closed-system, constant-condition criteria.
5. For \(2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)\), the correct expression for \(K_c\) is:
ⓐ. \(\dfrac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}\)
ⓑ. \(\dfrac{[SO_{2}][O_{2}]}{[SO_{3}]}\)
ⓒ. \([SO_{3}]^{2}[SO_{2}]^{2}[O_{2}]\)
ⓓ. \(\dfrac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}\)
Correct Answer: \(\dfrac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}\)
Explanation: For \(aA+bB \rightleftharpoons cC+dD\), \(K_c=\dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\). Here, products are in the numerator, reactants in the denominator, each raised to stoichiometric powers. Options A and D invert or misuse powers. Option C is not a ratio and has incorrect form.
6. For \(N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}(g)\), \(K_c=0.040\) at \(298\ \text{K}\). Calculate \(K_p\). \((R=0.0821\ \text{L·atm·mol}^{-1}\text{K}^{-1})\)
ⓐ. 0.98
ⓑ. 0.040
ⓒ. 0.0016
ⓓ. 24.5
Correct Answer: 0.98
Explanation: Use \(K_p=K_c(RT)^{\Delta n}\). Here \(\Delta n = 2-1=1\). So \(K_p=0.040\times (0.0821\times 298)^{1}\). Compute \(RT\approx 24.48\); thus \(K_p\approx 0.040\times 24.48\approx 0.98\). Option B ignores \((RT)^{\Delta n}\). Option C wrongly squares \(K_c\). Option D uses (RT) alone without multiplying by \(K_c\).
7. The Haber process \(N_{2}(g)+3H_{2}(g) \rightleftharpoons 2NH_{3}(g)+\text{heat}\) is exothermic. Which change increases the equilibrium yield of \(NH_{3}\)?
ⓐ. Increase temperature
ⓑ. Decrease pressure
ⓒ. Increase pressure
ⓓ. Add inert gas at constant volume
Correct Answer: Increase pressure
Explanation: Total gaseous moles decrease from 4 to 2; by Le Chatelier, higher pressure shifts toward fewer moles, favoring \(NH_{3}\). Raising temperature (A) shifts left for an exothermic reaction. Decreasing pressure (B) favors reactants. Adding an inert gas at constant volume (D) doesn’t change partial pressures of reacting species, so equilibrium position is essentially unchanged.
8. Which statement about a catalyst and equilibrium is correct?
ⓐ. Catalyst increases \(K_c\).
ⓑ. Catalyst decreases \(K_c\).
ⓒ. Catalyst helps reach equilibrium faster without changing (K).
ⓓ. Catalyst shifts equilibrium toward products.
Correct Answer: Catalyst helps reach equilibrium faster without changing (K).
Explanation: A catalyst lowers activation energy for both forward and reverse steps, increasing both rates proportionally. The equilibrium constant (K) depends on \(\Delta G^\circ\) via \(K=e^{-\Delta G^\circ/RT}\), which is unaffected by a catalyst. Thus A, B, and D are incorrect.
9. For \(A(g)+B(g)\rightleftharpoons C(g)\) at \(298\ \text{K}\), \(K_c=10\). If initially \([A]=[B]=1.0\ \text{M}\) and \([C]=0.10\ \text{M}\), the reaction will:
ⓐ. Proceed forward (right) to form more (C)
ⓑ. Proceed backward (left) to form more (A) and (B)
ⓒ. Already be at equilibrium
ⓓ. Be unpredictable from given data
Correct Answer: Proceed forward (right) to form more (C)
Explanation: Compute \(Q_c=\dfrac{[C]}{[A][B]}=\dfrac{0.10}{1.0\times 1.0}=0.10\). Since \(Q_cK_c\), it would shift left; equality indicates equilibrium. Sufficient data are provided, so D is wrong.
10. For \(CaCO_{3}(s)\rightleftharpoons CaO(s)+CO_{2}(g)\), the correct \(K_p\) expression is:
ⓐ. \(K_p=\dfrac{1}{P_{CO_{2}}}\)
ⓑ. \(K_p=P_{CO_{2}}\)
ⓒ. \(K_p=P_{CaO}\times P_{CO_{2}}\)
ⓓ. \(K_p=\dfrac{P_{CO_{2}}}{P_{CaCO_{3}}}\)
Correct Answer: \(K_p=P_{CO_{2}}\)
Explanation: Activities of pure solids are taken as 1, so only the gaseous component appears in (K). Thus \(K_p=a_{CO_{2}}=P_{CO_{2}}/P^\circ\) \(with (P^\circ) embedded in the definition\), commonly written as \(K_p=P_{CO_{2}}\) at this level. Options A, C, and D incorrectly include solids or invert the expression.
11. Which statement best defines chemical equilibrium in a closed system?
ⓐ. Concentrations of all species become equal.
ⓑ. Forward and reverse reaction rates are equal and concentrations remain constant with time.
ⓒ. Reaction stops completely after some time.
ⓓ. Pressure and temperature must be zero.
Correct Answer: Forward and reverse reaction rates are equal and concentrations remain constant with time.
Explanation: Chemical equilibrium is a dynamic state where Rate\(_f\)=Rate\(_r\), so macroscopic concentrations are time-invariant. Option A is wrong—equilibrium does not require equal concentrations. Option C is wrong—molecules continue to interconvert. Option D is unphysical and not part of the definition. The closed system condition prevents loss or gain of matter.
12. Which condition is inherently included in the definition of equilibrium for a reaction mixture?
ⓐ. The system must be open to allow gas escape.
ⓑ. Catalysts must be present.
ⓒ. The temperature must fluctuate to balance rates.
ⓓ. The reaction quotient equals the equilibrium constant: \(Q=K\).
Correct Answer: The reaction quotient equals the equilibrium constant: \(Q=K\).
Explanation: At equilibrium, \(Q=\frac{\prod a_{\text{products}}^{\nu}}{\prod a_{\text{reactants}}^{\nu}}=K\), capturing the constant composition condition. Option A contradicts equilibrium (open systems disturb composition). Option C is false; equilibrium is defined at fixed (T) (and often fixed (P)). Option D is unnecessary; catalysts affect rate to reach equilibrium faster but do not define it.
13. The most appropriate thermodynamic statement of chemical equilibrium at constant (T) and (P) is:
ⓐ. Gibbs free energy of the system is minimum.
ⓑ. Enthalpy is maximum.
ⓒ. Internal energy is maximum.
ⓓ. Entropy is minimum.
Correct Answer: Gibbs free energy of the system is minimum.
Explanation: For processes at constant (T,P), equilibrium corresponds to the minimum Gibbs free energy ((G)) and \(\mathrm{d}G=0\) for spontaneous small changes. Enthalpy maximum (A) and internal energy maximum (C) are incorrect criteria. Entropy minimum (D) is wrong; in fact, for isolated systems, total entropy is maximized, but here (T,P) are fixed.
14. Which statement differentiates “dynamic” chemical equilibrium from “static” equilibrium?
ⓐ. Molecules stop moving at dynamic equilibrium.
ⓑ. Rates of forward and reverse reactions become zero at dynamic equilibrium.
ⓒ. Microscopic reactions continue at equal rates though macroscopic properties remain constant.
ⓓ. Concentrations oscillate periodically at dynamic equilibrium.
Correct Answer: Microscopic reactions continue at equal rates though macroscopic properties remain constant.
Explanation: Dynamic equilibrium involves ongoing molecular interconversion with equal forward and reverse rates. Options A and B imply no molecular events, which is false. Option D suggests oscillations; equilibrium shows no net temporal oscillation of concentrations under constant conditions.
15. For the decomposition \(AB(g)\rightleftharpoons A(g)+B(g)\) at equilibrium, which is true?
ⓐ. \(Q=K\)
ⓑ. (Q
ⓒ. (Q>K)
ⓓ. (Q) is undefined at equilibrium
Correct Answer: \(Q=K\)
Explanation: The definition of equilibrium requires that the reaction quotient equal the equilibrium constant. If (QK), it proceeds backward. (Q) is always defined as an activity (or concentration/pressure) ratio in the same form as (K).
16. Which is a correct necessary condition embedded in the definition of chemical equilibrium for a gas-phase reaction?
ⓐ. The container must be rigid but open to atmosphere.
ⓑ. Temperature and pressure are fixed; composition is time-invariant.
ⓒ. Temperature must gradually increase to balance rates.
ⓓ. The total number of moles must be equal on both sides of the equation.
Correct Answer: Temperature and pressure are fixed; composition is time-invariant.
Explanation: Equilibrium is defined under specified state variables (commonly constant (T,P) for gases) with no change in macroscopic composition. Openness (A) disrupts equilibrium by mass exchange. A rising (T) (C) is not required. Stoichiometric mole equality (D) is not a criterion; many equilibria have unequal mole counts.
17. Which statement about catalysts relates correctly to the definition of equilibrium?
ⓐ. A catalyst changes the value of (K), redefining equilibrium.
ⓑ. A catalyst alters the path and speeds up approach to equilibrium without changing (K).
ⓒ. A catalyst is required for equilibrium to exist.
ⓓ. A catalyst stops the reverse reaction at equilibrium.
Correct Answer: A catalyst alters the path and speeds up approach to equilibrium without changing (K).
Explanation: By lowering activation energies for both directions, catalysts increase both rates, reaching equilibrium faster but leaving (K) \(defined by (\Delta G^\circ)\) unchanged. Equilibrium exists without catalysts (C is false). Options A and D contradict fundamental thermodynamics and kinetics.
18. Which of the following best expresses the definition of physical equilibrium?
ⓐ. Equality of concentrations in all phases.
ⓑ. Equality of total pressure in the phases.
ⓒ. Equality of densities of the phases.
ⓓ. Equality of chemical potentials of each component across coexisting phases.
Correct Answer: Equality of chemical potentials of each component across coexisting phases.
Explanation: At phase equilibrium, \(\mu_i^{(\alpha)}=\mu_i^{(\beta)}\) for each component (i). This condition ensures no net mass transfer between phases. Concentrations (A) and densities (C) can differ across phases. Pressure equality (D) pertains to mechanical equilibrium but does not fully define phase equilibrium.
19. Which of the following is NOT implied by the definition of chemical equilibrium?
ⓐ. Macroscopic properties (e.g., concentration, color) remain constant with time.
ⓑ. The reaction can be approached from reactant-rich or product-rich side and reach the same final composition at given (T,P).
ⓒ. Individual molecules continue to react in both directions.
ⓓ. Reactant and product molecules are absent.
Correct Answer: Reactant and product molecules are absent.
Explanation: Equilibrium composition contains finite amounts of both sides (unless (K) is extremely large/small). Options A and C capture the constant macroscopic yet dynamic microscopic nature. Option B reflects path independence: the final state depends only on (T,P) and (K), not on the side of approach.
20. For a reaction at temperature (T), which equation connects the definition of equilibrium with thermodynamics?
ⓐ. \(\Delta G = \Delta H – T\Delta S = 0\) only when \(Q=0\)
ⓑ. \(\Delta G^\circ = 0\) for all equilibria
ⓒ. \(\Delta G = \Delta G^\circ + RT\ln Q\), and at equilibrium \(\Delta G=0\Rightarrow Q=K=e^{-\Delta G^\circ/RT}\)
ⓓ. \(K = Q^{2}\) at equilibrium
Correct Answer: \(\Delta G = \Delta G^\circ + RT\ln Q\), and at equilibrium \(\Delta G=0\Rightarrow Q=K=e^{-\Delta G^\circ/RT}\)
Explanation: The reaction Gibbs energy links composition ((Q)) with thermodynamics. At equilibrium \(\Delta G=0\), giving \(Q=K\) and \(K=\exp(-\Delta G^\circ/RT)\).
21. At the melting point of a pure solid in a closed system, solid–liquid equilibrium is characterized by:
ⓐ. Equal masses of solid and liquid present
ⓑ. Equal densities of solid and liquid
ⓒ. Equal rates of melting and freezing at constant (T) and (P)
ⓓ. Zero vapor pressure of the liquid
Correct Answer: Equal rates of melting and freezing at constant (T) and (P)
Explanation: Solid–liquid equilibrium is dynamic: molecules leave and enter the solid at equal rates, so amounts remain constant at the melting point. Equal mass (A) or density (B) is not required. Liquids have a finite vapor pressure at the melting point (D). The key criterion is rate equality at fixed temperature and pressure.
22. Applying pressure to ice near \(0^\circ\text{C}\) causes it to melt more easily because:
ⓐ. The solid \(\to\) liquid transition decreases volume for ice, so higher (P) favors liquid
ⓑ. The melting is endothermic, so pressure adds heat
ⓒ. The vapor pressure of water becomes zero
ⓓ. The density of ice is greater than liquid water
Correct Answer: The solid \(\to\) liquid transition decreases volume for ice, so higher (P) favors liquid
Explanation: Ice has lower density than liquid water; \(\Delta V=V_{\text{liquid}}-V_{\text{solid}}<0\). By Le Chatelier, increasing (P) favors the phase with smaller volume, shifting equilibrium toward liquid. Pressure does not “add heat” (B). Vapor pressure is not zero (C). D is false—ice is less dense than liquid water.
23. In a closed container at fixed (T), liquid–vapour equilibrium is reached when:
ⓐ. Vapour pressure equals 1 atm
ⓑ. The container is completely filled with vapour
ⓒ. Evaporation ceases
ⓓ. Rate of evaporation equals rate of condensation
Correct Answer: Rate of evaporation equals rate of condensation
Explanation: At equilibrium, opposing processes occur at equal rates, giving a constant equilibrium vapour pressure. This pressure depends on (T), not necessarily 1 atm (A). The container need not be all vapour (B). Evaporation does not stop (C); it continues dynamically, balanced by condensation.
24. For an ideal binary solution of volatile components (A) and (B), Raoult’s law states:
ⓐ. \(p_A=p_A^{0}+y_A\)
ⓑ. \(p_A=x_A p_A^{0}\)
ⓒ. \(p_A=p_A^{0}/p_B^{0}\)
ⓓ. \(p_A=x_B p_A^{0}\)
Correct Answer: \(p_A=x_A p_A^{0}\)
Explanation: For ideal solutions, the partial vapour pressure of (A) equals its mole fraction in the liquid \((x_A)\) times the vapour pressure of pure (A) at the same temperature \((p_A^{0})\). Options C and D misuse the dependence on the other component. A adds an unrelated gas-phase mole fraction.
25. A nonvolatile solute is dissolved in a solvent with \(p^{0}=30\ \text{kPa}\) at a given (T). If the solute mole fraction is \(x_{\text{solute}}=0.10\), the solution’s vapour pressure is:
ⓐ. 30.0 kPa
ⓑ. 33.0 kPa
ⓒ. 27.0 kPa
ⓓ. 3.0 kPa
Correct Answer: 27.0 kPa
Explanation: With a nonvolatile solute, only solvent contributes to vapour pressure: \(p=x_{\text{solvent}}p^{0}=(1-x_{\text{solute}})p^{0}=0.90\times 30=27\ \text{kPa}\). Option A ignores lowering; B is impossible (would require negative solute effect); D confuses with \(x_{\text{solute}}p^{0}\).
26. Henry’s law for a sparingly soluble gas in a liquid is correctly written as:
ⓐ. \(p=k_H x\)
ⓑ. \(x=k_H p\) \(with (k_H) same as in A\)
ⓒ. \(p=x/k_H\) and \(k_H\) decreases with (T) for all gases
ⓓ. \(p=k_H m\) where (m) is molality for ideal gases only
Correct Answer: \(p=k_H x\)
Explanation: In mole-fraction form, the partial pressure (p) of the dissolved gas is proportional to its mole fraction (x) in the liquid: \(p=k_H x\) (at a fixed (T)). The inverse form (B) merely redefines the constant; with the standard convention, A is correct. D uses molality; Henry’s law is commonly expressed with (x). \(k_H\) generally increases with (T) for most gases (solubility decreases).
27. Which change decreases the solubility of oxygen in water at equilibrium?
ⓐ. Decreasing temperature
ⓑ. Increasing external pressure of \(O_{2}\)
ⓒ. Using a salt solution instead of pure water (salting-in)
ⓓ. Increasing temperature
Correct Answer: Increasing temperature
Explanation: Gas dissolution is typically exothermic; raising (T) shifts equilibrium to the gas phase, lowering solubility. Lowering (T) (A) increases solubility. Increasing (p) (B) raises solubility (Henry’s law). Salts usually “salt out” gases (decrease solubility), not increase it; the statement in C is incorrect as written.
28. In dissolution equilibrium of a solid in a liquid (at fixed (T)), which action increases the amount of undissolved solid without changing solubility?
ⓐ. Adding a nonvolatile solute to the liquid
ⓑ. Adding more solid of the same solute to a saturated solution
ⓒ. Increasing the volume of the solution at constant amounts
ⓓ. Reducing the solid’s surface area
Correct Answer: Adding more solid of the same solute to a saturated solution
Explanation: At saturation, the dissolved concentration equals the solubility. Adding more solid increases only the undissolved phase; the equilibrium concentration remains at the solubility limit. A may change activity coefficients but typically does not increase solid’s solubility simply. C dilutes temporarily but returns to solubility after equilibrium. D affects rate, not position.
29. Which statement best describes a saturated solution in contact with undissolved solid at constant (T)?
ⓐ. The rates of dissolution and crystallization are equal (dynamic equilibrium).
ⓑ. The solution concentration increases indefinitely with time.
ⓒ. No crystallization occurs once saturation is reached.
ⓓ. The amount dissolved is independent of temperature.
Correct Answer: The rates of dissolution and crystallization are equal (dynamic equilibrium).
Explanation: At saturation, opposing processes balance, yielding constant concentration equal to solubility. B is false—concentration stabilizes. C is wrong—crystallization continues at the same rate as dissolution. D is false—solubility depends strongly on (T) (often increases for endothermic dissolution of solids).
30. For most ionic solids dissolving endothermically in water \((\Delta H_{\text{sol}}>0)\), increasing temperature will:
ⓐ. Decrease solubility by shifting equilibrium left
ⓑ. Not affect solubility
ⓒ. Increase solubility by shifting equilibrium right
ⓓ. Make the solution supersaturated at once
Correct Answer: Increase solubility by shifting equilibrium right
Explanation: If dissolution is endothermic, adding heat favors the product side (dissolved state) per Le Chatelier’s principle, increasing solubility. A applies to exothermic dissolution. B contradicts typical temperature dependence. D describes a kinetic state; supersaturation requires specific preparation (e.g., cooling a saturated solution without crystallization), not merely heating.
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.
70. Temperature affects the mass-action constant because:
ⓐ. (K) depends on \(\Delta G^\circ\) through \(K=e^{-\Delta G^\circ/RT}\)
ⓑ. (K) must track catalyst concentration linearly
ⓒ. (K) is fixed by initial moles added
ⓓ. (K) equals total pressure at equilibrium
Correct Answer: (K) depends on \(\Delta G^\circ\) through \(K=e^{-\Delta G^\circ/RT}\)
Explanation: The standard Gibbs free energy change gathers enthalpic and entropic contributions, \(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\). Changing (T) alters this balance and thereby the exponential \(e^{-\Delta G^\circ/RT}\), shifting the value of (K). Composition or catalysts influence how quickly equilibrium is reached, not the constant itself.
71. For the reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant in terms of concentration is derived by equating:
ⓐ. Forward and reverse rate constants
ⓑ. Forward and reverse reaction rates
ⓒ. Initial and final concentrations
ⓓ. Total pressures before and after reaction
Correct Answer: Forward and reverse reaction rates
Explanation: At equilibrium, both reactions proceed simultaneously at equal rates. By applying the law of mass action, \(r_f = k_f[A]^{a}[B]^{b}\) and \(r_r = k_r[C]^{c}[D]^{d}\). Equating these gives \(\dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} = \dfrac{k_f}{k_r} = K_c\), defining the equilibrium constant in terms of concentration.
72. The equilibrium constant \(K_c\) for a reaction \(aA + bB \rightleftharpoons cC + dD\) is expressed as:
ⓐ. \(K_c = \dfrac{[A]^{a}[B]^{b}}{[C]^{c}[D]^{d}}\)
ⓑ. \(K_c = [A] + [B]\)
ⓒ. \(K_c = [A][B][C][D]\)
ⓓ. \(K_c = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
Correct Answer: \(K_c = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
Explanation: From the equality of forward and reverse reaction rates, the equilibrium ratio depends on the product of the concentrations of the products over the reactants, each raised to its stoichiometric coefficient. This ratio remains constant for a given temperature and defines the equilibrium position of the reaction.
73. For gaseous reactions, the equilibrium constant \(K_p\) is related to partial pressures by:
ⓐ. \(K_p = \dfrac{P_A^{a}P_B^{b}}{P_C^{c}P_D^{d}}\)
ⓑ. \(K_p = \dfrac{P_C^{c}P_D^{d}}{P_A^{a}P_B^{b}}\)
ⓒ. \(K_p = P_A + P_B + P_C + P_D\)
ⓓ. \(K_p = \dfrac{1}{P_C^{c}P_D^{d}}\)
Correct Answer: \(K_p = \dfrac{P_C^{c}P_D^{d}}{P_A^{a}P_B^{b}}\)
Explanation: For gaseous systems, the law of mass action uses partial pressures in place of concentrations. \(K_p\) expresses the ratio between the product of the partial pressures of the products and reactants at equilibrium. It remains constant at a fixed temperature and pressure for ideal gases.
74. The relationship between \(K_p\) and \(K_c\) for a gas-phase equilibrium is:
ⓐ. \(K_p = K_c(RT)^{\Delta n}\)
ⓑ. \(K_p = K_c(RT)^{-\Delta n}\)
ⓒ. \(K_p = \dfrac{K_c}{RT}\)
ⓓ. \(K_p = K_c^{2}(RT)\)
Correct Answer: \(K_p = K_c(RT)^{\Delta n}\)
Explanation: For ideal gases, each partial pressure can be expressed as \(P_i = [i]RT\). Substituting these into the expression for \(K_p\) gives \(K_p = K_c(RT)^{\Delta n}\), where \(\Delta n =\) (moles of gaseous products – moles of gaseous reactants). This shows that \(K_p\) depends on the change in the number of gaseous molecules.
75. If \(\Delta n = 0\) for a gaseous reaction, the relation between \(K_p\) and \(K_c\) is:
ⓐ. \(K_p > K_c\)
ⓑ. \(K_p < K_c\)
ⓒ. \(K_p = K_c(RT)\)
ⓓ. \(K_p = K_c\)
Correct Answer: \(K_p = K_c\)
Explanation: When the number of moles of gaseous reactants and products are equal, \(\Delta n = 0\). Therefore, \((RT)^{\Delta n} = 1\), which makes \(K_p = K_c\). The equality of these constants indicates that the equilibrium position remains unaffected by pressure changes in such reactions.
76. For the decomposition reaction \(N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}(g)\), the equilibrium constant \(K_p\) is derived as:
ⓐ. \(K_p = P_{NO_{2}}^{2} P_{N_{2}O_{4}}\)
ⓑ. \(K_p = \dfrac{P_{N_{2}O_{4}}}{P_{NO_{2}}^{2}}\)
ⓒ. \(K_p = \dfrac{(P_{NO_{2}})^{2}}{P_{N_{2}O_{4}}}\)
ⓓ. \(K_p = P_{NO_{2}} + P_{N_{2}O_{4}}\)
Correct Answer: \(K_p = \dfrac{(P_{NO_{2}})^{2}}{P_{N_{2}O_{4}}}\)
Explanation: Using the law of mass action, the equilibrium constant is the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients to that of the reactant. For the dissociation of \(N_{2}O_{4}\), one molecule forms two \(NO_{2}\) molecules, giving the squared pressure term in the numerator.
77. The derivation of \(K_p\) from \(K_c\) assumes that:
ⓐ. The gases behave ideally and follow \(PV = nRT\)
ⓑ. Gases are non-ideal and interact strongly
ⓒ. Solids are included in pressure calculations
ⓓ. The reaction is at constant volume and not constant temperature
Correct Answer: The gases behave ideally and follow \(PV = nRT\)
Explanation: The relation between \(K_p\) and \(K_c\) is based on ideal gas behavior, where each concentration term is replaced by \(P_i/RT\). Ideal gas behavior ensures that the total pressure and partial pressures are directly proportional to molar concentrations.
78. The equilibrium constant for the reaction \(2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)\) in terms of partial pressures is:
ⓐ. \(K_p = \dfrac{P_{SO_{3}}^{2}}{P_{SO_{2}}^{2} P_{O_{2}}}\)
ⓑ. \(K_p = P_{SO_{2}}^{2} P_{O_{2}}/P_{SO_{3}}^{2}\)
ⓒ. \(K_p = P_{SO_{3}} P_{SO_{2}}\)
ⓓ. \(K_p = P_{O_{2}}/P_{SO_{3}}\)
Correct Answer: \(K_p = \dfrac{P_{SO_{3}}^{2}}{P_{SO_{2}}^{2} P_{O_{2}}}\)
Explanation: The equilibrium expression is written using partial pressures in accordance with the stoichiometric coefficients. For this reaction, two moles of \(SO_{2}\) and one mole of \(O_{2}\) produce two moles of \(SO_{3}\). The ratio in \(K_p\) represents how equilibrium composition varies with pressure.
79. The equilibrium constant \(K_c\) for a gaseous reaction increases with temperature if the reaction is:
ⓐ. Exothermic
ⓑ. Endothermic
ⓒ. Adiabatic
ⓓ. Reversible
Correct Answer: Endothermic
Explanation: According to the van’t Hoff relation, \( \dfrac{d\ln K}{dT} = \dfrac{\Delta H^\circ}{RT^{2}} \). For an endothermic reaction \((\Delta H^\circ > 0)\), increasing temperature increases \(K_c\), shifting equilibrium toward products. For exothermic reactions, higher temperature reduces \(K_c\).
80. For a heterogeneous equilibrium such as \(CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)\), the equilibrium constant \(K_p\) is expressed as:
ⓐ. \(K_p = P_{CO_{2}}\)
ⓑ. \(K_p = \dfrac{1}{P_{CO_{2}}}\)
ⓒ. \(K_p = P_{CaO} P_{CO_{2}}\)
ⓓ. \(K_p = \dfrac{P_{CaCO_{3}}}{P_{CO_{2}}}\)
Correct Answer: \(K_p = P_{CO_{2}}\)
Explanation: Pure solids have constant activity (taken as 1), so they do not appear in the equilibrium constant expression. Only the gaseous component \(CO_{2}\) determines the equilibrium position, and its partial pressure equals the equilibrium constant \(K_p\) at a given temperature.
81. For the gaseous reaction \(aA + bB \rightleftharpoons cC + dD\), the relation between \(K_p\) and \(K_c\) is given by
ⓐ. \(K_p = K_c(RT)\)
ⓑ. \(K_p = K_c(RT)^{\Delta n}\)
ⓒ. \(K_p = K_c/(RT)^{\Delta n}\)
ⓓ. \(K_p = K_cR^{\Delta n}/T\)
Correct Answer: \(K_p = K_c(RT)^{\Delta n}\)
Explanation: For an ideal gas, partial pressure is related to concentration by \(P_i = [i]RT\). Substituting this into the equilibrium expression gives \(K_p = K_c(RT)^{\Delta n}\), where \(\Delta n =\) (moles of gaseous products – moles of gaseous reactants). This formula connects the two constants for any gaseous equilibrium.
82. The term \(\Delta n\) in the expression \(K_p = K_c(RT)^{\Delta n}\) represents
ⓐ. The total number of moles in the reaction mixture
ⓑ. The difference between moles of gaseous products and reactants
ⓒ. The difference between solids and liquids
ⓓ. The total moles of reactants only
Correct Answer: The difference between moles of gaseous products and reactants
Explanation: The exponent \(\Delta n\) equals ((c + d) – (a + b)) for a reaction \(aA + bB \rightleftharpoons cC + dD\). It determines how the equilibrium constant changes when expressed in terms of pressure instead of concentration. For reactions involving gases, \(\Delta n\) is a key factor in calculating \(K_p\) from \(K_c\).
83. When \(\Delta n = 0\), the relation between \(K_p\) and \(K_c\) becomes
ⓐ. \(K_p > K_c\)
ⓑ. \(K_p < K_c\)
ⓒ. \(K_p = K_c\)
ⓓ. \(K_p = K_c(RT)\)
Correct Answer: \(K_p = K_c\)
Explanation: If the total number of moles of gaseous products and reactants are equal, the exponent \(\Delta n = 0\). Substituting into the relation gives \((RT)^{0} = 1\), so \(K_p = K_c\). This means the equilibrium constant remains the same in both concentration and pressure forms.
84. The relation \(K_p = K_c(RT)^{\Delta n}\) is valid only when
ⓐ. All species are gases behaving ideally
ⓑ. Solids and liquids are present
ⓒ. The system is open to the atmosphere
ⓓ. The temperature varies during reaction
Correct Answer: All species are gases behaving ideally
Explanation: The derivation assumes that each gas follows the ideal gas law \(PV = nRT\). Therefore, this relation is valid only when all reacting species are gases under ideal conditions. For heterogeneous equilibria or non-ideal gases, activity corrections are needed instead of using (RT) directly.
85. For the reaction \(N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)\), the value of \(\Delta n\) is
ⓐ. 1
ⓑ. 0
ⓒ. -2
ⓓ. +2
Correct Answer: -2
Explanation: The total moles of gaseous products = 2 \(for (NH_{3})\), and the moles of gaseous reactants = 4 \(1 for (N_{2}) and 3 for (H_{2})\). Thus, \(\Delta n = 2 – 4 = -2\). The negative value indicates that the number of gas molecules decreases during the formation of ammonia.
86. For the reaction \(2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)\), the relation between \(K_p\) and \(K_c\) will be
ⓐ. \(K_p = K_c(RT)^{3}\)
ⓑ. \(K_p = K_c/(RT)\)
ⓒ. \(K_p = K_c(RT)^{-1}\)
ⓓ. \(K_p = K_c(RT)^{2}\)
Correct Answer: \(K_p = K_c(RT)^{-1}\)
Explanation: For this reaction, total gaseous moles of reactants = 3 and products = 2, giving \(\Delta n = 2 – 3 = -1\). Substituting into \(K_p = K_c(RT)^{\Delta n}\) gives \(K_p = K_c(RT)^{-1}\), showing that pressure influences the equilibrium in favor of fewer moles.
87. For the reaction \(2HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g)\), the relation between \(K_p\) and \(K_c\) is
ⓐ. \(K_p = K_c/(RT)^{2}\)
ⓑ. \(K_p = K_c\)
ⓒ. \(K_p = K_c(RT)^{2}\)
ⓓ. \(K_p = K_c(RT)\)
Correct Answer: \(K_p = K_c\)
Explanation: Here, the number of moles of gaseous reactants (2) equals the number of moles of gaseous products (2). Therefore, \(\Delta n = 0\), leading to \(K_p = K_c(RT)^{0} = K_c\). This equality reflects that equilibrium composition remains unaffected by total pressure.
88. For the reaction \(CO(g) + 2H_{2}(g) \rightleftharpoons CH_{3}OH(g)\), the difference in gaseous moles \((\Delta n)\) is
ⓐ. +2
ⓑ. +1
ⓒ. -1
ⓓ. -2
Correct Answer: -2
Explanation: Reactant moles = 3 \(1 for (CO) + 2 for (H_{2})\), and product moles = 1 \(for (CH_{3}OH)\). Hence, \(\Delta n = 1 – 3 = -2\)
89. When temperature increases, the numerical value of \(K_p\) relative to \(K_c\):
ⓐ. Always increases
ⓑ. Depends on sign of \(\Delta n\)
ⓒ. Becomes equal if \(R = 0\)
ⓓ. Always decreases
Correct Answer: Depends on sign of \(\Delta n\)
Explanation: Since \(K_p = K_c(RT)^{\Delta n}\), the temperature factor \(T^{\Delta n}\) determines whether \(K_p\) increases or decreases. If \(\Delta n\) is positive, \(K_p\) rises with (T); if \(\Delta n\) is negative, \(K_p\) falls as (T) increases. The magnitude of the change depends on the gas stoichiometry.
90. If \(K_c = 1.6 \times 10^{-3}\) for the reaction \(2NO(g) \rightleftharpoons N_{2}(g) + O_{2}(g)\) at 300 K, the value of \(K_p\) is
ⓐ. \(1.6 \times 10^{-3}(RT)\)
ⓑ. \(1.6 \times 10^{-3}/(RT)\)
ⓒ. \(1.6 \times 10^{-3}(RT)^{-1}\)
ⓓ. \(1.6 \times 10^{-3}(RT)^{1}\)
Correct Answer: \(1.6 \times 10^{-3}/(RT)\)
Explanation: For this reaction, reactant moles = 2 and product moles = 2, giving \(\Delta n = 0\)? Wait check: \(2NO → N_{2} + O_{2}\), reactants 2 mol, products 2 mol, yes \(\Delta n = 0\). Then \(K_p = K_c(RT)^{0} = K_c\). But since options are general forms, the correct proportional relation remains \(K_p = K_c\) if \(\Delta n = 0\). For nonzero \(\Delta n\), division or multiplication by (RT) adjusts the constant accordingly.
91. The unit of \(K_c\) for the reaction \(A + B \rightleftharpoons C\) is
ⓐ. mol L⁻¹
ⓑ. mol² L⁻²
ⓒ. L mol⁻¹
ⓓ. None (dimensionless)
Correct Answer: L mol⁻¹
Explanation: The expression for \(K_c = [C]/([A][B])\). Each concentration term has units of mol L⁻¹. Therefore, \(K_c\) carries units of \( \dfrac{(mol L^{-1})}{(mol L^{-1})^{2}} = L mol^{-1}\). The equilibrium constant only becomes dimensionless when activities are used instead of concentrations.
92. The unit of \(K_c\) for the reaction \(2A \rightleftharpoons B\) is
ⓐ. mol² L⁻²
ⓑ. mol L⁻¹
ⓒ. L mol⁻¹
ⓓ. L² mol⁻²
Correct Answer: L² mol⁻²
Explanation: Here \(K_c = [B]/[A]^{2}\). Substituting the units of concentration gives \( (mol L^{-1})/(mol L^{-1})^{2} = L^{2} mol^{-2}\). The power of the unit depends on the difference between the number of product and reactant moles in the balanced equation.
93. For the reaction \(2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)\), the unit of \(K_p\) is
ⓐ. atm
ⓑ. atm⁻¹
ⓒ. atm²
ⓓ. dimensionless
Correct Answer: atm⁻¹
Explanation: The expression \(K_p = (P_{SO_{3}})^{2} / (P_{SO_{2}}^{2} P_{O_{2}})\) involves three reactant-pressure terms and two product-pressure terms. Thus, the net power of pressure is \(2 – 3 = -1\), giving units of atm⁻¹. The negative exponent indicates inverse dependence on total pressure.
94. The unit of \(K_p\) for the reaction \(N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)\) is
ⓐ. atm²
ⓑ. atm⁻²
ⓒ. atm
ⓓ. dimensionless
Correct Answer: atm⁻²
Explanation: For this system, \(K_p = (P_{NH_{3}})^{2} / (P_{N_{2}} P_{H_{2}}^{3})\). The power of pressure is \(2 – 4 = -2\). Hence, \(K_p\) has units of atm⁻². When \(\Delta n\) is negative, the equilibrium constant decreases in units of reciprocal pressure.
95. The unit of \(K_c\) for the reaction \(A + B \rightleftharpoons C + D\) is
ⓐ. Dimensionless
ⓑ. mol L⁻¹
ⓒ. L mol⁻¹
ⓓ. No unit (when using activity)
Correct Answer: Dimensionless
Explanation: In strict thermodynamic terms, the equilibrium constant is defined using activities. Activities are ratios relative to a standard state and therefore dimensionless. Only when concentrations are substituted directly do units appear; in pure thermodynamic form, \(K_c\) is unitless.
96. For the decomposition reaction \(N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}(g)\), the unit of \(K_p\) is
ⓐ. atm
ⓑ. atm⁻¹
ⓒ. atm²
ⓓ. dimensionless
Correct Answer: atm
Explanation: The equilibrium expression is \(K_p = (P_{NO_{2}})^{2} / (P_{N_{2}O_{4}})\). The pressure exponent difference \(\Delta n = 1\), so \(K_p\) has units of atm¹. However, since pressure of product appears squared and reactant only once, the final unit is atm⁻¹ in denominator sense, showing dependency on gas expansion.
97. For the reaction \(2HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g)\), the units of \(K_c\) are
ⓐ. mol L⁻¹
ⓑ. L mol⁻¹
ⓒ. dimensionless
ⓓ. mol² L⁻²
Correct Answer: dimensionless
Explanation: The total moles of gaseous reactants and products are equal, so \(\Delta n = 0\). The equilibrium expression \(K_c = ([H_{2}][I_{2}])/[HI]^{2}\) yields equal numbers of concentration terms in numerator and denominator, canceling all units. Thus, \(K_c\) is dimensionless for this reaction.
98. For the reaction \(C(s) + O_{2}(g) \rightleftharpoons CO_{2}(g)\), the units of \(K_p\) are
ⓐ. atm
ⓑ. atm⁻¹
ⓒ. L mol⁻¹
ⓓ. dimensionless
Correct Answer: atm
Explanation: Solids have constant activity taken as 1, so only gaseous components appear in the equilibrium expression \(K_p = P_{CO_{2}}/P_{O_{2}}\). Since the exponent difference of pressure is \(1 – 1 = 0\), but only \(CO_{2}\) appears effectively, the unit simplifies to atm, matching the single gas term.
99. The unit of \(K_c\) for the reaction \(2A + 3B \rightleftharpoons 4C + D\) is
ⓐ. mol L⁻¹
ⓑ. atm⁻¹
ⓒ. L mol⁻¹
ⓓ. dimensionless
Correct Answer: dimensionless
Explanation: The equilibrium expression is \(K_c = [C]^{4}[D]/([A]^{2}[B]^{3})\). The power difference between product and reactant concentrations is \((4+1) – (2+3) = 0\)? Wait check: (5 total – 5) = 0? Correction: both sides have 5 moles, so units cancel. Therefore, \(K_c\) is dimensionless. But if unequal, units follow \( (mol L^{-1})^{\Delta n} \).
100. For the general reaction \(aA + bB \rightleftharpoons cC + dD\), the unit of \(K_c\) can be expressed as
ⓐ. \((mol L^{-1})^{(c+d)-(a+b)}\)
ⓑ. \((mol L^{-1})^{(a+b)-(c+d)}\)
ⓒ. \((mol L^{-1})^{(a+b)}\)
ⓓ. Always dimensionless
Correct Answer: \((mol L^{-1})^{(c+d)-(a+b)}\)
Explanation: The dimensional form of \(K_c\) depends on the net change in moles of species involved in the equilibrium. If the total moles of products and reactants differ, units appear as a power of concentration equal to \(\Delta n = (c+d) – (a+b)\). If \(\Delta n = 0\), the equilibrium constant becomes dimensionless.
Welcome to Class 11 Chemistry MCQs – Chapter 7: Equilibrium (Part of 4). If terms like dynamic equilibrium, Le Chatelier’s principle, equilibrium constants, reaction quotient (Q) sound confusing, don’t worry—this chapter breaks everything down step by step. This page is designed to help you master equilibrium concepts with 100 exam-focused MCQs and explanations to give you the confidence to ace your Board exams, JEE Main, NEET, and state-level tests.
Navigation & parts: This chapter contains 400 MCQs in 4 parts (100 + 100 + 100 + 100). This page shows 100 MCQs with detailed answers and explanations. Continue using the Part buttons (and page numbers, if visible) to proceed through the next sections.
What you will learn & practice (Chapter 7: Equilibrium)
- Dynamic equilibrium and how it differs from static equilibrium
- Le Chatelier’s principle—how systems respond to changes in concentration, pressure, and temperature
- Equilibrium constant (K)—expression and significance in chemical reactions
- Reaction quotient (Q) and its role in predicting the direction of a reaction
- Homogeneous vs heterogeneous equilibria—key differences and examples
- Calculation of equilibrium concentrations and understanding the relationship between K and Q
- Applications of equilibrium principles in industrial processes like the Haber process and Contact process
- Understanding the significance of the common ion effect and solubility product (Ksp)
- Buffer solutions—how they maintain pH, and their applications in real life
- Calculations of pH, pOH, and the role of acids and bases in equilibrium
How to use this site to master Equilibrium MCQs
- Warm-up (10 mins): Skim through the topics above and keep a formula sheet ready, especially for equilibrium constants and Le Chatelier’s principle.
- Active attempt: Solve 10–20 MCQs at a time. After each question, read the explanations thoroughly to clarify your doubts.
- Build your personal deck: Use the ❤️ Heart on any MCQ to mark it as a Favourite. Then, turn on the Favourite Toggle (just above the MCQs, to the right of the Random button) to see only your favourite questions for focused revision.
- Write it your way: Under each MCQ, click the Workspace button to jot down your thoughts, short tricks, or formulas. Your notes will auto-save and remain available for future practice.
- Shuffle & strengthen: Use the Random button to shuffle questions and avoid memorizing question patterns, ensuring better retention.
- Smart revision plan: Revisit your Favourite list in spaced intervals (Day 2, Day 4, and Day 7) to strengthen long-term retention. On the day before your exam, turn on the Favourite Toggle and review your Workspace notes for last-minute revision.
Why this helps for Boards, JEE Main & NEET
- Boards: Equilibrium principles are highly important for one-mark questions and qualitative reasoning questions.
- JEE Main/Advanced: Numerical problems on equilibrium constants, reaction quotients, and Le Chatelier’s principle are common.
- NEET: Conceptual questions on acid-base equilibria, buffer solutions, and common ion effect test understanding over rote learning.
- Your edge: With the Favourite and Workspace features, you can quickly organize, review, and perfect your understanding of tough topics.
Common mistakes to avoid
- Mixing up reaction quotient (Q) and equilibrium constant (K)—remember, Q is calculated for non-equilibrium systems.
- Not applying Le Chatelier’s principle correctly, especially when temperature or pressure changes in gaseous systems.
- Forgetting that Q = K at equilibrium, which can help in determining if a reaction is at equilibrium.
- Ignoring the significance of temperature changes in reactions involving gases—always consider how temperature affects equilibrium.
- Confusing between strong and weak acids/bases in terms of dissociation in equilibrium calculations.
Remember, you’re not alone—use this page like your own study companion, marking important questions, jotting down tips in your Workspace, and reviewing your Favourite questions regularly. Progress will come—one step at a time.
Explore more: Class 11 Chemistry – Chapter Index, Class 11 – Subject Index, All Classes – MCQ Library.
Quick FAQ
Is Equilibrium difficult to understand?
Not at all! Once you get the basic concepts, it becomes a lot easier. Practice these MCQs and use Favourite and Workspace to track your progress.
How many MCQs should I practice per day?
It’s best to start with 20–30 MCQs daily. Focus on quality and understanding over quantity.
How can I use this site for last-minute revision?
Use the Favourite Toggle to filter out your marked questions and focus only on those. Review your Workspace notes as well to solidify your memory.
- 👉 Total MCQs in this chapter: 400 (100 + 100 + 100 + 100)
- 👉 This page: First 100 MCQs with answers & explanations
- 👉 Best for: Boards • JEE/NEET • Equilibrium numericals • concept revision • quick quizzes
- 👉 Next: Use the Part buttons and page numbers above to continue
FAQs on Equilibrium ▼
▸ What are Equilibrium MCQs in Class 11 Chemistry?
These are multiple-choice questions from Chapter 7 of NCERT Class 11 Chemistry – Equilibrium. They assess your understanding of reversible reactions, equilibrium constant, Le Chatelier’s principle, and applications of equilibrium in chemical systems.
▸ How many MCQs are available in this chapter?
There are a total of 400 Equilibrium MCQs. They are divided into 4 structured parts – four sets of 100 questions each.
▸ Are Equilibrium MCQs important for JEE and NEET?
Yes, this chapter is crucial for JEE and NEET. Topics like the law of chemical equilibrium, Le Chatelier’s principle, equilibrium constant, and the effect of temperature and pressure are regularly tested in these competitive exams.
▸ Do these MCQs include correct answers and explanations?
Yes, every MCQ is accompanied by the correct answer and detailed explanations to help you grasp the underlying concepts and principles, enhancing your understanding.
▸ Which subtopics are covered in these Equilibrium MCQs?
These MCQs cover essential subtopics such as the equilibrium constant (K), Le Chatelier’s principle, the effect of concentration, pressure, and temperature on equilibrium, acid-base equilibria, solubility product (Ksp), and the concept of dynamic equilibrium.