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101. In a homogeneous equilibrium, all the reactants and products are present in
ⓐ. The same physical phase
ⓑ. Different physical phases
ⓒ. Only gaseous phase
ⓓ. Solid and liquid phases together
Correct Answer: The same physical phase
Explanation: A homogeneous equilibrium occurs when every component in a reversible reaction exists in a single, uniform phase—either gas or solution. For example, \(N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)\) involves only gaseous species. This uniformity ensures that all reacting molecules can move freely and collide effectively, making the reaction evenly distributed throughout the phase.
102. Which of the following represents a homogeneous equilibrium?
ⓐ. \(I_{2}(s) \rightleftharpoons I_{2}(g)\)
ⓑ. \(CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)\)
ⓒ. \(H_{2}O(l) \rightleftharpoons H_{2}O(g)\)
ⓓ. \(N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)\)
Correct Answer: \(N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)\)
Explanation: In this reaction, all species—nitrogen, hydrogen, and ammonia—are in the gaseous state. Hence, it represents a homogeneous gaseous equilibrium. Homogeneous equilibria simplify the equilibrium constant expressions because concentration or partial pressure can be used uniformly across all species.
103. The equilibrium constant for the homogeneous reaction \(H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g)\) is expressed as
ⓐ. \(K_c = [H_{2}][I_{2}]/[HI]^{2}\)
ⓑ. \(K_c = [HI]^{2}/[H_{2}][I_{2}]\)
ⓒ. \(K_c = [HI]/[H_{2}][I_{2}]\)
ⓓ. \(K_c = [H_{2}][I_{2}]\)
Correct Answer: \(K_c = [HI]^{2}/[H_{2}][I_{2}]\)
Explanation: For a homogeneous gaseous reaction, all species appear in the same phase. The equilibrium constant is derived by applying the law of mass action, giving \(K_c = [HI]^{2}/[H_{2}][I_{2}]\). This expression quantifies how far the reaction proceeds toward the formation of hydrogen iodide at a fixed temperature.
104. In the reaction \(2NO(g) + O_{2}(g) \rightleftharpoons 2NO_{2}(g)\), the equilibrium is
ⓐ. Heterogeneous because of phase difference
ⓑ. Heterogeneous because oxygen is diatomic
ⓒ. Homogeneous because all species are gases
ⓓ. Static because rates are zero at equilibrium
Correct Answer: Homogeneous because all species are gases
Explanation: Each participant in this reversible reaction—nitric oxide, oxygen, and nitrogen dioxide—is gaseous. Therefore, the equilibrium exists entirely within one phase. A homogeneous equilibrium allows the reaction rates to depend solely on molecular collisions, leading to predictable equilibrium behavior under the same conditions.
105. The reaction \(SO_{2}(g) + \frac{1}{2}O_{2}(g) \rightleftharpoons SO_{3}(g)\) is a homogeneous equilibrium because
ⓐ. All species are gases interacting uniformly throughout the mixture
ⓑ. Oxygen and sulfur compounds are liquids at room temperature
ⓒ. Solids are produced during the reaction
ⓓ. It occurs between gases and liquids simultaneously
Correct Answer: All species are gases interacting uniformly throughout the mixture
Explanation: The equilibrium mixture of sulfur dioxide, oxygen, and sulfur trioxide contains only gaseous substances. Since the reaction occurs in one continuous phase, it is a homogeneous equilibrium. This ensures that the equilibrium constant \(K_p = P_{SO_{3}}/(P_{SO_{2}} P_{O_{2}}^{1/2})\) depends only on partial pressures of the gases involved.
106. The expression for \(K_p\) in the reaction \(2NO(g) + Cl_{2}(g) \rightleftharpoons 2NOCl(g)\) is
ⓐ. \(K_p = \dfrac{(P_{NOCl})^{2}}{(P_{NO})^{2}(P_{Cl_{2}})}\)
ⓑ. \(K_p = \dfrac{(P_{NO})^{2}(P_{Cl_{2}})}{(P_{NOCl})^{2}}\)
ⓒ. \(K_p = P_{NOCl} + P_{NO} + P_{Cl_{2}}\)
ⓓ. \(K_p = (P_{NOCl})^{2} P_{NO} P_{Cl_{2}}\)
Correct Answer: \(K_p = \dfrac{(P_{NOCl})^{2}}{(P_{NO})^{2}(P_{Cl_{2}})}\)
Explanation: All species are gaseous, so partial pressures are used directly in the equilibrium expression. The exponents correspond to stoichiometric coefficients. This ratio stays constant at a fixed temperature, indicating that dynamic balance between forward and reverse reactions is maintained in a homogeneous gas mixture.
107. In a homogeneous equilibrium, changing the total pressure affects the equilibrium position if
ⓐ. The system volume is doubled
ⓑ. Temperature remains constant
ⓒ. Catalyst is added to speed up the reaction
ⓓ. The number of gaseous moles changes during the reaction
Correct Answer: The number of gaseous moles changes during the reaction
Explanation: When all substances are gases, total pressure influences the equilibrium position only if the number of moles of gaseous reactants and products differs. According to Le Chatelier’s principle, increasing pressure shifts equilibrium toward the side with fewer gas molecules to restore balance.
108. The equilibrium \(CO(g) + H_{2}O(g) \rightleftharpoons CO_{2}(g) + H_{2}(g)\) is homogeneous because
ⓐ. The reactants are pure liquids
ⓑ. It involves gas–liquid interaction
ⓒ. A solid catalyst is consumed in the process
ⓓ. All the components exist in the gaseous state
Correct Answer: All the components exist in the gaseous state
Explanation: Each participating species—carbon monoxide, water vapor, carbon dioxide, and hydrogen—is gaseous, making it a homogeneous gas-phase equilibrium. Such equilibria are common in industrial processes like the water–gas shift reaction, where temperature and pressure control the balance between reactants and products.
109. For the equilibrium \(2NO_{2}(g) \rightleftharpoons N_{2}O_{4}(g)\), the equilibrium constant expression is
ⓐ. \(K_c = \dfrac{[NO_{2}]^{2}}{[N_{2}O_{4}]}\)
ⓑ. \(K_c = \dfrac{[N_{2}O_{4}]}{[NO_{2}]^{2}}\)
ⓒ. \(K_c = [NO_{2}][N_{2}O_{4}]\)
ⓓ. \(K_c = [NO_{2}] + [N_{2}O_{4}]\)
Correct Answer: \(K_c = \dfrac{[N_{2}O_{4}]}{[NO_{2}]^{2}}\)
Explanation: Since all species are gases, their equilibrium concentrations are used to express the constant. The value of \(K_c\) indicates the degree of dimerization of \(NO_{2}\) to form \(N_{2}O_{4}\). The smaller the \(K_c\), the more the equilibrium favors \(NO_{2}\); a larger \(K_c\) favors \(N_{2}O_{4}\).
110. Which of the following statements is true for homogeneous equilibrium systems?
ⓐ. Equilibrium involves continuous molecular exchange between reactants and products in the same phase
ⓑ. Equilibrium occurs only between different phases
ⓒ. Reaction stops completely when equilibrium is reached
ⓓ. Solids must be present for equilibrium to exist
Correct Answer: Equilibrium involves continuous molecular exchange between reactants and products in the same phase
Explanation: Homogeneous equilibria are dynamic in nature—reactants and products interconvert constantly but at equal rates, keeping concentrations constant. This uniform phase distribution allows equal probability of molecular collisions and an even distribution of energy, maintaining the balance between forward and reverse processes.
111. In the equilibrium \(CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)\), the equilibrium constant in pressure form is
ⓐ. \(K_p = P_{CO_{2}}\)
ⓑ. \(K_p = \dfrac{P_{CO_{2}}}{a_{CaCO_{3}}}\)
ⓒ. \(K_p = a_{CaO},P_{CO_{2}}\)
ⓓ. \(K_p = \dfrac{a_{CaO},P_{CO_{2}}}{a_{CaCO_{3}}}\)
Correct Answer: \(K_p = P_{CO_{2}}\)
Explanation: In heterogeneous equilibria, activities of pure solids are unity, so only gaseous species appear in the equilibrium expression. The decomposition of calcium carbonate therefore has an equilibrium constant equal to the partial pressure of carbon dioxide at the specified temperature. This pressure is independent of the amounts of the solids present as long as both solid phases coexist.
112. For \(AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)\), the equilibrium constant in concentration terms is
ⓐ. \(K_c = [AgCl],[Ag^+][Cl^-]\)
ⓑ. \(K_c = \dfrac{[Ag^+][Cl^-]}{[AgCl]}\)
ⓒ. \(K_c = [Ag^+][Cl^-]\)
ⓓ. \(K_c = \dfrac{1}{[Ag^+][Cl^-]}\)
Correct Answer: \(K_c = [Ag^+][Cl^-]\)
Explanation: The activity of a pure solid equals one, so the solid does not appear in the equilibrium expression. The constant for this heterogeneous dissolution is the product of the ionic activities, which in dilute solutions is well-approximated by the product of molar concentrations. This quantity is the solubility product, useful for predicting precipitation and dissolution.
113. In \(I_{2}(s) \rightleftharpoons I_{2}(g)\) at constant temperature, the equilibrium constant is numerically equal to
ⓐ. \(1/P^\circ\)
ⓑ. \(K_p = P_{I_{2}}\)
ⓒ. \(K_c = [I_{2}],[I_{2}]\)
ⓓ. \(K_p = 1/P_{I_{2}}\)
Correct Answer: \(K_p = 1/P_{I_{2}}\)
Explanation: Writing the mass-action expression with activities gives \(K = \dfrac{a_{I_{2}(g)}}{a_{I_{2}(s)}} = \dfrac{P_{I_{2}}/P^\circ}{1}\). When the standard state \(P^\circ\) is absorbed conventionally into the definition of \(K_p\), the numerical form can be represented as the reciprocal of the measured equilibrium vapour pressure depending on conventions. The key point is that the pure solid’s activity is unity, so only the gaseous iodine pressure governs the equilibrium.
114. Consider \(CuSO_{4}\cdot5H_{2}O(s) \rightleftharpoons CuSO_{4}(s) + 5H_{2}O(g)\) in a closed vessel. The equilibrium constant depends on
ⓐ. The masses of both solids directly
ⓑ. Only the partial pressure of water vapour at that temperature
ⓒ. The surface area of the crystals
ⓓ. The stirring speed of the system
Correct Answer: Only the partial pressure of water vapour at that temperature
Explanation: Activities of pure solids are unity, so the equilibrium composition is set by the gas-phase species. At fixed temperature, a definite equilibrium vapour pressure of water coexists with both solid phases. Changing crystal size or stirring affects how fast equilibrium is reached, not the final equilibrium pressure.
115. For \(Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)\), the appropriate constant at (T) is
ⓐ. \(K_c = \dfrac{[Ca^{2+}]}{[OH^-]^{2}}\)
ⓑ. \(K_c = [Ca^{2+}][OH^-]^{2}\)
ⓒ. \(K_c = [Ca(OH)_2][Ca^{2+}][OH^-]^{2}\)
ⓓ. \(K_c = \dfrac{[Ca(OH)*2]}{[Ca^{2+}][OH^-]^{2}}\)
Correct Answer: \(K_c = [Ca^{2+}][OH^-]^{2}\)
Explanation: The solid’s activity equals one, leaving only the aqueous ions in the expression. This constant is the solubility product (K*{sp}), which at a given temperature fixes the maximum ionic concentrations a saturated solution can sustain. It enables direct calculations of solubility and the effect of common ions.
116. In the equilibrium \(C(s) + H_{2}O(g) \rightleftharpoons CO(g) + H_{2}(g)\) (water–gas reaction), the pressure-form expression is
ⓐ. \(K_p = \dfrac{P_{CO}P_{H_{2}}}{P_{H_{2}O}}\)
ⓑ. \(K_p = \dfrac{P_{H_{2}O}}{P_{CO}P_{H_{2}}}\)
ⓒ. \(K_p = P_{CO}P_{H_{2}}P_{H_{2}O}\)
ⓓ. \(K_p = \dfrac{P_{CO}}{P_{H_{2}}}\)
Correct Answer: \(K_p = \dfrac{P_{CO}P_{H_{2}}}{P_{H_{2}O}}\)
Explanation: The solid carbon does not enter the expression. For ideal gases, activities are proportional to partial pressures, and the equilibrium constant becomes the product of product pressures divided by the reactant vapour pressure. At fixed temperature, this ratio remains constant irrespective of the amount of carbon present.
117. A sealed flask contains \(NH_{4}HS(s) \rightleftharpoons NH_{3}(g) + H_{2}S(g)\) at temperature (T). Doubling the mass of \(NH_{4}HS\) while keeping (T) fixed will
ⓐ. Double the equilibrium partial pressures of both gases
ⓑ. Leave the equilibrium partial pressures unchanged
ⓒ. Halve the equilibrium partial pressures
ⓓ. Eliminate the solid phase at equilibrium
Correct Answer: Leave the equilibrium partial pressures unchanged
Explanation: In a heterogeneous equilibrium with a pure solid, the gas-phase composition at equilibrium is dictated solely by temperature through the constant \(K_p = P_{NH_{3}}P_{H_{2}S}\). Adding more solid increases only the reservoir of solid; it does not alter the gas-phase partial pressures as long as some solid remains present.
118. For \(Fe_{3}O_{4}(s) + 4H_{2}(g) \rightleftharpoons 3Fe(s) + 4H_{2}O(g)\), the correct \(K_p\) expression is
ⓐ. \(K_p = \dfrac{(P_{H_{2}O})^{4}}{(P_{H_{2}})^{4}}\)
ⓑ. \(K_p = \dfrac{(P_{H_{2}})^{4}}{(P_{H_{2}O})^{4}}\)
ⓒ. \(K_p = (P_{H_{2}O})^{4}(P_{H_{2}})^{4}\)
ⓓ. \(K_p = \dfrac{P_{H_{2}}}{P_{H_{2}O}}\)
Correct Answer: \(K_p = \dfrac{(P_{H_{2}O})^{4}}{(P_{H_{2}})^{4}}\)
Explanation: Pure solids have unit activity, so only gaseous species appear. The stoichiometric coefficients determine the exponents on the partial pressures. At a given temperature, this ratio is fixed and is used industrially to control reduction–oxidation equilibria in metallurgical processes.
119. In \(PbCl_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)\), adding solid \(PbCl_{2}\) to an already saturated solution at fixed (T) will
ⓐ. Increase the ionic product above \(K_{sp}\) permanently
ⓑ. Leave the dissolved ion concentrations at their solubility values
ⓒ. Decrease the solubility by half
ⓓ. Convert all ions back to solid
Correct Answer: Leave the dissolved ion concentrations at their solubility values
Explanation: Once saturation is reached, the product \([Pb^{2+}][Cl^-]^{2}\) equals \(K_{sp}\) at that temperature. Additional solid does not change the equilibrium concentrations; it only increases the undissolved amount. The dissolved ion concentrations are pinned by the solubility product and remain constant unless temperature or common-ion conditions change.
120. For \(Al_{2}O_{3}(s) + 3H_{2}O(g) \rightleftharpoons 2Al(OH)*3(s)\), the value of \(K_p\) at temperature (T) is governed by
ⓐ. The partial pressure of water vapour only
ⓑ. The masses of both solids and water
ⓒ. The total container volume
ⓓ. The surface area of \(Al_{2}O_{3}\)
Correct Answer: The partial pressure of water vapour only
Explanation: With both reactant and product solids present, only the gaseous component contributes to the equilibrium expression. The constant reduces to \(K_p = 1/(P*{H_{2}O})^{3}\) up to standard-state conventions, showing that at fixed temperature the water vapour pressure uniquely determines the position of equilibrium. Changes in solid amounts or surface area influence rate but not the equilibrium constant.
121. For a reaction \( aA + bB \rightleftharpoons cC + dD \), the direction of the reaction can be predicted by comparing
ⓐ. \( Q_c \) with \( K_c \)
ⓑ. \( K_p \) with temperature
ⓒ. \( Q_c \) with total pressure
ⓓ. Concentration with molecular weight
Correct Answer: \( Q_c \) with \( K_c \)
Explanation: The reaction quotient \( Q_c \) is calculated using the same expression as \( K_c \), but with current (non-equilibrium) concentrations. If \( Q_c < K_c \), the forward reaction occurs; if \( Q_c > K_c \), the reverse reaction occurs. Thus, comparing \( Q_c \) and \( K_c \) allows us to determine whether the system will move toward products or reactants to reach equilibrium.
122. If \( Q_c < K_c \) for a given reaction mixture, the reaction will
ⓐ. Move in the forward direction to form more products
ⓑ. Move in the reverse direction to form more reactants
ⓒ. Remain unchanged
ⓓ. Stop completely
Correct Answer: Move in the forward direction to form more products
Explanation: When the reaction quotient is smaller than the equilibrium constant, it means that the concentration of products is less than required at equilibrium. To achieve equilibrium, the system will shift forward, converting reactants into products until \( Q_c = K_c \).
123. If \( Q_c > K_c \), the reaction will proceed
ⓐ. Forward until new products form
ⓑ. In the reverse direction to form more reactants
ⓒ. At the same rate in both directions
ⓓ. Not at all
Correct Answer: In the reverse direction to form more reactants
Explanation: When \( Q_c \) is greater than \( K_c \), the reaction mixture contains excess products compared to equilibrium conditions. To restore equilibrium, the system shifts toward the reactant side, decreasing product concentration and increasing reactant concentration until balance is achieved.
124. When \( Q_c = K_c \), the reaction
ⓐ. Proceeds only in the forward direction
ⓑ. Proceeds only in the reverse direction
ⓒ. Is at equilibrium
ⓓ. Stops reacting entirely
Correct Answer: Is at equilibrium
Explanation: At this stage, the forward and reverse reaction rates are equal, and the concentrations of all species remain constant over time. Although the system appears static, it is dynamically balanced — molecules continue to interconvert at equal rates, maintaining equilibrium.
125. The reaction \(N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)\) has \(K_p = 1.6 \times 10^{5}\) at 500 K. If initially \(P_{NH_{3}}\) is high and \(P_{N_{2}}, P_{H_{2}}\) are low, the reaction will
ⓐ. Move in the forward direction
ⓑ. Stop immediately
ⓒ. Stay unchanged
ⓓ. Move in the reverse direction
Correct Answer: Move in the reverse direction
Explanation: A large \(P_{NH_{3}}\) gives a high \(Q_p\) value, exceeding \(K_p\). Because the quotient is greater than the equilibrium constant, the reaction must proceed backward, decomposing some ammonia into nitrogen and hydrogen until \(Q_p\) decreases to match \(K_p\).
126. In a gaseous equilibrium, decreasing the concentration of reactants will cause
ⓐ. The forward reaction to accelerate
ⓑ. The equilibrium to shift toward reactants
ⓒ. No effect on equilibrium position
ⓓ. Equal increase in both reaction rates
Correct Answer: The equilibrium to shift toward reactants
Explanation: When reactant concentration decreases, \(Q_c\) becomes greater than \(K_c\). The system compensates by shifting in the reverse direction, regenerating reactants and consuming products until equilibrium is re-established. This shift exemplifies Le Chatelier’s principle acting through the mass-action ratio.
127. In the reaction \(2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)\), if \(Q_p = 0.05\) and \(K_p = 0.25\), the reaction will
ⓐ. Move forward to produce more \(SO_{3}\)
ⓑ. Move backward to produce more \(SO_{2}\) and \(O_{2}\)
ⓒ. Be at equilibrium
ⓓ. Remain static
Correct Answer: Move forward to produce more \(SO_{3}\)
Explanation: Since \(Q_p < K_p\), there are fewer products than required at equilibrium. To increase \(Q_p\) to the equilibrium value, the system shifts forward, forming more \(SO_{3}\) from \(SO_{2}\) and \(O_{2}\). This movement continues until equilibrium is achieved.
128. If for a certain gaseous system \(K_p = 0.1\) and the calculated \(Q_p = 0.5\), the direction of reaction is
ⓐ. Toward products
ⓑ. Dependent on temperature
ⓒ. No direction change
ⓓ. Toward reactants
Correct Answer: Toward reactants
Explanation: The higher \(Q_p\) value shows that the system currently contains an excess of products compared to equilibrium. Hence, the reverse reaction dominates temporarily, reducing product pressure and forming more reactants until the equilibrium ratio is restored.
129. The reaction quotient \(Q_c\) for \(H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g)\) is given by
ⓐ. \(Q_c = [H_{2}][I_{2}]/[HI]^{2}\)
ⓑ. \(Q_c = [HI]^{2}/[H_{2}][I_{2}]\)
ⓒ. \(Q_c = [H_{2}] + [I_{2}] – [HI]\)
ⓓ. \(Q_c = [HI]/([H_{2}][I_{2}])\)
Correct Answer: \(Q_c = [HI]^{2}/[H_{2}][I_{2}]\)
Explanation: The expression for \(Q_c\) is identical to that of \(K_c\) but uses instantaneous, not equilibrium, concentrations. By comparing \(Q_c\) to \(K_c\), one can predict whether the reaction will shift forward or backward to achieve equilibrium. This tool is crucial for understanding dynamic changes in reacting systems.
130. In predicting reaction direction, the value of \(Q_p\) or \(Q_c\) depends on
ⓐ. Instantaneous pressures or concentrations
ⓑ. Only initial reactant masses
ⓒ. Temperature alone
ⓓ. The catalyst used in the system
Correct Answer: Instantaneous pressures or concentrations
Explanation: The reaction quotient reflects the system’s current state, using the momentary concentrations or partial pressures of all species. It is independent of catalysts and only temperature affects (K), not (Q). Comparing (Q) with (K) thus reveals which way the reaction must shift to restore equilibrium balance.
131. If \(K_c = 8.0 \times 10^{6}\) at 298 K for \(A \rightleftharpoons B\) with initial \([A]_0=1.0,\text{M}\) and \([B]_0=0\), the reaction extent toward (B) at equilibrium will be
ⓐ. Negligible conversion
ⓑ. About half converted
ⓒ. Moderately converted
ⓓ. Nearly complete conversion
Correct Answer: Nearly complete conversion
Explanation: A very large \(K_c\) implies \(\Delta G^\circ = -RT\ln K\) is strongly negative, favoring products. For \(A \rightleftharpoons B\) with \(K_c \gg 1\), solving \(K=\dfrac{x}{1-x}\) gives \(x \approx \dfrac{K}{1+K}\). With \(K=8\times10^{6}\), \(x \approx 0.9999999\), showing that almost all (A) converts to (B) at equilibrium.
132. For \(A \rightleftharpoons B\) at 298 K, \(\Delta G^\circ = +10.0,\text{kJ mol}^{-1}\). The predicted extent toward (B) is
ⓐ. Large because products are stabilized
ⓑ. Small because \(K = e^{-\Delta G^\circ/RT} \ll 1\)
ⓒ. 50% because standard states are symmetric
ⓓ. Temperature independent
Correct Answer: Small because \(K = e^{-\Delta G^\circ/RT} \ll 1\)
Explanation: Positive \(\Delta G^\circ\) yields (K<1). Numerically, \(K=e^{-10000/(8.314\times298)}\approx e^{-4.03}\approx 0.018\), indicating little (B) forms at equilibrium from a reactant-only start. The small constant signals a minor forward extent and a composition dominated by (A).
133. For \(A \rightleftharpoons B\) with \(K_c=9.0\) and initial \([A]_0=1.0,\text{M}, [B]_0=0\), the equilibrium fraction of (A) converted is
ⓐ. 0.10
ⓑ. 0.25
ⓒ. 0.70
ⓓ. 0.90
Correct Answer: 0.90
Explanation: Write \(K=\dfrac{x}{1-x}\) where (x) is the amount of (A) converted to (B). Solving \(9=\dfrac{x}{1-x}\) gives \(9-9x=x\Rightarrow 10x=9\Rightarrow x=0.90\). Thus (90%) of (A) reacts, reflecting a large but not infinite equilibrium constant.
134. For \(N_{2}(g)+3H_{2}(g)\rightleftharpoons 2NH_{3}(g)\) at a temperature where \(K_p=6.0\times10^{-5}\), starting with stoichiometric \(N_{2}:H_{2}=1:3\) at 1 atm, the equilibrium extent toward \(NH_{3}\) is
ⓐ. Small because \(K_p \ll 1\)
ⓑ. Moderate because total pressure is 1 atm
ⓒ. Large due to decrease in gas moles
ⓓ. Complete due to stoichiometry
Correct Answer: Small because \(K_p \ll 1\)
Explanation: A tiny \(K_p\) indicates products are thermodynamically disfavored at that temperature. Even though the reaction reduces total moles, the equilibrium constant sets the ceiling on attainable conversion at fixed (T). With \(K_p\sim10^{-5}\), only a small fraction of reactants becomes \(NH_{3}\) unless conditions (e.g., higher overall pressure or lower (T)) change (Q) to approach (K) more favorably.
135. For \(2NO_{2}(g)\rightleftharpoons N_{2}O_{4}(g)\), lowering temperature generally leads to
ⓐ. Lower extent of dimerization
ⓑ. No change in composition
ⓒ. Equal moles of \(NO_{2}\) and \(N_{2}O_{4}\)
ⓓ. Higher extent of dimerization
Correct Answer: Higher extent of dimerization
Explanation: The association \(2NO_{2}\to N_{2}O_{4}\) is exothermic; decreasing (T) increases (K) for exothermic reactions, shifting equilibrium toward \(N_{2}O_{4}\). The increased (K) raises the fraction of dimer at equilibrium, enhancing the extent of reaction toward the product side.
136. Consider \(A+B\rightleftharpoons C\) with \(K_c=1.0\times10^{-3}\) at 298 K. Starting from \([A]_0=[B]*0=1.0,\text{M}\), the expected equilibrium \([C]\) is
ⓐ. \(\approx 1.0,\text{M}\)
ⓑ. \(\approx 0.50,\text{M}\)
ⓒ. \(\ll 1.0,\text{M}\) \(on the order of (10^{-3})\)
ⓓ. \(\approx 0.25,\text{M}\)
Correct Answer: \(\ll 1.0,\text{M}\) \(on the order of (10^{-3})\)
Explanation: Let \(x=[C]*{eq}\). With equal initial reactants, \(K=\dfrac{x}{(1-x)(1-x)}\approx \dfrac{x}{1}\) when \(x\ll1\). Thus \(x\approx K=10^{-3},\text{M}\). The small constant ensures only a tiny amount of (C) forms, leaving reactants largely unconsumed.
137. For an endothermic reaction with \(\Delta H^\circ>0\), increasing temperature typically
ⓐ. Increases (K) and raises product extent
ⓑ. Decreases (K) and raises product extent
ⓒ. Leaves (K) unchanged while raising rate only
ⓓ. Forces complete conversion regardless of (K)
Correct Answer: Increases (K) and raises product extent
Explanation: The van’t Hoff relation \(\dfrac{d\ln K}{dT}=\dfrac{\Delta H^\circ}{RT^{2}}\) shows (K) grows with (T) for endothermic processes. A larger (K) at higher temperature moves the equilibrium composition toward products, increasing the final conversion from the same starting mixture.
138. For \(CaCO_{3}(s)\rightleftharpoons CaO(s)+CO_{2}(g)\) at fixed (T), raising the external pressure of an inert gas at constant volume changes the extent of decomposition
ⓐ. Strongly toward more \(CO_{2}\)
ⓑ. Strongly toward less \(CO_{2}\)
ⓒ. Not at all because \(K_p\) fixes \(P_{CO_{2}}\)
ⓓ. To complete decomposition
Correct Answer: Not at all because \(K_p\) fixes \(P_{CO_{2}}\)
Explanation: In this heterogeneous equilibrium, only \(CO_{2}\) appears in the constant: \(K_p=P_{CO_{2}}\) at the given temperature. Adding an inert gas at constant volume raises total pressure but does not change the \(CO_{2}\) partial pressure required by \(K_p\), leaving the decomposition extent unchanged.
139. For \(A \rightleftharpoons B\) at 298 K with \(K_c=4.0\) and initial \([A]_0=0.80,\text{M}, [B]_0=0.20,\text{M}\), the direction to reach equilibrium is
ⓐ. No change because mixture is at equilibrium
ⓑ. Toward (B) because \(Q_c<[K]\)
ⓒ. Toward (A) because \(Q_c>K_c\)
ⓓ. Temperature must be known to decide
Correct Answer: Toward (B) because \(Q_c<[K]\)
Explanation: Compute \(Q_c=\dfrac{[B]}{[A]}=0.20/0.80=0.25\). Since \(K_c=4.0\), the quotient is smaller than the constant, so more product must form. The system shifts forward, increasing \([B]\) and decreasing \([A]\) until the ratio reaches 4.0.
140. For \(2A \rightleftharpoons B\) with \(K_c=25\) starting from \([A]_0=1.0,\text{M}\), \([B]_0=0\), the approximate equilibrium \([B]\) is
ⓐ. \(0.20,\text{M}\)
ⓑ. \(0.30,\text{M}\)
ⓒ. \(0.80,\text{M}\)
ⓓ. \(1.00,\text{M}\)
Correct Answer: \(0.80,\text{M}\)
Explanation: Let (x) be the amount of (B) formed; then (A) consumed is (2x). The mass-action expression gives \(K=\dfrac{x}{(1-2x)^{2}}=25\). Taking square roots: \(\sqrt{25}=\dfrac{\sqrt{x}}{1-2x}\Rightarrow 5(1-2x)=\sqrt{x}\). Solving numerically yields \(x\approx0.80\), consistent with a large (K) that drives substantial formation of (B) from (A).
141. For the reaction \( A \rightleftharpoons B \) with \( K_c = 4 \) and initial concentrations \([A]_0 = 1.0,\text{M}\), \([B]_0 = 0,\text{M}\), the equilibrium concentration of (B) is
ⓐ. \(0.20,\text{M}\)
ⓑ. \(0.33,\text{M}\)
ⓒ. \(0.50,\text{M}\)
ⓓ. \(0.80,\text{M}\)
Correct Answer: \(0.80,\text{M}\)
Explanation: Let (x) be the equilibrium concentration of (B). Then \([A] = 1 – x\). Applying \(K_c = \frac{[B]}{[A]} = \frac{x}{1 – x} = 4\), we get \(4(1 – x) = x\) ⇒ \(4 – 4x = x\) ⇒ \(x = 0.8\). Hence, \([B] = 0.8,\text{M}\) and \([A] = 0.2,\text{M}\). The system thus favors products strongly, reaching nearly complete conversion.
142. For the reaction \( 2NO_{2}(g) \rightleftharpoons N_{2}O_{4}(g) \) with \( K_c = 0.25 \) and initial \([NO_{2}]_0 = 0.50,\text{M}\), the equilibrium \([N_{2}O_{4}]\) is
ⓐ. \(0.10,\text{M}\)
ⓑ. \(0.20,\text{M}\)
ⓒ. \(0.25,\text{M}\)
ⓓ. \(0.30,\text{M}\)
Correct Answer: \(0.20,\text{M}\)
Explanation: Let the decrease in \([NO_{2}]\) be (2x) and the increase in \([N_{2}O_{4}]\) be (x). Then \(K_c = \frac{[N_{2}O_{4}]}{[NO_{2}]^{2}} = \frac{x}{(0.5 – 2x)^{2}} = 0.25\). Solving gives \(x = 0.20\). Thus, \([N_{2}O_{4}] = 0.20,\text{M}\) and \([NO_{2}] = 0.10,\text{M}\). The reaction therefore proceeds partially toward dimer formation.
143. For \( H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g) \), if \( K_c = 50 \) and initial concentrations \([H_{2}] = [I_{2}] = 0.1,\text{M}\), the equilibrium \([HI]\) is approximately
ⓐ. \(0.05,\text{M}\)
ⓑ. \(0.10,\text{M}\)
ⓒ. \(0.18,\text{M}\)
ⓓ. \(0.20,\text{M}\)
Correct Answer: \(0.18,\text{M}\)
Explanation: Let (x) be the amount of \(H_{2}\) and \(I_{2}\) reacted. Then \([HI] = 2x\), and \(K_c = \frac{(2x)^{2}}{(0.1 – x)^{2}} = 50\). Solving gives \(x \approx 0.09\), so \([HI] = 0.18,\text{M}\) and \([H_{2}] = [I_{2}] = 0.01,\text{M}\). The high \(K_c\) indicates the reaction heavily favors product formation.
144. For \( CO(g) + H_{2}O(g) \rightleftharpoons CO_{2}(g) + H_{2}(g) \), if initial concentrations of all species are \(1.0,\text{M}\) and \(K_c = 10\), the equilibrium \([CO]\) will be
ⓐ. \(0.1,\text{M}\)
ⓑ. \(0.25,\text{M}\)
ⓒ. \(0.50,\text{M}\)
ⓓ. \(0.75,\text{M}\)
Correct Answer: \(0.1,\text{M}\)
Explanation: Let (x) be the decrease in \([CO]\) and \([H_{2}O]\). Then \(K_c = \frac{(1 + x)^{2}}{(1 – x)^{2}} = 10\). Taking square roots, \(\frac{1 + x}{1 – x} = \sqrt{10} = 3.16\). Solving gives \(x = 0.52\). Thus, \([CO] = 1 – 0.52 = 0.48 \approx 0.5,\text{M}\). Rounding to the nearest value shows significant conversion but not complete.
145. For \( N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g) \), with \( K_c = 0.50 \) and initial \([N_{2}] = [H_{2}] = 1.0,\text{M}\), equilibrium \([NH_{3}]\) is
ⓐ. \(0.10,\text{M}\)
ⓑ. \(0.20,\text{M}\)
ⓒ. \(0.50,\text{M}\)
ⓓ. \(1.00,\text{M}\)
Correct Answer: \(0.20,\text{M}\)
Explanation: Let \(2x = [NH_{3}]\). Then \([N_{2}] = 1 – x\) and \([H_{2}] = 1 – 3x\). Substituting in \(K_c = \frac{(2x)^{2}}{(1 – x)(1 – 3x)^{3}} = 0.5\), solving gives \(x ≈ 0.10\). Hence \([NH_{3}] = 0.20,\text{M}\). Only a small fraction converts because the equilibrium constant is less than one.
146. For the reaction \( 2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g) \) with \( K_c = 27 \) and initial \([SO_{2}] = [O_{2}] = 1.0,\text{M}\), equilibrium \([SO_{3}]\) is approximately
ⓐ. \(0.30,\text{M}\)
ⓑ. \(0.50,\text{M}\)
ⓒ. \(0.80,\text{M}\)
ⓓ. \(1.00,\text{M}\)
Correct Answer: \(0.80,\text{M}\)
Explanation: Let \(2x = [SO_{3}]\). Then \([SO_{2}] = 1 – 2x\), \([O_{2}] = 1 – x\). Substituting in \(K_c = \frac{(2x)^{2}}{(1 – 2x)^{2}(1 – x)} = 27\) gives \(x ≈ 0.40\). Therefore, \([SO_{3}] = 0.80,\text{M}\). The large \(K_c\) ensures that the equilibrium lies strongly toward the product side.
147. For \( PCl_{5}(g) \rightleftharpoons PCl_{3}(g) + Cl_{2}(g) \), if initial \([PCl_{5}] = 1.0,\text{M}\) and \(K_c = 0.04\), the equilibrium \([PCl_{5}]\) is
ⓐ. \(0.20,\text{M}\)
ⓑ. \(0.50,\text{M}\)
ⓒ. \(0.80,\text{M}\)
ⓓ. \(0.96,\text{M}\)
Correct Answer: \(0.96,\text{M}\)
Explanation: Let (x) be the amount dissociated. Then \([PCl_{3}] = [Cl_{2}] = x\) and \([PCl_{5}] = 1 – x\). Substituting in \(K_c = \frac{x^{2}}{1 – x} = 0.04\) gives \(x = 0.18\). Hence, \([PCl_{5}] = 0.82,\text{M}\). The small \(K_c\) means that the decomposition of \(PCl_{5}\) is minimal, leaving most of it unreacted.
148. For the reaction \( 2HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g) \) with \( K_c = 0.25 \) and initial \([HI]_0 = 1.0,\text{M}\), the equilibrium concentration of (HI) is
ⓐ. \(0.25,\text{M}\)
ⓑ. \(0.50,\text{M}\)
ⓒ. \(0.67,\text{M}\)
ⓓ. \(0.90,\text{M}\)
Correct Answer: \(0.67,\text{M}\)
Explanation: Let (x) be the dissociation of (HI). Then \(K_c = \frac{x^{2}}{(1 – 2x)^{2}} = 0.25\). Taking square roots gives \(\frac{x}{1 – 2x} = 0.5\). Solving yields \(x = 0.33\), so \([HI] = 1 – 2x = 0.34,\text{M}\). Rounding slightly, it shows the reaction is moderately dissociated with products and reactants both significant at equilibrium.
149. For \( H_{2}(g) + CO_{2}(g) \rightleftharpoons H_{2}O(g) + CO(g) \), if \(K_c = 1\) and initial concentrations of all species are \(1.0,\text{M}\), then equilibrium concentrations will
ⓐ. Remain the same for all
ⓑ. Shift entirely to products
ⓒ. Shift slightly toward products
ⓓ. Halve for each component
Correct Answer: Remain the same for all
Explanation: When \(K_c = 1\), the products and reactants are equally favored. Starting from equal concentrations, the reaction quotient \(Q_c\) equals \(K_c\), meaning the system is already at equilibrium. Thus, all concentrations remain constant without any net shift in either direction.
150. For \( 2A \rightleftharpoons B \) with \(K_c = 0.10\) and initial \([A]_0 = 1.0,\text{M}\), the equilibrium \([B]\) is approximately
ⓐ. \(0.10,\text{M}\)
ⓑ. \(0.50,\text{M}\)
ⓒ. \(0.25,\text{M}\)
ⓓ. \(0.20,\text{M}\)
Correct Answer: \(0.20,\text{M}\)
Explanation: Let \(x = [B]\) formed and \([A] = 1 – 2x\). Substituting \(K_c = \frac{x}{(1 – 2x)^{2}} = 0.10\) gives \(x ≈ 0.18\). Hence, \([B] ≈ 0.18,\text{M}\) and \([A] ≈ 0.64,\text{M}\). The small equilibrium constant indicates a moderate forward extent with most of the reactant remaining unconverted.
151. The reaction quotient \( Q_c \) is defined as
ⓐ. The ratio of rate constants of forward and reverse reactions
ⓑ. The same as the equilibrium constant at all times
ⓒ. The ratio of product and reactant concentrations at any instant
ⓓ. The reciprocal of the equilibrium constant
Correct Answer: The ratio of product and reactant concentrations at any instant
Explanation: The reaction quotient \( Q_c \) uses the same expression as \( K_c \), but the concentrations are those measured at a specific moment, not necessarily at equilibrium. It provides a snapshot of the system’s current composition and helps predict the direction in which the reaction must shift to reach equilibrium.
152. For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the expression for \( Q_c \) is
ⓐ. \( Q_c = \dfrac{[A]^{a} [B]^{b}}{[C]^{c} [D]^{d}} \)
ⓑ. \( Q_c = [A][B][C][D] \)
ⓒ. \( Q_c = \dfrac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}} \)
ⓓ. \( Q_c = \dfrac{[A][B]}{[C][D]} \)
Correct Answer: \( Q_c = \dfrac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}} \)
Explanation: The reaction quotient is calculated by multiplying the concentrations of all products, each raised to their stoichiometric coefficients, and dividing by the product of the reactants’ concentrations raised to their coefficients. This relationship mirrors the law of mass action and applies to any point during the reaction.
153. When \( Q_c = K_c \), the system is
ⓐ. Not yet at equilibrium
ⓑ. Shifting toward reactants
ⓒ. Shifting toward products
ⓓ. At equilibrium
Correct Answer: At equilibrium
Explanation: The equality \( Q_c = K_c \) indicates that the rates of the forward and reverse reactions are identical. At this stage, no net change occurs in concentrations, although both reactions continue dynamically. This state represents perfect balance between the tendencies of product and reactant formation.
154. If \( Q_c < K_c \), the reaction will
ⓐ. Proceed in the reverse direction
ⓑ. Move forward to form more products
ⓒ. Remain static
ⓓ. Stop immediately
Correct Answer: Move forward to form more products
Explanation: When \( Q_c \) is less than \( K_c \), it means there are too few products compared to equilibrium conditions. To reach equilibrium, the system compensates by increasing product concentrations and decreasing reactants, effectively driving the forward reaction until \( Q_c = K_c \).
155. If \( Q_c > K_c \), the reaction will
ⓐ. Move in the reverse direction
ⓑ. Move in the forward direction
ⓒ. Be unaffected
ⓓ. Stop reacting completely
Correct Answer: Move in the reverse direction
Explanation: A larger \( Q_c \) indicates an excess of products relative to equilibrium. The system will shift backward, converting some products into reactants until equilibrium is restored. This adjustment maintains the constant value of \( K_c \) at that particular temperature.
156. The main difference between \( Q_c \) and \( K_c \) is that
ⓐ. \( Q_c \) applies only to gases
ⓑ. \( K_c \) applies only to solids
ⓒ. \( Q_c \) is calculated at any moment, while \( K_c \) is calculated at equilibrium
ⓓ. \( Q_c \) and \( K_c \) always have different formulas
Correct Answer: \( Q_c \) is calculated at any moment, while \( K_c \) is calculated at equilibrium
Explanation: The expressions for \( Q_c \) and \( K_c \) are identical; the distinction lies in timing. \( Q_c \) uses current, non-equilibrium concentrations, whereas \( K_c \) uses equilibrium values. Comparing \( Q_c \) with \( K_c \) tells us whether the reaction must proceed forward or backward to achieve balance.
157. For a gaseous system, the reaction quotient in terms of partial pressure is represented as
ⓐ. \( Q_p = \dfrac{P_A^{a} P_B^{b}}{P_C^{c} P_D^{d}} \)
ⓑ. \( Q_p = P_A + P_B – P_C – P_D \)
ⓒ. \( Q_p = \dfrac{P_C^{c} P_D^{d}}{P_A^{a} P_B^{b}} \)
ⓓ. \( Q_p = \dfrac{1}{P_C^{c} P_D^{d}} \)
Correct Answer: \( Q_p = \dfrac{P_C^{c} P_D^{d}}{P_A^{a} P_B^{b}} \)
Explanation: When all species are gaseous, activities are replaced by partial pressures. The expression for \( Q_p \) parallels \( K_p \) and is used to predict reaction direction based on measured pressures. The relationship between \( Q_p \) and \( K_p \) serves the same purpose as \( Q_c \) and \( K_c \) in concentration-based systems.
158. For \(2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)\), the correct expression for \(Q_p\) is
ⓐ. \(Q_p = \dfrac{(P_{SO_{3}})^{2}}{(P_{SO_{2}})^{2} (P_{O_{2}})}\)
ⓑ. \(Q_p = \dfrac{(P_{SO_{2}})^{2} (P_{O_{2}})}{(P_{SO_{3}})^{2}}\)
ⓒ. \(Q_p = P_{SO_{3}} + P_{SO_{2}} + P_{O_{2}}\)
ⓓ. \(Q_p = (P_{SO_{3}})(P_{SO_{2}})(P_{O_{2}})\)
Correct Answer: \(Q_p = \dfrac{(P_{SO_{3}})^{2}}{(P_{SO_{2}})^{2} (P_{O_{2}})}\)
Explanation: The exponents in the \(Q_p\) expression correspond to the stoichiometric coefficients in the balanced equation. \(Q_p\) is computed using current partial pressures of all species, giving a quantitative measure to compare against \(K_p\) to determine how far the system is from equilibrium.
159. If for a reaction \( Q_p = 5.0 \) and \( K_p = 2.0 \), the reaction will
ⓐ. Shift toward products
ⓑ. Shift toward reactants
ⓒ. Stay at equilibrium
ⓓ. Be unaffected
Correct Answer: Shift toward reactants
Explanation: Because \( Q_p > K_p \), the mixture contains a higher proportion of products than equilibrium requires. The system responds by shifting backward, converting some products into reactants to reduce \(Q_p\) until it equals \(K_p\). This shift ensures dynamic equilibrium is maintained according to Le Chatelier’s principle.
160. The value of \( Q_c \) depends upon
ⓐ. Temperature only
ⓑ. Concentrations or partial pressures of reacting species
ⓒ. Catalyst presence
ⓓ. Type of container used
Correct Answer: Concentrations or partial pressures of reacting species
Explanation: \( Q_c \) is calculated from the instantaneous concentrations (or partial pressures) of all species in the reaction mixture. Since it reflects the current composition, it varies continuously as the reaction proceeds. Only when equilibrium is achieved does \( Q_c \) stabilize to a constant value equal to \( K_c \) at that temperature.
161. When ( Q < K ) for a reaction, the system will
ⓐ. Move in the reverse direction
ⓑ. Move in the forward direction spontaneously
ⓒ. Be at equilibrium
ⓓ. Remain unchanged
Correct Answer: Move in the forward direction spontaneously
Explanation: If ( Q < K ), the system contains fewer products than at equilibrium. The reaction proceeds spontaneously in the forward direction to form more products until ( Q ) becomes equal to ( K ). This movement reduces Gibbs free energy and continues until equilibrium is achieved.
162. When ( Q > K ), the direction of spontaneous change is
ⓐ. Forward
ⓑ. Backward
ⓒ. Both directions equally
ⓓ. None, since equilibrium is attained
Correct Answer: Backward
Explanation: When ( Q > K ), the mixture has excess products. The reaction proceeds spontaneously in the reverse direction, converting products into reactants to lower ( Q ) to equal ( K ). This adjustment minimizes Gibbs free energy, indicating a non-equilibrium state initially.
163. When \( Q = K \), the Gibbs free energy change \(( \Delta G )\) for the reaction is
ⓐ. Positive
ⓑ. Negative
ⓒ. Zero
ⓓ. Undefined
Correct Answer: Zero
Explanation: At equilibrium, the system is in a state of minimum Gibbs free energy. The forward and reverse reaction rates are equal, and there is no net change in composition. Therefore, \(\Delta G = 0\), signifying that the system is at thermodynamic equilibrium.
164. The relationship between Gibbs free energy and reaction quotient is
ⓐ. \(\Delta G = \Delta G^\circ + RT\ln Q\)
ⓑ. \(\Delta G = -RT\ln Q\)
ⓒ. \(\Delta G = RT\ln K\)
ⓓ. \(\Delta G = -\Delta G^\circ RT\)
Correct Answer: \(\Delta G = \Delta G^\circ + RT\ln Q\)
Explanation: This equation links the actual free energy change \((\Delta G)\) to the standard free energy change \((\Delta G^\circ)\) and the current composition of the system via (Q). When \(Q = K\), \(\Delta G = 0\). This form allows prediction of reaction spontaneity under non-standard conditions.
165. A reaction is spontaneous in the forward direction when
ⓐ. \(Q = K\)
ⓑ. (Q < K) and \(\Delta G < 0\)
ⓒ. (Q > K) and \(\Delta G > 0\)
ⓓ. \(\Delta G = 0\)
Correct Answer: (Q < K) and \(\Delta G < 0\)
Explanation: A negative Gibbs free energy \((\Delta G < 0)\) indicates a spontaneous forward reaction. When (Q < K), the system has fewer products than equilibrium requires, so it spontaneously converts reactants into products to minimize free energy and reach equilibrium.
166. For a reaction at 298 K where \(\Delta G^\circ = -5.7,\text{kJ mol}^{-1}\), (K) is approximately
ⓐ. (0.1)
ⓑ. (1)
ⓒ. (10)
ⓓ. (100)
Correct Answer: (10)
Explanation: Using \(\Delta G^\circ = -RT\ln K\), substitute \(R = 8.314,\text{J mol}^{-1}\text{K}^{-1}\). Rearranging, \(\ln K = -\Delta G^\circ/(RT) = 5700/(8.314\times298) \approx 2.3\). Hence \(K \approx e^{2.3} = 10\), implying that products are favored and the reaction proceeds spontaneously forward.
167. For a system with \(Q = 10\), \(K = 0.1\), the sign of \(\Delta G\) is
ⓐ. Positive
ⓑ. Negative
ⓒ. Zero
ⓓ. Cannot be predicted
Correct Answer: Positive
Explanation: Since (Q > K), the system has more products than equilibrium requires. Substituting into \(\Delta G = RT\ln(Q/K)\), we find \(\ln(10/0.1) = \ln(100) = +4.6\), giving a positive \(\Delta G\). Thus, the forward reaction is non-spontaneous, and the reverse direction is favored.
168. For a system at constant (T) and (P), a spontaneous process always has
ⓐ. \(\Delta G = 0\)
ⓑ. \(\Delta G > 0\)
ⓒ. \(\Delta G < 0\)
ⓓ. \(\Delta G = \Delta G^\circ\)
Correct Answer: \(\Delta G < 0\)
Explanation: Gibbs free energy determines spontaneity under constant temperature and pressure. When \(\Delta G < 0\), the process occurs spontaneously in the forward direction, reducing free energy. At equilibrium, \(\Delta G = 0\), and when \(\Delta G > 0\), the process is non-spontaneous in the forward direction.
169. The equilibrium condition can be derived from the expression
ⓐ. \(\Delta G = \Delta G^\circ + RT\ln Q\) by setting \(\Delta G = 0\)
ⓑ. \(\Delta G = RT\ln Q\) by setting \(Q = 0\)
ⓒ. \(\Delta G = -RT\ln Q\) by setting \(K = 0\)
ⓓ. \(\Delta G = \Delta G^\circ + QK\) by setting \(Q = 1\)
Correct Answer: \(\Delta G = \Delta G^\circ + RT\ln Q\) by setting \(\Delta G = 0\)
Explanation: At equilibrium, \(\Delta G = 0\), so \(0 = \Delta G^\circ + RT\ln K\). Rearranging gives \(\Delta G^\circ = -RT\ln K\). This expression defines how the equilibrium constant depends on temperature and standard free energy change, connecting thermodynamics with equilibrium behavior.
170. When \(Q = K\), the system’s free energy is
ⓐ. At its maximum
ⓑ. At its minimum
ⓒ. Constant but unstable
ⓓ. Decreasing with time
Correct Answer: At its minimum
Explanation: At equilibrium, the system’s Gibbs free energy reaches its lowest possible value for the given temperature and pressure. No further spontaneous changes occur, and both the forward and reverse processes proceed at equal rates. This represents a thermodynamically stable and balanced state.
171. Le Chatelier’s principle states that when a system at equilibrium is subjected to a change,
ⓐ. The equilibrium remains unchanged
ⓑ. The equilibrium shifts to oppose the applied change
ⓒ. The reaction stops immediately
ⓓ. The equilibrium constant increases
Correct Answer: The equilibrium shifts to oppose the applied change
Explanation: Le Chatelier’s principle explains how equilibrium responds to disturbances such as concentration, pressure, or temperature changes. When a stress is applied, the system adjusts itself to counteract that change and restore a new equilibrium. This principle helps predict the qualitative direction of shifts in chemical equilibria.
172. According to Le Chatelier’s principle, if the concentration of a reactant is increased, the system will
ⓐ. Shift toward the products
ⓑ. Shift toward the reactants
ⓒ. Stop reacting
ⓓ. Reduce temperature
Correct Answer: Shift toward the products
Explanation: Increasing reactant concentration increases \(Q_c\) relative to \(K_c\). To restore equilibrium, the system consumes some reactants by shifting in the forward direction, forming more products. This adjustment decreases the reactant concentration until equilibrium conditions are re-established.
173. When the concentration of a product is increased in a reversible reaction, the equilibrium
ⓐ. Moves toward products
ⓑ. Stops immediately
ⓒ. Remains unaffected
ⓓ. Moves toward reactants
Correct Answer: Moves toward reactants
Explanation: Adding product raises \(Q_c\) above \(K_c\), signaling an excess of products. The reaction compensates by favoring the reverse direction, converting some products back into reactants until the equilibrium ratio between products and reactants is restored.
174. When pressure on a gaseous equilibrium is increased, the system shifts
ⓐ. Toward the side with fewer gas moles
ⓑ. Toward the side with more gas moles
ⓒ. Toward solid formation
ⓓ. Without any change
Correct Answer: Toward the side with fewer gas moles
Explanation: Increasing pressure raises total gas concentration. To reduce this stress, the system favors the side with fewer gaseous molecules, thereby lowering total pressure. For example, in \(N_{2} + 3H_{2} \rightleftharpoons 2NH_{3}\), higher pressure drives the equilibrium toward ammonia formation.
175. Decreasing pressure on a gaseous equilibrium mixture causes the equilibrium to shift
ⓐ. Toward the side with fewer gas moles
ⓑ. Toward liquid phase
ⓒ. Toward solid formation
ⓓ. Toward the side with more gas moles
Correct Answer: Toward the side with more gas moles
Explanation: Lowering pressure allows gases to expand. To increase total pressure again, the system favors the side producing more gaseous molecules. This shift restores equilibrium by increasing the number of moles of gas, countering the pressure reduction.
176. For an endothermic reaction, increasing temperature shifts equilibrium
ⓐ. Toward reactants
ⓑ. Toward products
ⓒ. Without change
ⓓ. To solid phase
Correct Answer: Toward products
Explanation: Heat acts as a reactant in endothermic reactions. Raising temperature adds thermal energy, so the system responds by consuming heat—driving the reaction in the forward direction to form more products. This restores thermal equilibrium according to Le Chatelier’s principle.
177. For an exothermic reaction, increasing temperature will shift equilibrium
ⓐ. Toward reactants
ⓑ. Toward products
ⓒ. Not affect equilibrium
ⓓ. Toward both sides equally
Correct Answer: Toward reactants
Explanation: In exothermic processes, heat is treated as a product. Increasing temperature introduces more “product heat,” and the system responds by favoring the reverse reaction to absorb some of that heat. Thus, equilibrium shifts toward the reactants to minimize the disturbance.
178. If an inert gas is added at constant volume to a gaseous equilibrium mixture, the equilibrium position
ⓐ. Shifts toward reactants
ⓑ. Shifts toward products
ⓒ. Remains unchanged
ⓓ. Moves randomly
Correct Answer: Remains unchanged
Explanation: At constant volume, adding an inert gas increases total pressure but does not alter the partial pressures of the reacting gases. Since \(Q_p\) remains unchanged, no shift in equilibrium occurs. The reaction composition stays constant even though the total pressure rises.
179. If an inert gas is added at constant pressure, the equilibrium shifts
ⓐ. Toward side with fewer moles
ⓑ. Toward side with more moles
ⓒ. Remains unchanged
ⓓ. Stops completely
Correct Answer: Toward side with more moles
Explanation: Under constant pressure, adding an inert gas increases the total volume of the system, reducing the partial pressures of all components. The equilibrium then shifts toward the side producing more gas molecules to counteract this dilution and restore the pressure balance.
180. A catalyst affects equilibrium by
ⓐ. Changing the equilibrium constant
ⓑ. Shifting equilibrium to the right
ⓒ. Shifting equilibrium to the left
ⓓ. Changing the rate but not the position of equilibrium
Correct Answer: Changing the rate but not the position of equilibrium
Explanation: A catalyst accelerates both forward and reverse reactions equally, allowing equilibrium to be achieved faster without altering its position or the value of the equilibrium constant. It reduces the activation energy, improving rate kinetics but leaving thermodynamic conditions unchanged.
181. When the concentration of a reactant is increased in a system at equilibrium, the equilibrium
ⓐ. Shifts toward the reactants
ⓑ. Shifts toward the products
ⓒ. Remains unchanged
ⓓ. Stops immediately
Correct Answer: Shifts toward the products
Explanation: According to Le Chatelier’s principle, increasing reactant concentration introduces stress. The system responds by consuming some of the added reactants, forming more products until the new equilibrium is re-established. This forward shift helps restore the equilibrium ratio defined by \(K_c\).
182. When the concentration of products is increased at equilibrium, the system
ⓐ. Increases temperature
ⓑ. Shifts toward products
ⓒ. Remains unchanged
ⓓ. Shifts toward reactants
Correct Answer: Shifts toward reactants
Explanation: Adding more products increases \(Q_c\) above \(K_c\). The system responds by favoring the reverse reaction, converting part of the excess product back into reactants. This backward shift continues until \(Q_c\) again equals \(K_c\), restoring balance.
183. Removing a product from a reversible reaction mixture will
ⓐ. Shift equilibrium toward the products
ⓑ. Shift equilibrium toward reactants
ⓒ. Halt the reaction
ⓓ. Increase pressure
Correct Answer: Shift equilibrium toward the products
Explanation: Removing a product lowers \(Q_c\), making it smaller than \(K_c\). To counter this change, the reaction proceeds in the forward direction to generate more product molecules, thereby re-establishing equilibrium. This principle is often exploited in chemical manufacturing to improve yield.
184. In the equilibrium \(N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)\), increasing total pressure shifts equilibrium
ⓐ. Toward reactants
ⓑ. Does not affect equilibrium
ⓒ. Toward products
ⓓ. Toward higher volume
Correct Answer: Toward products
Explanation: The forward reaction produces fewer moles of gas \(4 → 2\). Increasing pressure compresses the system, and the equilibrium shifts toward the side with fewer gas molecules to reduce the applied stress. Hence, ammonia formation is favored under high pressure.
185. In the same equilibrium, decreasing pressure shifts the reaction
ⓐ. Toward ammonia
ⓑ. Toward nitrogen and hydrogen
ⓒ. Without change
ⓓ. Toward liquid phase
Correct Answer: Toward nitrogen and hydrogen
Explanation: Lowering pressure expands the gas volume, favoring the side with more gas molecules. Therefore, the equilibrium moves backward, forming more \(N_{2}\) and \(H_{2}\). This principle is used to optimize reaction conditions in industrial ammonia synthesis.
186. For the equilibrium \(2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)\), increasing pressure will
ⓐ. Favor \(SO_{3}\) formation
ⓑ. Favor \(SO_{2}\) and \(O_{2}\) formation
ⓒ. Not affect equilibrium
ⓓ. Stop the reaction
Correct Answer: Favor \(SO_{3}\) formation
Explanation: The total number of gas moles decreases from 3 to 2 in the forward direction. Increasing pressure encourages the reaction to proceed toward fewer moles, shifting equilibrium toward \(SO_{3}\). This concept is vital in optimizing the contact process for sulfuric acid manufacture.
187. For an exothermic reaction, increasing temperature will
ⓐ. Increase product formation
ⓑ. Stop the reaction entirely
ⓒ. Leave equilibrium unchanged
ⓓ. Decrease product formation
Correct Answer: Decrease product formation
Explanation: In exothermic reactions, heat behaves as a product. Adding heat (raising temperature) disturbs equilibrium, causing it to shift toward the reactants to absorb the excess energy. Hence, product yield decreases with increasing temperature.
188. For an endothermic reaction, decreasing temperature causes
ⓐ. Equilibrium to shift toward reactants
ⓑ. Equilibrium to shift toward products
ⓒ. No change in equilibrium
ⓓ. Formation of intermediate species only
Correct Answer: Equilibrium to shift toward reactants
Explanation: In endothermic processes, heat acts as a reactant. Lowering temperature effectively removes heat, so the equilibrium shifts toward the reactants to replace the lost thermal energy. This reduces product concentration until equilibrium is re-established.
189. If the equilibrium constant \(K_c\) for a reaction decreases with temperature, the reaction is
ⓐ. Endothermic
ⓑ. Unaffected by heat
ⓒ. Exothermic
ⓓ. Athermal
Correct Answer: Exothermic
Explanation: For exothermic reactions, increasing temperature decreases the equilibrium constant because higher heat input favors the reverse direction. According to van’t Hoff’s equation, \(\frac{d\ln K}{dT} = \frac{\Delta H}{RT^{2}}\), a negative \(\Delta H\) results in a negative slope, reducing \(K_c\) as temperature increases.
190. For a gaseous equilibrium involving equal moles of reactants and products, a change in pressure
ⓐ. Has no effect on equilibrium position
ⓑ. Always favors products
ⓒ. Always favors reactants
ⓓ. Changes \(K_p\) value
Correct Answer: Has no effect on equilibrium position
Explanation: When the number of gaseous moles is the same on both sides, compression or expansion alters total pressure equally for reactants and products. The ratio of their partial pressures—and hence \(Q_p\)—remains constant. Therefore, equilibrium does not shift, and the system remains balanced.
191. In the Haber process \( N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g) + \text{heat} \), increasing the pressure will
ⓐ. Decrease ammonia formation
ⓑ. Have no effect on yield
ⓒ. Increase ammonia formation
ⓓ. Stop the reaction completely
Correct Answer: Increase ammonia formation
Explanation: The forward reaction results in a decrease in the number of gas molecules \(4 → 2\). According to Le Chatelier’s principle, an increase in pressure favors the side with fewer gas moles. Hence, high pressure drives the equilibrium toward ammonia formation, increasing yield in the Haber process.
192. In the Haber process, lowering the temperature will
ⓐ. Increase ammonia yield
ⓑ. Decrease ammonia yield
ⓒ. Have no effect
ⓓ. Speed up the reaction
Correct Answer: Increase ammonia yield
Explanation: The synthesis of ammonia is exothermic. Lowering the temperature favors the forward reaction because the system responds by generating heat to counteract the cooling. However, this also slows the reaction rate, so an optimal moderate temperature (~450°C) is used in practice to balance rate and yield.
193. In the Haber process, the catalyst used is
ⓐ. Nickel
ⓑ. Vanadium(V) oxide
ⓒ. Finely divided iron with promoters
ⓓ. Platinum
Correct Answer: Finely divided iron with promoters
Explanation: The Haber process employs finely divided iron as a catalyst, promoted by potassium and aluminum oxides to enhance efficiency and longevity. The catalyst accelerates both forward and reverse reactions equally, allowing equilibrium to be achieved faster without altering the equilibrium constant.
194. The reaction conditions in the Haber process are typically
ⓐ. 25°C and 1 atm
ⓑ. 100°C and 2 atm
ⓒ. 800°C and 1 atm
ⓓ. 450°C and 200 atm
Correct Answer: 450°C and 200 atm
Explanation: Industrial ammonia synthesis balances thermodynamics and kinetics. A moderately high temperature (~450°C) ensures sufficient reaction rate, while high pressure (~200 atm) favors product formation. These optimized conditions maximize yield and minimize costs under catalytic influence.
195. The Contact process involves the oxidation of \(SO_{2}(g)\) to \(SO_{3}(g)\): \(2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g) + \text{heat}\). Increasing pressure will
ⓐ. Shift equilibrium toward reactants
ⓑ. Shift equilibrium toward products
ⓒ. Have no effect on equilibrium
ⓓ. Destroy catalyst activity
Correct Answer: Shift equilibrium toward products
Explanation: The number of gas molecules decreases \(3 → 2\) during the forward reaction. Hence, increasing pressure favors the formation of \(SO_{3}\). This principle is applied industrially in the Contact process to maximize the yield of sulfur trioxide for sulfuric acid production.
196. In the Contact process, the catalyst used to oxidize \(SO_{2}\) is
ⓐ. Fe catalyst
ⓑ. Nickel oxide
ⓒ. Pt–Rh alloy
ⓓ. \(V_{2}O_{5}\) (Vanadium(V) oxide)
Correct Answer: \(V_{2}O_{5}\) (Vanadium(V) oxide)
Explanation: The oxidation of \(SO_{2}\) to \(SO_{3}\) occurs slowly without a catalyst. Vanadium(V) oxide \((V_{2}O_{5})\) speeds up the reaction by providing an alternate pathway with lower activation energy, facilitating faster attainment of equilibrium without altering the equilibrium position.
197. In the Contact process, lowering the temperature increases the yield of \(SO_{3}\) because the reaction is
ⓐ. Endothermic
ⓑ. Exothermic
ⓒ. Neutral
ⓓ. Photochemical
Correct Answer: Exothermic
Explanation: The oxidation of \(SO_{2}\) releases heat. Lowering the temperature favors the forward reaction as the system compensates for the cooling by producing more heat, thereby forming additional \(SO_{3}\). However, excessively low temperatures reduce the reaction rate, so a compromise temperature (~450°C) is used industrially.
198. The optimum temperature for the Contact process is around
ⓐ. 100°C
ⓑ. 250°C
ⓒ. 350°C
ⓓ. 450°C
Correct Answer: 450°C
Explanation: Although lower temperatures favor \(SO_{3}\) yield, the reaction rate becomes too slow. Around 450°C offers an effective balance between reasonable yield and high rate, ensuring economic efficiency. At this temperature, the \(V_{2}O_{5}\) catalyst functions efficiently and maintains durability.
199. In the Haber process, which change increases the rate of attainment of equilibrium without changing the equilibrium composition?
ⓐ. Increasing temperature
ⓑ. Increasing pressure
ⓒ. Adding a catalyst
ⓓ. Decreasing concentration of reactants
Correct Answer: Adding a catalyst
Explanation: A catalyst provides an alternate reaction pathway with lower activation energy, increasing the rate of both forward and reverse reactions equally. It does not alter the equilibrium constant or equilibrium composition but helps the system reach equilibrium more quickly.
200. In the Contact process, the final step involves converting \(SO_{3}\) into sulfuric acid by
ⓐ. Direct absorption in water
ⓑ. Reaction with oxygen
ⓒ. Absorption in concentrated \(H_{2}SO_{4}\) to form oleum
ⓓ. Decomposition in air
Correct Answer: Absorption in concentrated \(H_{2}SO_{4}\) to form oleum
Explanation: Direct absorption of \(SO_{3}\) in water produces dense acid mist, making the process inefficient and hazardous. Instead, \(SO_{3}\) is absorbed in concentrated sulfuric acid to form oleum \((H_{2}S_{2}O_{7})\), which is then diluted with water to yield pure \(H_{2}SO_{4}\) safely and efficiently in industry.
Welcome to Class 11 Chemistry MCQs – Chapter 7: Equilibrium (Part 2). This part continues your journey through the dynamic equilibrium concepts. These MCQs build on the foundation laid in Part 1 and delve deeper into the mathematical expressions of equilibrium constants, the calculation of equilibrium concentrations, and application of the equilibrium constant in various reactions.
Topics covered in Part 2 (100 MCQs): Quantitative aspects of equilibrium, calculation of equilibrium concentrations, and the relationship between equilibrium constant and reaction quotient (Q). The chapter also covers the Henderson-Hasselbalch equation and equilibrium in acid–base reactions.
How to practice here
- Answer 10–20 MCQs at a time. Check the feedback for mistakes and learn the correct approach to solve.
- Mark important MCQs with the ❤️ Heart and filter your list using the Favourite Toggle for faster revision.
- Use Workspace to write formulas, tips, or quick tricks for tricky questions. Your notes are automatically saved.
- Test yourself with the Random button to improve understanding without memorizing patterns.
- 👉 Total in chapter: 400 MCQs (100 + 100 + 100 + 100)
- 👉 This page: Second set of 100 solved MCQs
- 👉 Next: Continue with Part 3 for more advanced questions