JEE Chemistry Equilibrium MCQs Set 34

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Multiple Choice Questions

Question 1: In the dissociation of 4 mol of \( \ce{PCl5(g) \)} in a 1 litre container, 0.8 mol of \( \ce{PCl3 \)} is produced. The equilibrium constant \( K_c \) of \( \ce{PCl5 <=> PCl3 + Cl2 \)} is:
(a) 0.2
(b) 0.1
(c) 0.4
(d) 1
Show Answer & Explanation

Answer: (a) 0.2

Explanation:
1. At equilibrium: \( \ce{PCl5 \)} = 4 - 0.8 = 3.2 M, \( \ce{PCl3 \)} = 0.8 M, \( \ce{Cl2 \)} = 0.8 M (V = 1 L).
2. \( K_c = (0.8)(0.8)/3.2 = \) 0.2.

Teacher's Note:
1. Use equilibrium concentrations (moles / volume).
2. A volume of 1 L makes moles equal to concentrations.

Question 2: In the system \( \ce{P(s) <=> 2Q(g) + R(g) \)}, if the concentration of Q at equilibrium is doubled, the concentration of R at equilibrium becomes:
(a) Double its original concentration
(b) \( \dfrac14 \) of its original concentration
(c) \( \dfrac12 \) of its original concentration
(d) 4 times its original concentration
Show Answer & Explanation

Answer: (b) \( \dfrac14 \) of its original concentration

Explanation:
1. \( K_c = [Q]^2[R] \); the solid P is omitted.
2. If [Q] doubles, [Q]2 becomes 4 times, so [R] must fall to 1/4 to keep Kc constant.
3. [R] becomes 1/4 of its original value.

Teacher's Note:
1. Pure solids do not appear in the equilibrium expression.
2. K is fixed at a given temperature.

Question 3: 1 mol of \( \ce{XY(g) \)} and 0.2 mol of \( \ce{Y(g) \)} are mixed in a 1 L vessel. At equilibrium 0.6 mol of \( \ce{Y(g) \)} is present. K for \( \ce{XY(g) <=> X(g) + Y(g) \)} is:
(a) 0.04
(b) 0.06
(c) 0.36
(d) 0.40
Show Answer & Explanation

Answer: (d) 0.40

Explanation:
1. Y rose from 0.2 to 0.6, so 0.4 mol XY dissociated: XY = 0.6, X = 0.4, Y = 0.6.
2. \( K = (0.4)(0.6)/0.6 = \) 0.40.

Teacher's Note:
1. Initial Y is not zero, so use the change in Y.
2. Moles equal concentration here because V = 1 L.

Question 4: In which of the following reactions is the reaction almost complete?
(a) \( K_{eq}=10 \)
(b) \( K_{eq}=1 \)
(c) \( K_{eq}=10^{5 \)}
(d) \( K_{eq}=10^{-2 \)}
Show Answer & Explanation

Answer: (c) \( K_{eq}=10^{5 \)}

Explanation:
1. A very large K means products dominate at equilibrium.
2. \( K=10^5 \) is by far the largest, so the reaction is almost complete.

Teacher's Note:
1. K >> 1 means products favoured; K << 1 means reactants favoured.
2. K = 1 means comparable amounts.

Question 5: For the equilibrium \( \ce{XY3(s) <=> X^{3+}(aq) + 3Y^-(aq) \)}, if the concentration of \( \ce{Y^- \)} is decreased to half, the equilibrium concentration of \( \ce{X^{3+ \)}} will increase by:
(a) 2 times
(b) 4 times
(c) 8 times
(d) 16 times
Show Answer & Explanation

Answer: (c) 8 times

Explanation:
1. \( K_{sp} = [X^{3+}][Y^-]^3 \).
2. If [Y-] becomes 1/2, [Y-]3 becomes 1/8, so [X3+] must rise 8 times.
3. Increase = 8 times.

Teacher's Note:
1. The exponent of the concentration is the stoichiometric coefficient.
2. Solubility product stays constant at fixed temperature.

Question 6: For the reaction \( \ce{X2(g) + Y2(g) + Z2(g) <=> 2XYZ(g) \)}, at equilibrium the concentrations are \( \ce{X2 \)} = 3 M, \( \ce{Y2 \)} = 6 M, \( \ce{Z2 \)} = 9 M and XYZ = 6 M. The value of \( K_c \) is:
(a) \( 27/2 \)
(b) \( 2/9 \)
(c) 36
(d) 24
Show Answer & Explanation

Answer: (b) \( 2/9 \)

Explanation:
1. \( K_c = [XYZ]^2/([X_2][Y_2][Z_2]) = 36/(3\times6\times9) \).
2. \( K_c = 36/162 = \) \( 2/9 \).

Teacher's Note:
1. The coefficient 2 becomes the exponent of [XYZ].
2. Temperature given (527°C) is not needed for Kc here.

Question 7: A gaseous mixture is made by taking 3 mol of \( \ce{H2 \)} and 1 mol of CO. If the total pressure of the mixture is 2 atm, the partial pressure of hydrogen is:
(a) \( 3/2 \) atm
(b) \( 1/2 \) atm
(c) 1 atm
(d) 2 atm
Show Answer & Explanation

Answer: (a) \( 3/2 \) atm

Explanation:
1. Mole fraction of H2 = 3/(3 + 1) = 3/4.
2. Partial pressure = 3/4 x 2 = 3/2 atm.

Teacher's Note:
1. Dalton's law: p = mole fraction x total pressure.
2. Partial pressures add up to the total.

Question 8: The value of \( K_c \) at 900 K for \( \ce{N2(g) + O2(g) <=> 2NO(g) \)} is 5. If the equilibrium concentration of \( \ce{N2 \)} and \( \ce{O2 \)} is 0.5 M, the value of \( K_p \) is:
(a) 0.02
(b) 0.2
(c) 5
(d) \( 5/RT \)
Show Answer & Explanation

Answer: (c) 5

Explanation:
1. Δng = 2 - 2 = 0.
2. \( K_p = K_c(RT)^{0} = K_c = \) 5.

Teacher's Note:
1. When Δng = 0, Kp = Kc.
2. The concentration data are not needed.

Question 9: The equilibrium constant \( K_c \) for \( \ce{1/2 F2(g) <=> F(g) \)} at 842°C is \( 7.90\times10^{-3 \)}. \( K_p \) at the same temperature is:
(a) \( 8.64\times10^{-5 \)}
(b) \( 8.26\times10^{-4 \)}
(c) \( 7.90\times10^{-3 \)}
(d) \( 7.56\times10^{-2 \)}
Show Answer & Explanation

Answer: (d) \( 7.56\times10^{-2 \)}

Explanation:
1. Δng = 1 - 1/2 = 1/2; T = 842 + 273 = 1115 K.
2. \( K_p = K_c(RT)^{1/2} = 7.90\times10^{-3}\times\sqrt{0.0821\times1115} = 7.90\times10^{-3}\times9.57 \).
3. \( K_p = \) \( 7.56\times10^{-2 \)}.

Teacher's Note:
1. Use R = 0.0821 L atm/(mol K) with T in kelvin.
2. Δng can be a fraction.

Question 10: For \( \ce{NH4HS(s) <=> NH3(g) + H2S(g) \)}, the equilibrium pressure at 25°C is 0.660 atm. Kp is:
(a) 0.109
(b) 0.218
(c) 1.89
(d) 2.18
Show Answer & Explanation

Answer: (a) 0.109

Explanation:
1. Total pressure 0.660 atm is shared equally: each gas has 0.330 atm.
2. \( K_p = 0.330\times0.330 = \) 0.109.

Teacher's Note:
1. Only gases appear in Kp.
2. The two gases form in equal moles.

Question 11: At a certain temperature only 50% of HI is dissociated at equilibrium in \( \ce{2HI(g) <=> H2(g) + I2(g) \)}. The equilibrium constant for the reaction is:
(a) 0.25
(b) 1.0
(c) 3.0
(d) 0.5
Show Answer & Explanation

Answer: (a) 0.25

Explanation:
1. Start with 1 mol HI: HI = 0.5, H2 = 0.25, I2 = 0.25 (Δng = 0, volume cancels).
2. \( K = (0.25)(0.25)/(0.5)^2 = \) 0.25.

Teacher's Note:
1. 50% dissociation of 1 mol leaves 0.5 mol.
2. Volume cancels when Δng = 0.

Question 12: For the reaction \( \ce{CO(g) + Cl2(g) <=> COCl2(g) \)}, the value of \( K_c/K_p \) is equal to:
(a) \( \sqrt{RT \)}
(b) RT
(c) \( 1/RT \)
(d) 1
Show Answer & Explanation

Answer: (b) RT

Explanation:
1. Δng = 1 - 2 = -1, so \( K_p = K_c(RT)^{-1 \)}.
2. \( K_c/K_p = \) RT.

Teacher's Note:
1. Kp = Kc(RT)Δn.
2. Be careful: the ratio asked is Kc/Kp.

Question 13: \( K_c \) for \( \ce{2HCl(g) <=> H2(g) + Cl2(g) \)} is \( 2\times10^{-2 \)} at 25°C. The equilibrium constant for \( \ce{H2(g) + Cl2(g) <=> 2HCl(g) \)} is:
(a) \( 2\times10^{-2 \)}
(b) 50
(c) \( 5\times10^{2 \)}
(d) None of these
Show Answer & Explanation

Answer: (b) 50

Explanation:
1. The second reaction is the reverse of the first, so its constant is the reciprocal.
2. \( K' = 1/(2\times10^{-2}) = \) 50.

Teacher's Note:
1. Reversing a reaction inverts K.
2. Multiplying coefficients by n raises K to the power n.

Question 14: K1 and K2 are equilibrium constants for: \( \ce{XeF6 + H2O <=> XeOF4 + 2HF \)} (K1) and \( \ce{XeO4 + XeF6 <=> XeOF4 + XeO3F2 \)} (K2). The equilibrium constant for \( \ce{XeO4 + 2HF <=> XeO3F2 + H2O \)} is:
(a) \( K_1K_2 \)
(b) \( K_1/K_2^2 \)
(c) \( K_2/K_1 \)
(d) \( K_1/K_2 \)
Show Answer & Explanation

Answer: (c) \( K_2/K_1 \)

Explanation:
1. Subtract reaction 1 from reaction 2: \( \ce{XeO4 + XeF6 + XeOF4 + 2HF \)} on the left minus ... gives \( \ce{XeO4 + 2HF <=> XeO3F2 + H2O \)}.
2. Subtracting reactions divides their constants: K = K2/K1.
3. K = \( K_2/K_1 \).

Teacher's Note:
1. Adding reactions multiplies K; subtracting divides.
2. Write the target reaction as (reaction 2) - (reaction 1).

Question 15: For which of the following gaseous reactions is Kp less than Kc at 298 K?
(a) \( \ce{N2O4 <=> 2NO2 \)}
(b) \( \ce{2HI <=> H2 + I2 \)}
(c) \( \ce{2SO2 + O2 <=> 2SO3 \)}
(d) \( \ce{N2 + O2 <=> 2NO \)}
Show Answer & Explanation

Answer: (c) \( \ce{2SO2 + O2 <=> 2SO3 \)}

Explanation:
1. Kp = Kc(RT)Δn is smaller than Kc only when Δng < 0.
2. Δng: (1) +1, (2) 0, (3) -1, (4) 0.
3. The reaction is \( \ce{2SO2 + O2 <=> 2SO3 \)}.

Teacher's Note:
1. RT is greater than 1 when expressed in L atm units.
2. Kp > Kc when Δng > 0.

Question 16: At 25°C the equilibrium constants K1 and K2 of \( \ce{2NH3 <=> N2 + 3H2 \)} (K1) and \( \ce{1/2 N2 + 3/2 H2 <=> NH3 \)} (K2) are related as:
(a) \( K_1=K_2 \)
(b) \( K_2=1/K_1^2 \)
(c) \( K_1=1/K_2^2 \)
(d) \( K_1=1/K_2 \)
Show Answer & Explanation

Answer: (c) \( K_1=1/K_2^2 \)

Explanation:
1. Reaction 2 is the reverse of reaction 1 with coefficients halved.
2. So \( K_2 = (1/K_1)^{1/2 \)}, giving \( K_1 = 1/K_2^2 \).
3. Relation: \( K_1 = 1/K_2^2 \).

Teacher's Note:
1. Reverse: K to 1/K. Halve: K to its square root.
2. Combine both operations.

Question 17: For \( \ce{2SO3(g) <=> 2SO2(g) + O2(g) \)}, the value of \( K_c/K_p \) is:
(a) \( (RT)^1 \)
(b) \( (RT)^{-1 \)}
(c) \( \sqrt{RT \)}
(d) \( (RT)^{-2 \)}
Show Answer & Explanation

Answer: (b) \( (RT)^{-1 \)}

Explanation:
1. Δng = 3 - 2 = 1, so \( K_p = K_c(RT) \).
2. \( K_c/K_p = 1/RT = \) \( (RT)^{-1 \)}.

Teacher's Note:
1. Find Δng from gaseous species only.
2. Rearrange Kp = Kc(RT)Δn.

Question 18: For \( \ce{N2 + 3H2 <=> 2NH3 \)} (ΔH negative), which relation holds?
(a) \( K_p = K_c \)
(b) \( K_p = K_cRT \)
(c) \( K_p = K_c(RT)^{-2 \)}
(d) \( K_p = K_c(RT)^{-1 \)}
Show Answer & Explanation

Answer: (c) \( K_p = K_c(RT)^{-2 \)}

Explanation:
1. Δng = 2 - 4 = -2.
2. \( K_p = K_c(RT)^{-2 \)}, i.e. option (3).

Teacher's Note:
1. The sign of ΔH does not matter for this relation.
2. Δng = moles of gaseous products - moles of gaseous reactants.

Question 19: Equilibrium constants: \( \ce{2CO + 2H2O <=> 2CO2 + 2H2 \)} (K1); \( \ce{CH4 + H2O <=> CO + 3H2 \)} (K2); \( \ce{CH4 + 2H2O <=> CO2 + 4H2 \)} (K3). Which relation is correct?
(a) \( K_3=K_1/K_2 \)
(b) \( K_3=K_1^2/K_2^2 \)
(c) \( K_3=K_1K_2 \)
(d) \( K_3=\sqrt{K_1}\cdot K_2 \)
Show Answer & Explanation

Answer: (d) \( K_3=\sqrt{K_1}\cdot K_2 \)

Explanation:
1. Halve reaction 1: \( \ce{CO + H2O <=> CO2 + H2 \)} has constant \( \sqrt{K_1 \)}.
2. Add it to reaction 2: \( \ce{CH4 + 2H2O <=> CO2 + 4H2 \)}.
3. \( K_3 = \sqrt{K_1}\cdot K_2 \), i.e. option (4).

Teacher's Note:
1. Adding reactions multiplies constants.
2. Halving a reaction takes the square root of K.

Question 20: One mole of \( \ce{X2 \)} is mixed with one mole of \( \ce{Y2 \)} in a 1 litre flask. If 0.5 mole of \( \ce{Y2 \)} is obtained at equilibrium, Kp for \( \ce{X2(g) + Y2(g) <=> 2XY(g) \)} is:
(a) 12
(b) 9
(c) 4
(d) 36
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. 0.5 mol of Y2 reacted, so X2 = 0.5, Y2 = 0.5, XY = 1.0.
2. \( K_c = (1)^2/(0.5\times0.5) = 4 \), and Δng = 0 so Kp = Kc.
3. Kp = 4.

Teacher's Note:
1. Volume cancels when Δng = 0.
2. Each mole of Y2 reacted gives 2 mol XY.

Question 21: For \( \ce{2NO2(g) + 1/2 O2(g) <=> N2O5(g) \)} the equilibrium constant is Kp. The equilibrium constant for \( \ce{2N2O5(g) <=> 4NO2(g) + O2(g) \)} would be:
(a) \( K_p^2 \)
(b) \( 2/K_p \)
(c) \( 1/K_p^2 \)
(d) \( 1/\sqrt{K_p \)}
Show Answer & Explanation

Answer: (c) \( 1/K_p^2 \)

Explanation:
1. The required reaction is the reverse of the given one multiplied by 2.
2. Reverse: 1/Kp; multiply by 2: (1/Kp)2.
3. K = \( 1/K_p^2 \).

Teacher's Note:
1. Doubling a reaction squares K.
2. Reversing inverts K.

Question 22: In \( \ce{X(g) + Y(g) <=> 2Z(g) \)}, 2 mol of X, 1 mol of Y and 1 mol of Z are placed in a 10 L vessel and allowed to reach equilibrium. If the final concentration of Z is 0.2 M, Kc is:
(a) 1.6
(b) \( 80/3 \)
(c) \( 16/3 \)
(d) None of these
Show Answer & Explanation

Answer: (c) \( 16/3 \)

Explanation:
1. Z at equilibrium = 0.2 x 10 = 2 mol, i.e. 1 mol more than initially, so 0.5 mol each of X and Y reacted.
2. X = 1.5 mol, Y = 0.5 mol, Z = 2 mol (concentrations 0.15, 0.05, 0.2 M).
3. \( K_c = (0.2)^2/(0.15\times0.05) = \) \( 16/3 \).

Teacher's Note:
1. Convert moles to concentrations by dividing by 10 L.
2. Reaction stoichiometry: 1 X + 1 Y gives 2 Z.

Question 23: For the reversible reaction \( \ce{PCl3(g) + Cl2(g) <=> PCl5(g) \)} + heat, the equilibrium will shift in the forward direction:
(a) By increasing the concentration of \( \ce{PCl5 \)}
(b) By decreasing the pressure
(c) By decreasing the concentration of \( \ce{PCl3 \)} and \( \ce{Cl2 \)}
(d) By increasing the pressure and decreasing the temperature
Show Answer & Explanation

Answer: (d) By increasing the pressure and decreasing the temperature

Explanation:
1. Forward direction reduces gas moles (2 to 1) and releases heat.
2. Higher pressure and lower temperature both favour the forward reaction (Le Chatelier).
3. Answer: increase pressure and decrease temperature.

Teacher's Note:
1. Pressure change shifts toward fewer gas moles when increased.
2. Exothermic reactions are favoured by cooling.

Question 24: For an endothermic reaction, Kp and Kp' are the equilibrium constants at temperatures T1 and T2 respectively. If T2 > T1, then:
(a) \( K_p \gt K_p' \)
(b) \( K_p \lt K_p' \)
(c) \( K_p=K_p' \)
(d) \( K_p=1/K_p' \)
Show Answer & Explanation

Answer: (b) \( K_p \lt K_p' \)

Explanation:
1. For an endothermic reaction, K increases with temperature (van 't Hoff).
2. Since T2 > T1, Kp' is larger: \( K_p \lt K_p' \).

Teacher's Note:
1. Endothermic: K rises with T. Exothermic: K falls with T.
2. Heat acts like a reactant in endothermic reactions.

Question 25: Which reaction gives more product as a result of an increase in pressure?
(a) \( \ce{H2O + CO <=> H2 + CO2 \)}
(b) \( \ce{H2 + Br2 <=> 2HBr \)}
(c) \( \ce{2SO2 + O2 <=> 2SO3 \)}
(d) \( \ce{2HI <=> H2 + I2 \)}
Show Answer & Explanation

Answer: (c) \( \ce{2SO2 + O2 <=> 2SO3 \)}

Explanation:
1. Higher pressure shifts equilibrium to the side with fewer gas moles.
2. Only (3) has fewer gas moles on the product side (3 to 2).
3. The reaction is \( \ce{2SO2 + O2 <=> 2SO3 \)}.

Teacher's Note:
1. Reactions with Δng = 0 are unaffected by pressure.
2. Compare total gas moles on each side.

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