JEE Chemistry Equilibrium MCQs Set 05

Answer: (a) \( 10^{-6} \)

Explanation:
1. \( K_c = \dfrac{[\ce{Ag(NH3)2+}]}{[\ce{Ag+}][\ce{NH3}]^2} \).
2. Substituting: \( K_c = \dfrac{10^{-1}}{10^{-1} \times (10^{3})^2} = 10^{-6} \).

Teacher's Note:
1. The coefficient 2 on \( \ce{NH3} \) becomes a power of 2 in \( K_c \).
2. Do not forget to square \( 10^3 \).

Answer: (d) \( 5 \times 10^{-5} \times R^{-1} \)

Explanation:
1. \( \Delta n_g = (3+1) - (1+2) = 1 \).
2. \( K_p = K_c (RT)^{\Delta n} \) gives \( K_c = K_p/(RT) = 0.05/(1000R) \).
3. So \( K_c = 5 \times 10^{-5} R^{-1} \).

Teacher's Note:
1. Count only gaseous moles in \( \Delta n \).
2. \( K_c = K_p (RT)^{-\Delta n} \).

Answer: (d) 1

Explanation:
1. Solid carbon is left out of the expression.
2. \( K_p = p_{CO}^2 / p_{CO_2} = 2^2/4 \).
3. \( K_p = 1 \).

Teacher's Note:
1. Pure solids and liquids have constant activity and do not appear in K.
2. Square the pressure of CO because its coefficient is 2.

Answer: (c) \( \ce{H2(g) + Cl2(g) <=> 2HCl(g)} \)

Explanation:
1. \( K_p = K_c(RT)^{\Delta n_g} \), so the two are equal only when \( \Delta n_g = 0 \).
2. For \( \ce{H2 + Cl2 <=> 2HCl} \), \( \Delta n_g = 2 - 2 = 0 \).
3. The others have \( \Delta n_g = +1, -2, -1 \).

Teacher's Note:
1. Equal numbers of gas moles on both sides means \( K_p = K_c \).
2. Check this first in any conversion question.

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

Explanation:
1. The condition gives \( \log(K_p/K_c) = -\log RT \), so \( K_p/K_c = (RT)^{-1} \) and \( \Delta n_g = -1 \).
2. For \( \ce{PCl5 <=> PCl3 + Cl2} \), \( \Delta n_g = +1 \).
3. For \( \ce{2SO2 + O2 <=> 2SO3} \), \( \Delta n_g = -1 \). For \( \ce{N2 + 3H2 <=> 2NH3} \), \( \Delta n_g = -2 \).
4. Only the second reaction fits.

Teacher's Note:
1. Take logs of \( K_p = K_c(RT)^{\Delta n} \) to relate the expression to \( \Delta n \).
2. \( \Delta n = -1 \) is the key.

Answer: (a) \( K_2 = \dfrac{1}{K_1^4} \)

Explanation:
1. Reversing the first reaction gives the constant \( 1/K_1 \).
2. Multiplying all coefficients by 4 raises the constant to the fourth power.
3. So \( K_2 = (1/K_1)^4 = 1/K_1^4 \).

Teacher's Note:
1. Reverse means reciprocal; multiplying coefficients by n means raising K to the power n.
2. Combine both rules in order.

Answer: (d) 48

Explanation:
1. When reactions are added, their equilibrium constants multiply.
2. \( K = 2 \times 4 \times 6 = 48 \).

Teacher's Note:
1. Adding reactions multiplies K values.
2. Subtracting (reversing) a reaction divides.

Answer: (d) 0.03

Explanation:
1. For \( \ce{A <=> D} \), \( K = 3 \times 5 \times 2 = 30 \).
2. The reverse reaction \( \ce{D <=> A} \) has \( K = 1/30 \approx 0.03 \).

Teacher's Note:
1. Find the constant for the forward overall reaction first.
2. Then take the reciprocal for the reverse.

Answer: (a) endothermic

Explanation:
1. K increases when the temperature rises from 500 K to 700 K.
2. By Le Chatelier's principle and the van't Hoff equation, K rises with T only for an endothermic reaction.

Teacher's Note:
1. A rise of K with temperature indicates an endothermic reaction.
2. K says nothing about the speed of the reaction.

Answer: (b) K

Explanation:
1. The equilibrium constant depends only on temperature.
2. The temperature is unchanged, so the value stays K.

Teacher's Note:
1. Do not recalculate K when only the starting concentrations change.
2. The equilibrium composition will change, not K.

Answer: (d) \( \dfrac{1}{2\sqrt{2}} \) times its original concentration

Explanation:
1. \( K_c = [B]^2[C]^3 \) with A being a solid.
2. If [C] is doubled, \( [C]^3 \) becomes 8 times larger.
3. To keep K constant, \( [B]^2 \) must become 1/8 of its value, so [B] becomes \( 1/\sqrt{8} = 1/(2\sqrt{2}) \) times.

Teacher's Note:
1. Keep K fixed and solve for the changed concentration.
2. Cube the factor for C.

Answer: (c) 16

Explanation:
1. \( K_p \) depends only on temperature.
2. Compressing the vessel shifts the equilibrium position, but \( K_p \) remains 16.

Teacher's Note:
1. Pressure and volume changes alter the composition, not K.
2. Only temperature changes K.

Answer: (c) \( 4 \times 10^{4} \)

Explanation:
1. A catalyst speeds up the forward and backward reactions equally.
2. It does not change the equilibrium composition, so K is unchanged at \( 4 \times 10^{4} \).

Teacher's Note:
1. Catalysts change the time to reach equilibrium, not the position of equilibrium.
2. K stays the same.

Answer: (b) d, b

Explanation:
1. \( K_p/K_c = (RT)^{\Delta n} \), which increases with \( \Delta n \) because \( RT > 1 \).
2. \( \Delta n \) values: (a) +1, (b) −1, (c) +2, (d) +3.
3. The maximum is for (d) and the minimum for (b).

Teacher's Note:
1. Compare \( \Delta n_g \) values to rank \( K_p/K_c \).
2. Remember \( RT \) in L atm units is greater than 1.

Answer: (c) 0.25

Explanation:
1. Degree of dissociation = moles dissociated / initial moles.
2. \( \alpha = 0.5/2 = 0.25 \).

Teacher's Note:
1. The degree of dissociation is always between 0 and 1.
2. That removes options (b) and (d) immediately.

Answer: (a) 2.4

Explanation:
1. \( \ce{CO2} \) dissociated = 0.4 × 2 = 0.8 mol, leaving 1.2 mol.
2. CO formed = 0.8 mol and \( \ce{O2} \) formed = 0.4 mol.
3. Total = 1.2 + 0.8 + 0.4 = 2.4 mol.

Teacher's Note:
1. Write the mole table for reactant and each product.
2. Add them all for the total.

Answer: (a) n/2

Explanation:
1. Let x mol L⁻¹ of C form. Then B at equilibrium = n − x.
2. The condition [C] = [B] gives x = n − x, so x = n/2.
3. D is formed in equal amount to C, so [D] = n/2.

Teacher's Note:
1. Use a table of initial, change and equilibrium values.
2. Equate the two quantities given in the question.

Answer: (b) 6

Explanation:
1. \( \ce{PCl5} \) left = 4(1 − 0.5) = 2 mol.
2. \( \ce{PCl3} \) = 2 mol and \( \ce{Cl2} \) = 2 mol.
3. Total = 2 + 2 + 2 = 6 mol, which is \( 4(1+\alpha) \).

Teacher's Note:
1. For \( \ce{A <=> B + C} \) the total moles are \( n(1+\alpha) \).
2. Remember this shortcut.

Answer: (b) P/15

Explanation:
1. Dissociated = 2 mol. At equilibrium: \( \ce{PCl5} \) = 6, \( \ce{PCl3} \) = 2, \( \ce{Cl2} \) = 2, total = 10 mol.
2. \( K_p = \dfrac{(2P/10)(2P/10)}{6P/10} = \dfrac{0.04P}{0.6} \).
3. \( K_p = P/15 \).

Teacher's Note:
1. Convert moles to partial pressures using mole fractions.
2. Compare with \( K_p = \alpha^2 P/(1-\alpha^2) \) with \( \alpha = 0.25 \), which gives the same value.

Answer: (c) 2.4 atm

Explanation:
1. \( K_p = (0.3 \times 0.2)/0.6 = 0.1 \) atm.
2. New \( p_{PCl_3} = 0.6 \) and \( p_{Cl_2} = 0.4 \).
3. At constant \( K_p \): \( p_{PCl_5} = (0.6 \times 0.4)/0.1 = 2.4 \) atm.

Teacher's Note:
1. First get K from the given data.
2. Then use it with the new partial pressures.

Answer: (c) 0.5

Explanation:
1. \( \ce{PCl3} \) and \( \ce{Cl2} \) are formed in equal moles, so \( x_{Cl_2} = 0.25 \).
2. \( p_{Cl_2} = 0.25 \times 2.0 = 0.5 \) atm.

Teacher's Note:
1. Partial pressure = mole fraction × total pressure.
2. Equal stoichiometry gives equal mole fractions.

Answer: (a) \( \left(\dfrac{1}{18}\right)^2 \)

Explanation:
1. NO dissociated = 0.4 mol. At equilibrium: NO = 3.6, \( \ce{N2} \) = 0.2, \( \ce{O2} \) = 0.2 mol.
2. \( \Delta n_g = 0 \), so \( K_p = K_c \) and the volume cancels.
3. \( K = \dfrac{0.2 \times 0.2}{3.6^2} = \dfrac{1}{324} = \left(\dfrac{1}{18}\right)^2 \).

Teacher's Note:
1. When \( \Delta n = 0 \) the volume cancels and mole ratios can be used directly.
2. Do the arithmetic in fractions.

Answer: (d) 4/9

Explanation:
1. At equilibrium: C = D = 0.4 mol and A = B = 0.6 mol.
2. \( K = \dfrac{0.4 \times 0.4}{0.6 \times 0.6} = \dfrac{4}{9} \).

Teacher's Note:
1. Volume cancels because the moles of gas on both sides are equal.
2. Always subtract the amount reacted from the initial amount.

Answer: (b) 4/3

Explanation:
1. Let x mol of ester form. \( K_c = \dfrac{x^2}{(4-x)^2} = 4 \), so \( \dfrac{x}{4-x} = 2 \) and x = 8/3.
2. Acid left = 4 − 8/3 = 4/3 mol.

Teacher's Note:
1. The volume cancels since \( \Delta n = 0 \).
2. Take the square root of both sides when a perfect square appears.