JEE Chemistry Equilibrium MCQs Set 26

Answer: (a) 0.75 mol

Explanation:
1. \( K = x^2/(1-x)^2 = 9 \), so \( x/(1-x) = 3 \).
2. \( x = 0.75 \) mol.

Teacher's Note:
1. Set up the moles at equilibrium with a single unknown x, express concentrations using the volume, and substitute in K.
2. Check that the final concentrations are positive and satisfy the expression.

Answer: (c) 0.4

Explanation:
1. 0.5 mol of \( \ce{N2} \) reacts, producing 1.0 mol of NO.
2. \( [\ce{NO}] = 1.0/2.5 = 0.4 \) M.

Teacher's Note:
1. Set up the moles at equilibrium with a single unknown x, express concentrations using the volume, and substitute in K.
2. Check that the final concentrations are positive and satisfy the expression.

Answer: (b) atm\(^{-2}\)

Explanation:
1. \( K_p = p_{CH_4}p_{H_2S}^2/(p_{CS_2}p_{H_2}^4) \), with 3 pressure terms above and 5 below.
2. The unit is atm\(^{3-5}\) = atm\(^{-2}\).

Teacher's Note:
1. Write the equilibrium expression from the balanced equation and leave out pure solids and liquids.
2. Always substitute equilibrium concentrations, never initial ones.

Answer: (d) \( \dfrac{4}{3P^2} \)

Explanation:
1. Since N2 and H2 stay in 1 : 3 ratio, \( p_{H_2} = P \).
2. \( p_{NH_3} = 2P - P/3 - P = 2P/3 \).
3. \( K_p = (2P/3)^2/[(P/3)(P)^3] = 4/(3P^2) \).

Teacher's Note:
1. Write the moles of each species at equilibrium in terms of the degree of dissociation, then find the total moles.
2. Partial pressure equals mole fraction times total pressure.

Answer: (c) 0.25 atm

Explanation:
1. With \( p_{SO_2} = p_{SO_3} \), \( K_p = 1/p_{O_2} = 4 \).
2. \( p_{O_2} = 0.25 \) atm.

Teacher's Note:
1. Write the moles of each species at equilibrium in terms of the degree of dissociation, then find the total moles.
2. Partial pressure equals mole fraction times total pressure.

Answer: (b) \( \dfrac{bc\,P}{a(a+b+c)} \)

Explanation:
1. \( p_i = n_i P/(a+b+c) \).
2. \( K_p = \dfrac{b P}{a+b+c}\cdot\dfrac{cP}{a+b+c}\Big/\dfrac{aP}{a+b+c} = \dfrac{bcP}{a(a+b+c)} \).

Teacher's Note:
1. Write the moles of each species at equilibrium in terms of the degree of dissociation, then find the total moles.
2. Partial pressure equals mole fraction times total pressure.

Answer: (c) 0.025 atm

Explanation:
1. \( p_{NO} = 2p_{O_2} = 0.5 \) atm.
2. \( K_p = (0.5)^2(0.25)/p_{NO_2}^2 = 100 \), so \( p_{NO_2}^2 = 6.25\times10^{-4} \).
3. \( p_{NO_2} = 0.025 \) atm.

Teacher's Note:
1. Write the moles of each species at equilibrium in terms of the degree of dissociation, then find the total moles.
2. Partial pressure equals mole fraction times total pressure.

Answer: (c) 0.14 M and 49/150

Explanation:
1. Moles of \( \ce{PCl3} \) = \( \ce{Cl2} \) = 0.7, so \( [\ce{PCl3}] = 0.7/5 = 0.14 \) M.
2. \( [\ce{PCl5}] = 0.06 \) M, \( K_c = 0.14^2/0.06 = 0.3267 = 49/150 \).

Teacher's Note:
1. Set up the moles at equilibrium with a single unknown x, express concentrations using the volume, and substitute in K.
2. Check that the final concentrations are positive and satisfy the expression.

Answer: (b) \( 4\times800R \)

Explanation:
1. \( \Delta n = 2 - 1 = 1 \) and T = 800 K.
2. \( K_p = K_c(RT) = 4\times800R \).

Teacher's Note:
1. Use Kp = Kc(RT)^(change in gas moles) and take R in units that match the pressure and concentration units.
2. Count only gaseous species when finding the change in moles.

Answer: (b) \( n(\ce{SO3}) + n(\ce{CO2}) = 4 \) and \( n(\ce{SO2})/n(\ce{CO}) < 1 \)

Explanation:
1. \( Q = (2\times2)/(2\times2) = 1 > 1/9 \), so the reaction goes backward.
2. Backward: \( \ce{SO2} \) and \( \ce{CO2} \) decrease, \( \ce{SO3} \) and CO increase, so \( n(\ce{SO2})/n(\ce{CO}) < 1 \).
3. \( n(\ce{SO3}) + n(\ce{CO2}) \) stays 4 and total moles stay 8 (\( \Delta n = 0 \)).

Teacher's Note:
1. Compare the reaction quotient Q with K: Q < K means the net reaction goes forward and Q > K means it goes backward.
2. Q has the same form as K but uses the concentrations present at that moment.

Answer: (a) backward

Explanation:
1. \( Q_p = 3^2/(1\times2^3) = 1.125 \).
2. \( Q_p \gg K_p \), so the net reaction is backward.

Teacher's Note:
1. Compare the reaction quotient Q with K: Q < K means the net reaction goes forward and Q > K means it goes backward.
2. Q has the same form as K but uses the concentrations present at that moment.

Answer: (d) must proceed in the forward direction

Explanation:
1. \( [\ce{A}] = 2/3 \), \( [\ce{B}] = 1/3 \), \( [\ce{C}] = 1 \) M.
2. \( Q = 1^3/[(2/3)^2(1/3)] = 6.75 < 49 \).
3. So the reaction proceeds forward.

Teacher's Note:
1. Compare the reaction quotient Q with K: Q < K means the net reaction goes forward and Q > K means it goes backward.
2. Q has the same form as K but uses the concentrations present at that moment.

Answer: (b) increases with time

Explanation:
1. Initially there is no product, but Q rises as products are formed.
2. Q increases with time until it equals K.

Teacher's Note:
1. Compare the reaction quotient Q with K: Q < K means the net reaction goes forward and Q > K means it goes backward.
2. Q has the same form as K but uses the concentrations present at that moment.

Answer: (a) \( 6.25\times10^{4} \) atm

Explanation:
1. Reversing the equation and doubling it gives \( K' = 1/K^2 \).
2. \( K' = 1/(4\times10^{-3})^2 = 6.25\times10^4 \) atm.

Teacher's Note:
1. Reverse the reaction and K becomes 1/K; multiply the equation by n and K is raised to the power n; add equations and multiply the K values.
2. Combine the given equilibria so that the unwanted species cancel.

Answer: (b) 1/4

Explanation:
1. Let the initial moles be 1 each: ester and water = 1/3, acid and alcohol = 2/3.
2. \( K_c = (1/3)^2/(2/3)^2 = 1/4 \).

Teacher's Note:
1. Write the equilibrium expression from the balanced equation and leave out pure solids and liquids.
2. Always substitute equilibrium concentrations, never initial ones.

Answer: (b) 32.5 g

Explanation:
1. \( K_c = [\ce{CO2}] = 0.05 \) mol/L, so moles of \( \ce{CO2} \) = 0.05\(\times\)6.5 = 0.325 mol.
2. Mass of \( \ce{CaCO3} \) = 0.325\(\times\)100 = 32.5 g; below this it decomposes completely.

Teacher's Note:
1. In a heterogeneous equilibrium pure solids and liquids do not appear in K, so Kp is a product of gas partial pressures.
2. Use the stoichiometric ratio of the gases formed to find each partial pressure.

Answer: (c) \( \sqrt{\dfrac{K_p}{4P + K_p}} \)

Explanation:
1. \( K_p = 4\alpha^2P/(1-\alpha^2) \).
2. \( K_p - K_p\alpha^2 = 4\alpha^2P \), so \( \alpha^2 = K_p/(4P + K_p) \).

Teacher's Note:
1. Write the moles of each species at equilibrium in terms of the degree of dissociation, then find the total moles.
2. Partial pressure equals mole fraction times total pressure.

Answer: (c) \( \dfrac{\alpha^3P_0}{(2+\alpha)(1-\alpha)^2} \)

Explanation:
1. Start with 1 mol: \( \ce{SO3} = 1-\alpha \), \( \ce{SO2} = \alpha \), \( \ce{O2} = \alpha/2 \), total = \( 1+\alpha/2 \).
2. \( K_p = \dfrac{\alpha^2(\alpha/2)}{(1-\alpha)^2(1+\alpha/2)}P_0 = \dfrac{\alpha^3P_0}{(2+\alpha)(1-\alpha)^2} \).

Teacher's Note:
1. Write the moles of each species at equilibrium in terms of the degree of dissociation, then find the total moles.
2. Partial pressure equals mole fraction times total pressure.

Answer: (a) \( d_{20} > d_{45} > d_{65} > d_{80} \)

Explanation:
1. Observed vapour density \( d = D/(1+\alpha) \).
2. More dissociation gives more moles and a lower density, so the order decreases as \( \alpha \) increases.

Teacher's Note:
1. Write the moles of each species at equilibrium in terms of the degree of dissociation, then find the total moles.
2. Partial pressure equals mole fraction times total pressure.

Answer: (d) \( \alpha \propto 1/\sqrt{P} \)

Explanation:
1. \( K_p = \alpha^2P/(1-\alpha^2) \approx \alpha^2P \) for small \( \alpha \).
2. \( \alpha = \sqrt{K_p/P} \propto P^{-1/2} \).

Teacher's Note:
1. Write the moles of each species at equilibrium in terms of the degree of dissociation, then find the total moles.
2. Partial pressure equals mole fraction times total pressure.

Answer: (a) \( \alpha = 2 - \dfrac{2D}{d} \)

Explanation:
1. Start with 1 mol of B: at equilibrium \( 1-\alpha \) mol B and \( \alpha/2 \) mol \( \ce{B2} \), total \( 1-\alpha/2 \).
2. Vapour density \( d = D/(1-\alpha/2) \), so \( 1 - \alpha/2 = D/d \).
3. \( \alpha = 2 - 2D/d \).

Teacher's Note:
1. Write the moles of each species at equilibrium in terms of the degree of dissociation, then find the total moles.
2. Partial pressure equals mole fraction times total pressure.

Answer: (d) 4.23 g/L

Explanation:
1. Molar mass of \( \ce{PCl5} \) = 208.5; average molar mass = 208.5/1.5 = 139.
2. \( d = PM/RT = 2\times139/(0.0821\times800) = 4.23 \) g/L.

Teacher's Note:
1. Write the moles of each species at equilibrium in terms of the degree of dissociation, then find the total moles.
2. Partial pressure equals mole fraction times total pressure.

Answer: (b) \( -\Delta G^\circ = RT\ln K \)

Explanation:
1. The standard free energy change is \( \Delta G^\circ = -RT\ln K \).

Teacher's Note:
1. The van't Hoff relation shows that K increases with temperature for an endothermic reaction and decreases for an exothermic one.
2. Use the standard free energy relation to connect K with energy changes.

Answer: (b) \( \Delta H < 0 \)

Explanation:
1. K falls as the temperature rises, so the forward reaction is exothermic and \( \Delta H < 0 \).

Teacher's Note:
1. The van't Hoff relation shows that K increases with temperature for an endothermic reaction and decreases for an exothermic one.
2. Use the standard free energy relation to connect K with energy changes.

Answer: (b) endothermic

Explanation:
1. K increases with temperature, which is the behaviour of an endothermic reaction.

Teacher's Note:
1. The van't Hoff relation shows that K increases with temperature for an endothermic reaction and decreases for an exothermic one.
2. Use the standard free energy relation to connect K with energy changes.