JEE Chemistry Equilibrium MCQs Set 27

Answer: (b) \( (1/T_2 - 1/T_1) \) is positive

Explanation:
1. For \( T_2 > T_1 \), \( 1/T_2 < 1/T_1 \), so \( (1/T_2 - 1/T_1) \) is negative; the statement that it is positive is false.
2. For an endothermic reaction \( \Delta H > 0 \) and \( K_2 > K_1 \).

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: (a) increasing the amount of CO

Explanation:
1. A catalyst has no effect on the amount at equilibrium, and volume change does not shift the equilibrium because \( \Delta n = 0 \).
2. Adding CO (a reactant) shifts the equilibrium forward and increases \( \ce{CO2} \).

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Answer: (d) Less \( \ce{NH3} \) is produced

Explanation:
1. At constant pressure the volume rises, lowering all partial pressures.
2. The equilibrium shifts to the side with more moles (reactants), so less \( \ce{NH3} \) forms. \( K_p \) depends only on temperature.

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Answer: (d) there is no effect on the equilibrium

Explanation:
1. In a closed (constant volume) container the concentrations of all species are unchanged by the added helium.
2. The equilibrium and \( K_p \) are therefore unaffected.

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Answer: (c) high pressure and high temperature

Explanation:
1. Diamond is denser (smaller volume), so high pressure favours it.
2. Graphite to diamond is endothermic, so high temperature favours diamond (and helps the rate).

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Answer: (d) the pressure is constant and \( \Delta n_g \neq 0 \)

Explanation:
1. At constant volume an inert gas changes no partial pressure.
2. At constant pressure it increases the volume, which shifts the equilibrium only if \( \Delta n_g \neq 0 \).

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Answer: (d) More ice melts if the pressure is increased

Explanation:
1. Water is denser than ice, so melting reduces the volume.
2. Increasing the pressure shifts the equilibrium towards water, so more ice melts.

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Answer: (c) 35%

Explanation:
1. Relative humidity = (partial pressure/vapour pressure)\(\times\)100.
2. \( 19.355/55.3\times100 = 35\% \).

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: (a) 67.85%

Explanation:
1. \( p_{H_2O} = \sqrt{4\times10^{-4}} = 0.02 \) atm = 15.2 torr.
2. Relative humidity = 15.2/22.4 = 67.85%.

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: (b) inversely proportional to the square of \( [\ce{B-}] \)

Explanation:
1. \( K_1 = [\ce{A+}][\ce{B-}]/[\ce{AB}] \) and \( K_2 = [\ce{AB2-}]/([\ce{AB}][\ce{B-}]) \).
2. Eliminating [AB]: \( K_1/K_2 = [\ce{A+}][\ce{B-}]^2/[\ce{AB2-}] \).
3. So \( [\ce{A+}]/[\ce{AB2-}] = (K_1/K_2)/[\ce{B-}]^2 \), inversely proportional to the square of \( [\ce{B-}] \).

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) the amount of \( \ce{PCl5} \) will be less

Explanation:
1. Added CO shifts the second equilibrium backward and consumes \( \ce{Cl2} \).
2. Lower \( \ce{Cl2} \) shifts the first equilibrium forward, so \( \ce{PCl5} \) decreases.

Teacher's Note:
1. For simultaneous equilibria sharing a gas, the common species has one partial pressure that satisfies every expression.
2. Write each Kp, express every unknown through the common pressure and solve.

Answer: (d) \( \log K_1 + \log K_2 + \log K_3 = \log K \)

Explanation:
1. Adding the three equations gives the required one.
2. So \( K = K_1K_2K_3 \), i.e. \( \log K = \log K_1 + \log K_2 + \log K_3 \).

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) \( 11.3\times10^{-7} \) atm

Explanation:
1. \( p_{O_2} \approx 8 \) atm. \( K_p = p_{O_2}^3/p_{O_3}^2 \).
2. \( p_{O_3}^2 = 512/(4\times10^{14}) = 1.28\times10^{-12} \).
3. \( p_{O_3} = 1.13\times10^{-6} \) 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) the degree of dissociation of \( \ce{A2B} \) is 2/3

Explanation:
1. Moles: \( \ce{A2B} = 4(1-\alpha) \), \( \ce{A2} = 4\alpha \), \( \ce{B2} = 2\alpha \), total \( 4+2\alpha \).
2. \( K_p = \alpha^3P/[(1-\alpha)^2(2+\alpha)] = P \) gives \( \alpha^3 = 2 - 3\alpha + \alpha^3 \), so \( \alpha = 2/3 \).
3. Total moles = 16/3, \( \ce{A2B} = 4/3 \), \( \ce{A2} = 8/3 \), \( \ce{B2} = 4/3 \); only the first statement is true.

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) 2.5 L

Explanation:
1. \( \ce{CO2} \) pressure can rise only to \( K_p = 0.8 \) atm before \( \ce{CaCO3} \) forms.
2. Isothermally \( P_1V_1 = P_2V_2 \): \( V = 0.2\times10/0.8 = 2.5 \) L.

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: (a) \( \sqrt2 \)

Explanation:
1. \( K_p = p_{HCl}^2/p_{H_2O} \) is constant.
2. If \( p_{H_2O} \) doubles, \( p_{HCl}^2 \) doubles, so \( p_{HCl} \) increases by \( \sqrt2 \).

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: (d) \( 1.18\times10^{-4} \)

Explanation:
1. \( \log(K_{1300}/K_{1200}) = \dfrac{597400}{2.303\times8.314}\left(\dfrac1{1200} - \dfrac1{1300}\right) = 31200\times6.41\times10^{-5} = 2.0 \).
2. \( K_{1200} = 0.0118/100 = 1.18\times10^{-4} \).

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: (c) the relative humidity is always 100% at equilibrium at constant temperature

Explanation:
1. At equilibrium in the presence of liquid, the vapour pressure equals the saturation value at that temperature, so the relative humidity is 100%.
2. Adding inert gas raises the total pressure, and any other gas also has a fixed saturation pressure at fixed T; hence the other statements are wrong.

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: (b) \( \ce{[Cr(dien)2]^{3+}(aq) + 3H2O(l) + 3Cl-(aq) <=> [Cr(H2O)3Cl3](aq) + 2dien(aq)} \)

Explanation:
1. On dilution an equilibrium shifts towards the side with more solute particles.
2. In the first reaction the left side has 4 dissolved particles and the right side 3, so dilution shifts it backward.
3. In the other three, dilution favours the ionised (forward) side.

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Answer: (d) the dissociation of both \( \ce{PCl5} \) and \( \ce{N2O4} \) increases

Explanation:
1. At constant pressure the volume increases, which lowers the partial pressures.
2. Both reactions have \( \Delta n > 0 \), so both shift forward, increasing the dissociation.

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Answer: (a) 101/18

Explanation:
1. \( K_c = (0.03)^2/(0.01\times0.02) = 4.5 \).
2. New NO = 4 mol, so 1 mol more NO forms using 0.5 mol each of \( \ce{N2} \) and \( \ce{O2} \): \( \ce{N2} \) = 0.5 mol.
3. \( 4.5 = 16/[0.5\times(1.5+y)] \) gives \( 1.5 + y = 7.11 \), so \( y = 5.61 = 101/18 \).

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) \( \ce{CaCl2.6H2O} \) acts as an efflorescent substance

Explanation:
1. \( p_{H_2O}^6 = 6.4\times10^{-17} \) gives \( p = 2.0\times10^{-3} \) atm = 1.52 torr.
2. The actual 1.14 torr is lower, so \( Q < K_p \) and the hydrate loses water: it is efflorescent.

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: (a) 7

Explanation:
1. \( (2x)^2/(0.5-x)^2 = 49 \) gives \( 2x/(0.5-x) = 7 \), so \( x = 0.389 \) mol.
2. HI = 0.778 mol and \( \ce{I2} \) = 0.111 mol, ratio = 7 (the same as \( \sqrt{49} \)).

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) 4

Explanation:
1. For each mole decomposed, the gas moles rise by \( 2 + 0.5 - 1 = 1.5 \).
2. \( 600 + 1.5\times600x = 960 \) gives \( x = 0.4 \), so 10x = 4.

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) 4

Explanation:
1. Each gas has pressure \( P/2 \) when the moles are equal.
2. \( K_p = p/p^4 = 1/p^3 = 0.125 \) gives \( p = 2 \) atm.
3. Total pressure = 4 atm.

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.