JEE Chemistry Equilibrium MCQs Set 28

Answer: (b) 78

Explanation:
1. At the point where the solid is just fully converted, \( p_{NH_3} = \sqrt{9} = 3 \) atm.
2. Free \( \ce{NH3} \) = PV/RT = 3\(\times\)5/(0.0821\(\times\)313) = 0.584 mol.
3. \( \ce{NH3} \) bound in the solid = 2\(\times\)0.1 = 0.2 mol. Total = 0.784 mol, and 100\(\times\) = 78.

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) 5

Explanation:
1. Increases: (b) backward shift raises \( \ce{H2O} \); (d) decreasing volume shifts backward (fewer moles), raising \( \ce{Cl2} \); (e) \( p_{Cl_2} \) rises as the volume decreases; (g) K rises for an endothermic reaction; (h) forward shift raises [HCl].
2. Not increased: (a), (c), (f), (i) and (j).
3. The count is 5.

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) 4

Explanation:
1. Let \( p_A = a \). Then \( p_C = 500/a \) and \( p_D = 2000/a \).
2. From the stoichiometry \( p_A = p_C + p_D \), so \( a^2 = 2500 \) and \( a = 50 \).
3. \( p_C = 10 \), \( p_D = 40 \), total = 100 atm, and 100/25 = 4.

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: (a) 58

Explanation:
1. Originally \( p_{NH_3} = 2p \), \( p_{CO_2} = p \), total = 3p, \( K_p = 4p^3 \).
2. New \( p_{NH_3} = 3p \) gives \( p_{CO_2} = 4p^3/9p^2 = 4p/9 \).
3. New total = 3p + 4p/9 = 31p/9; ratio to 3p is 31/27, so a + b = 58.

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) A, B and D

Explanation:
1. \( K_c \) depends only on temperature.
2. It is independent of initial concentrations, pressure and catalyst.

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) A, C and D

Explanation:
1. The fourth reaction = reverse of first + second + 3(third).
2. \( K_4 = K_2K_3^3/K_1 \), which is (D).
3. Rearranging gives \( K_1 = K_2K_3^3/K_4 \) (A) and \( K_2 = K_4K_1/K_3^3 \) (C); (B) is wrong.

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: (d) A, B, C and D

Explanation:
1. \( K = k_f/k_b = 9 \).
2. \( x^2/(1-x)^2 = 9 \) gives \( x = 0.75 \) mol; \( [\ce{C}]/[\ce{B}] = 0.75/0.25 = 3 \) and \( [\ce{D}] = 0.75/5 = 0.15 \) mol/L.
3. \( \Delta n = 0 \), so \( K_p = K_c \).

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: (a) A, C and D

Explanation:
1. \( K_p = p_{SO_2}^2p_{O_2} \) depends only on temperature, and with solids only the two gases are formed in fixed ratio.
2. So the partial pressures are the same after a volume change as long as some \( \ce{CaSO4} \) remains: (B) is wrong and (C) is right.
3. A higher T favours the endothermic forward reaction (A), and He at constant pressure lowers the partial pressures and shifts forward (D).

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) high pressure and low temperature

Explanation:
1. Dissolution is exothermic, so a low temperature favours it.
2. It lowers the volume, so a high pressure favours it.

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: (b) low pressure and high temperature

Explanation:
1. The volume rises on melting, so low pressure favours it.
2. Melting needs heat (endothermic), so high temperature favours it.

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) -50

Explanation:
1. The slope of \( \log K \) against \( 1/T \) is \( -\Delta H/(2.303R) \).
2. \( = -230/(2.303\times1.987) = -50 \).

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) temperature and pressure

Explanation:
1. The reaction changes the number of gas moles, so pressure shifts it; temperature changes K.
2. A catalyst does not change the equilibrium amount.

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: (b) \( 1.44\times10^{-5}\,(0.082\times773)^2 \)

Explanation:
1. \( \Delta n = -2 \), so \( K_p = K_c(RT)^{-2} \).
2. \( K_c = K_p(RT)^2 \) with T = 773 K and R = 0.082 L atm/mol K, i.e. \( 1.44\times10^{-5}\times(0.082\times773)^2 \).

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: (d) \( K_p \) remains constant with changes in P and x

Explanation:
1. \( K_p \) depends only on temperature.
2. When P changes, x adjusts so that \( K_p \) is unchanged.

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) A, B and D

Explanation:
1. \( \Delta H \) varies with temperature and does not depend on a catalyst.
2. K depends only on temperature; it is independent of the initial amount of solid.
3. K is not changed by changing the pressure at a given T; (C) is incorrect.

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) \( \dfrac{8\beta^2}{4-\beta^2} \)

Explanation:
1. Moles: \( \ce{X2} = 1 - \beta/2 \), X = \( \beta \), total = \( 1 + \beta/2 \).
2. \( K_p = \dfrac{[2\beta/(2+\beta)]^2\times2^2}{[(2-\beta)/(2+\beta)]\times2} \), which simplifies to \( \dfrac{8\beta^2}{4-\beta^2} \).

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) \( (RT)^{1/2} \)

Explanation:
1. \( \Delta n = 1 - 1.5 = -1/2 \), so \( K_p = K_c(RT)^{-1/2} \).
2. \( K_c/K_p = (RT)^{1/2} \).

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: (c) lowering the temperature and increasing the pressure

Explanation:
1. The reaction is exothermic, so low temperature favours it.
2. Gas moles decrease, so high pressure favours it.

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: (b) \( K_c = 1/[\ce{O2}]^5 \)

Explanation:
1. Pure solids do not appear in K.
2. Only \( \ce{O2} \) is in the denominator: \( K_c = 1/[\ce{O2}]^5 \).

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) 1/RT

Explanation:
1. \( \Delta n = 1 - 2 = -1 \).
2. \( K_p/K_c = (RT)^{-1} = 1/RT \).

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) 50

Explanation:
1. The required equation is the reverse of half the given one.
2. \( K = (1/4\times10^{-4})^{1/2} = (2500)^{1/2} = 50 \).

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: (d) \( K_p \) is greater than \( K_c \)

Explanation:
1. \( \Delta n = 3 - 2 = +1 \), so \( K_p = K_c(RT) \).
2. Since RT > 1 (in L atm/mol), \( K_p > K_c \).

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) adding \( \ce{F2} \)

Explanation:
1. Adding a reactant shifts the equilibrium forward.
2. Increasing the volume, removing \( \ce{Cl2} \) or raising the temperature each shift it backward.

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: (b) 0.11

Explanation:
1. Let \( p_{H_2S} = x \); \( p_{NH_3} = 0.5 + x \). Total = 0.5 + 2x = 0.84, so x = 0.17.
2. \( K_p = (0.67)(0.17) = 0.114 \approx 0.11 \).

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) \( K_3 = K_1K_2 \)

Explanation:
1. Reaction (c) = (a) + (b).
2. So \( K_3 = K_1K_2 \).

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.