JEE Chemistry Equilibrium MCQs Set 29

Answer: (a) 1 : 36

Explanation:
1. For \( \ce{X <=> 2Y} \), \( K_{p_1} = 4\alpha^2P_1/(1-\alpha^2) \). For \( \ce{Z <=> P + Q} \), \( K_{p_2} = \alpha^2P_2/(1-\alpha^2) \).
2. \( K_{p_1}/K_{p_2} = 4P_1/P_2 = 1/9 \), so \( P_1 : P_2 = 1 : 36 \).

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) 1.8 atm

Explanation:
1. Let x atm of \( \ce{CO2} \) react: \( p_{CO_2} = 0.5 - x \), \( p_{CO} = 2x \); total = 0.5 + x = 0.8, so x = 0.3.
2. \( K_p = (0.6)^2/0.2 = 1.8 \) 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.

Answer: (d) -1/2

Explanation:
1. \( x = \Delta n_g = 1 - 1.5 = -1/2 \).

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: (d) reverse direction because \( Q > K_c \)

Explanation:
1. \( K_c = e^{-\Delta G^\circ/RT} = e^{-2494.2/(8.314\times300)} = e^{-1} = 0.37 \).
2. \( Q = (2)(1/2)/(1/2)^2 = 4 \).
3. \( Q > K_c \), so the reaction goes in the reverse direction.

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) 0.25

Explanation:
1. Start with 1 mol HI: HI = 0.5, \( \ce{H2} \) = 0.25, \( \ce{I2} \) = 0.25 mol.
2. \( K = (0.25\times0.25)/(0.5)^2 = 0.25 \).

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) The equilibrium remains unchanged

Explanation:
1. At constant volume the concentrations, and hence Q, do not change.
2. The equilibrium is 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: (d) \( 5.82\times10^{-2} \) atm

Explanation:
1. \( K_p = (2p)^2p = 4p^3 = 2.9\times10^{-5} \) gives \( p = 1.94\times10^{-2} \) atm.
2. Total = 3p = \( 5.82\times10^{-2} \) 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.

Answer: (a) 3.11 g/L

Explanation:
1. Average molar mass = 92/1.2 = 76.67 g/mol.
2. \( d = PM/RT = 76.67/(0.0821\times300) = 3.11 \) 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) a shift of the equilibrium in the forward direction

Explanation:
1. Water is denser than ice, so ice to water decreases the volume.
2. Higher pressure shifts the equilibrium forward (more water forms).

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) \( \ce{2NO2(g) <=> 2NO(g) + O2(g)} \)

Explanation:
1. A volume increase favours the side with more gas moles.
2. \( \ce{2NO2 <=> 2NO + O2} \) has 2 moles of gas on the left and 3 on the right, so products are favoured.
3. The other reactions have equal or fewer gas moles on the product 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: (a) 2.5

Explanation:
1. At equilibrium: CO = 1 mol, \( \ce{Cl2} \) = 2 mol, \( \ce{COCl2} \) = 1 mol in 5 L.
2. \( K_c = (1/5)/[(1/5)(2/5)] = 2.5 \).

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) 4763

Explanation:
1. \( K_p = 4\alpha^2P/(1-\alpha^2) = 0.16/0.96 = 1/6 \).
2. \( \Delta G^\circ = -RT\ln K = RT\ln6 = 8.314\times320\times(0.693+1.098) \approx 4763 \) J/mol.

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: (d) addition of an inert gas at constant pressure

Explanation:
1. Decomposition of \( \ce{N2O4} \) is the backward reaction (more moles, endothermic).
2. Constant-pressure inert gas lowers partial pressures and favours it; lowering T or raising P favours \( \ce{N2O4} \) formation.

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) \( K_2 = K_1^{-3} \)

Explanation:
1. The second equation is the reverse of 3 times the first.
2. \( K_2 = (1/K_1)^3 = K_1^{-3} \).

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: (d) 0.242 atm\(^2\)

Explanation:
1. Moles of \( \ce{NH4SH} \) = 5.1/51 = 0.1; 30% decomposes: 0.03 mol each of \( \ce{NH3} \) and \( \ce{H2S} \).
2. \( p = nRT/V = 0.03\times0.082\times600/3 = 0.492 \) atm each.
3. \( K_p = 0.492^2 = 0.242 \) atm\(^2\).

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) 1

Explanation:
1. HI = 3 mol, so 1.5 mol each of \( \ce{H2} \) and \( \ce{I2} \) are used up, leaving 3 mol each.
2. \( K_c = (3)^2/(3\times3) = 1 \) (the volume cancels).

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) 0.02

Explanation:
1. The dissociation is the reverse of the synthesis, so \( K = 1/50 = 0.02 \).

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: (a) 64

Explanation:
1. K depends only on temperature and does not change with the volume.

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: (c) 0.75 and 0.55 mol/L

Explanation:
1. 1.0 mol/L of NO needs 0.5 mol/L each of \( \ce{N2} \) and \( \ce{O2} \).
2. Initial \( \ce{N2} \) = 0.25 + 0.5 = 0.75 and \( \ce{O2} \) = 0.05 + 0.5 = 0.55 mol/L.

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) \( 1/18^2 \)

Explanation:
1. At equilibrium: NO = 3.6 mol, \( \ce{N2} \) = \( \ce{O2} \) = 0.2 mol.
2. \( \Delta n = 0 \), so \( K_p = K_c = (0.2\times0.2)/3.6^2 = 1/324 = 1/18^2 \).

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: (b) \( \dfrac{n}{n-1}\left(\dfrac{d_f - d_i}{d_f}\right) = \alpha \)

Explanation:
1. At constant P and T the volume is proportional to the moles, so density is inversely proportional to moles.
2. Final moles = \( 1 - \alpha + \alpha/n = 1 - \alpha(n-1)/n \).
3. \( d_i/d_f = 1 - \alpha(n-1)/n \), so \( \alpha = \dfrac{n}{n-1}\cdot\dfrac{d_f - d_i}{d_f} \).

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) \( P^2/4 \) atm\(^2\)

Explanation:
1. Each gas has pressure P/2.
2. \( K_p = (P/2)^2 = P^2/4 \) atm\(^2\).

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) exothermic

Explanation:
1. K decreases when the temperature rises, so the forward reaction is exothermic.

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: (d) 163 kJ/mol

Explanation:
1. \( \Delta G^\circ = -RT\ln K_p = -8.314\times298\times\ln(2.47\times10^{-29}) \).
2. \( \ln K = -65.9 \), so \( \Delta G^\circ = 2478\times65.9 = 1.63\times10^5 \) J = 163 kJ/mol.

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) catalyst

Explanation:
1. A catalyst changes only the rate, not the equilibrium.
2. Pressure and volume change shift the equilibrium because the gas moles change.

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.