JEE Chemistry Equilibrium MCQs Set 25

Answer: (c) 4

Explanation:
1. At equilibrium the forward and backward rates are equal, so \( K = k_f/k_b \).
2. \( K = 2\times10^{-4} / 5\times10^{-5} = 4 \).

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: (c) 0.0446 mol/L

Explanation:
1. 5.6 L at S.T.P. is 5.6/22.4 = 0.25 mol.
2. Active mass is concentration: 0.25 mol / 5.6 L = 0.0446 mol/L (equal to 1/22.4).

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: (a) 0.1 mol

Explanation:
1. With \( [\ce{SO3}] = [\ce{SO2}] \), \( K_c = 1/[\ce{O2}] = 100 \).
2. \( [\ce{O2}] = 0.01 \) mol/L, so in 10 L the amount is 0.1 mol.

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: (a) 0.4 mol

Explanation:
1. \( K_c = [\ce{SO3}]^2/([\ce{SO2}]^2[\ce{O2}]) = 4/[\ce{O2}] = 100 \), so \( [\ce{O2}] = 0.04 \) M.
2. In 10 L this is 0.4 mol.

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) 16.7%

Explanation:
1. Take 1 mol of each; let x mol of each react, so \( K = (2x)^2/(1-x)^2 = 16 \).
2. \( 2x/(1-x) = 4 \), giving \( x = 2/3 \).
3. Moles of \( \ce{A2} \) left = 1/3 out of 2 mol total, which is 16.7%.

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) 0.33 M

Explanation:
1. If the volume is V, \( K_c = (2/V)^2 / [(2/V)^3(2/V)] = V^2/4 = 9 \), so V = 6 L.
2. \( [\ce{C}] = 2/6 = 0.33 \) M.

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) 12 atm

Explanation:
1. Total moles = 3, so \( p_{N_2O_4} = 3 \) atm and \( p_{NO_2} = 6 \) atm.
2. \( K_p = 6^2/3 = 12 \) 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: (a) 0.5 M and 12.3 atm

Explanation:
1. \( K_c = k_f/k_b = 0.5 \).
2. \( x^2/(1-x) = 0.5 \) gives \( x^2 + 0.5x - 0.5 = 0 \), so \( x = 0.5 \) and \( [\ce{A}] = [\ce{B}] = [\ce{C}] = 0.5 \) M.
3. \( K_p = K_c RT = 0.5\times0.0821\times300 = 12.3 \) atm.

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) 0.0347

Explanation:
1. \( \Delta n = -2 \), so \( K_p = K_c (RT)^{-2} \).
2. \( RT = 0.082\times500 = 41 \); \( K_p = 58.3/41^2 = 0.0347 \) atm\(^{-2}\).

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) \( Q = 2.13\times10^3 \); backward

Explanation:
1. Concentrations: \( \ce{N2} = 0.075 \), \( \ce{H2} = 0.1 \), \( \ce{NH3} = 0.4 \) M.
2. \( Q = 0.4^2/(0.075\times0.1^3) = 2133 \).
3. \( Q > K_c \), so the net reaction goes 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: (a) 0.306 M

Explanation:
1. \( Q = 0.3^2/0.04^2 = 56.25 < 81 \), so the reaction goes forward.
2. \( (0.3+x)^2/(0.04-x)^2 = 81 \) gives \( 0.3 + x = 9(0.04 - x) \), so \( x = 0.006 \).
3. \( [\ce{NO}] = 0.306 \) M.

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) \( K_1K_2 \)

Explanation:
1. Adding the two equations gives \( \ce{N2 + 2O2 <=> 2NO2} \).
2. When equations are added, the equilibrium constants multiply: \( K = K_1K_2 \).

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.

Answer: (b) 4 mol/L

Explanation:
1. \( K = 0.667^2/0.333^2 = 4 \).
2. For X: \( (8/3)^2/(X-8/3)^2 = 4 \), so \( (8/3)/(X-8/3) = 2 \).
3. \( X = 4 \) mol/L.

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: (a) \( \dfrac{2}{3\sqrt3}\,P^{3/2} \)

Explanation:
1. The gases are formed in the ratio 1 : 1/2, so \( p_{NO_2} = 2P/3 \) and \( p_{O_2} = P/3 \).
2. \( K_p = (2P/3)(P/3)^{1/2} = \dfrac{2}{3\sqrt3}P^{3/2} \).

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: (d) 0.267 atm

Explanation:
1. \( K_p = 4\alpha^2P/(1-\alpha^2) = 4(0.0625)(1)/(0.9375) = 0.267 \) 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: (a) 63%

Explanation:
1. \( \alpha^2 = K_p/(4P + K_p) = 0.267/(0.4+0.267) = 0.4 \).
2. \( \alpha = 0.632 \), i.e. about 63%.

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) 0.2 and 0.333 atm

Explanation:
1. Observed molar mass = 2\(\times\)38.33 = 76.66; \( \alpha = (92-76.66)/76.66 = 0.2 \).
2. \( K_p = 4\alpha^2P/(1-\alpha^2) = 4(0.04)(2)/0.96 = 0.333 \) 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) \( 1.86\times10^{12} \)

Explanation:
1. \( \Delta G^\circ = -98320 - 298(-95) = -70010 \) J/mol.
2. \( \log K = 70010/(2.303\times8.314\times298) = 12.27 \).
3. \( K = 10^{12}\times10^{0.27} = 1.86\times10^{12} \).

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) 0.34 M

Explanation:
1. \( K_c = 2\times2/1 = 4 \). On doubling the volume the concentrations become 0.5, 1, 1, so \( Q = 2 < K \) and the reaction goes forward.
2. \( (1+x)^2/(0.5-x) = 4 \) gives \( x^2 + 6x - 1 = 0 \), so \( x = (-6+6.324)/2 = 0.162 \).
3. \( [\ce{A}] = 0.5 - 0.162 = 0.34 \) 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: (a) 50%

Explanation:
1. \( p_{H_2O}^5 = 2.43\times10^{-13} \) gives \( p_{H_2O} = 3.0\times10^{-3} \) atm = 2.28 torr.
2. Relative humidity at equilibrium = 2.28/4.56 = 50%.
3. Below this humidity the hydrate loses water (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: (c) 78.1 mm

Explanation:
1. Let \( p \) be the pressure of \( \ce{H2S} \). Then \( p_{A'} = 625/p \) and \( p_{B'} = 900/p \).
2. Mass balance: \( p = p_{A'} + p_{B'} = 1525/p \), so \( p = 39.05 \) mm.
3. Total = \( p + p_{A'} + p_{B'} = 2p = 78.1 \) mm.

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: (b) achieves an equilibrium state and proceeds in both directions

Explanation:
1. A reversible reaction can proceed in both directions.
2. Because both directions occur, it reaches a state of equilibrium.

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) the rates of forward and backward reactions are equal, the net rate is zero and measurable properties are constant

Explanation:
1. At equilibrium both rates are equal, so the net rate is zero.
2. Measurable properties such as concentration and colour stay constant, although the reaction is dynamic.

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) \( \dfrac{k_1a - k_2b}{k_1+k_2} \)

Explanation:
1. At equilibrium \( k_1(a-x) = k_2(b+x) \).
2. \( k_1a - k_2b = x(k_1+k_2) \), so \( x = (k_1a - k_2b)/(k_1+k_2) \).

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.