JEE Chemistry Equilibrium MCQs Set 08

Answer: (c) P-1 and 5; Q-2 and 3; R-1 and 4

Explanation:
1. P is endothermic, so high temperature favours it, and \( \Delta n = 0 \) so pressure has no effect: 1 and 5.
2. Q is exothermic (low temperature favours it) and \( \Delta n = -1 \) so high pressure favours it: 2 and 3.
3. R is endothermic (high temperature) and \( \Delta n = +3 \) so low pressure favours it: 1 and 4.

Teacher's Note:
1. Decide temperature from the sign of the heat change and pressure from the change in number of gas moles.
2. If the moles of gas do not change, pressure has no effect.

Answer: (d) \( a > (l + m) \)

Explanation:
1. Increasing the volume favours the side with more moles of gas.
2. If dissociation decreases, the reactant side has more moles, so \( a > (l + m) \).

Teacher's Note:
1. A volume increase works like a pressure decrease and favours the side with more moles.
2. Dissociation falls only when the reactant side is larger.

Answer: (b) the rate of the forward reaction equals the rate of the backward reaction

Explanation:
1. Equilibrium is a dynamic state in which the forward and backward rates are equal.
2. Reaction does not go to completion, and the concentrations need not be equal; they are only constant.

Teacher's Note:
1. Equal rates, not equal concentrations, define equilibrium.
2. Concentrations stay constant but not necessarily equal.

Answer: (c) Low temperature, high pressure and high concentration of reactants

Explanation:
1. \( \ce{N2 + 3H2 <=> 2NH3} \) is exothermic, so low temperature favours the product.
2. \( \Delta n = -2 \), so high pressure favours the product.
3. A high concentration of reactants also drives the reaction forward.

Teacher's Note:
1. Exothermic reaction means low temperature gives more product.
2. Fewer gas moles on the product side means high pressure helps.

Answer: (b) \( K \ll 1 \)

Explanation:
1. \( K = [\text{products}]/[\text{reactants}] \).
2. If the reactants dominate at equilibrium, K is much less than 1, but not exactly zero.

Teacher's Note:
1. A large K means products dominate, a small K means reactants dominate.
2. K is never exactly zero for a reaction that can occur.

Answer: (c) the temperature

Explanation:
1. The equilibrium constant depends only on temperature.
2. It does not depend on initial concentrations, pressure, volume or a catalyst.

Teacher's Note:
1. K changes only when temperature changes.
2. Pressure and concentration change the position of equilibrium, not K.

Answer: (a) 0.168 atm

Explanation:
1. Let \( p_{NO_2} = x \), then \( p_{N_2O_4} = 0.5 - x \).
2. \( K_p = x^2/(0.5 - x) = 0.66 \), so \( x^2 + 0.66x - 0.33 = 0 \).
3. This gives \( x = 0.3327 \) atm, so \( p_{N_2O_4} = 0.5 - 0.3327 = 0.168 \) atm.

Teacher's Note:
1. Write the partial pressures in terms of one unknown using the total pressure.
2. Solve the quadratic and subtract from the total.

Answer: (a) The amount of A increases

Explanation:
1. Increasing pressure favours the side with fewer gas moles.
2. A is the side with one mole, so the equilibrium shifts backward and the amount of A increases.

Teacher's Note:
1. High pressure shifts the equilibrium to the side with fewer moles of gas.
2. Count gas moles on both sides first.

Answer: (a) 0.0625

Explanation:
1. \( K_c = \dfrac{[BA_2]}{[A]^2[B]} \).
2. \( K_c = \dfrac{2}{4^2 \times 2} = \dfrac{2}{32} = 0.0625 \).

Teacher's Note:
1. Square the concentration of A because its coefficient is 2.
2. Substitute carefully and divide.

Answer: (d) \( -\dfrac{1}{2} \)

Explanation:
1. \( x = \Delta n_g = 1 - (1 + \tfrac{1}{2}) = -\tfrac{1}{2} \).

Teacher's Note:
1. x is the change in moles of gas, products minus reactants.
2. Use the coefficients exactly as written.

Answer: (c) 0.35

Explanation:
1. At equilibrium \( [PCl_3] = [Cl_2] = 0.1 \).
2. \( K_c = \dfrac{(0.1)(0.1)}{[PCl_5]} = 0.04 \), so \( [PCl_5] = 0.25 \).
3. Initial \( \ce{PCl5} \) = 0.25 + 0.1 = 0.35 mol L⁻¹.

Teacher's Note:
1. Do not forget that the dissociated PCl5 must be added back to the equilibrium amount.
2. Initial = left over + dissociated.

Answer: (b) \( \dfrac{2}{3} \)

Explanation:
1. \( K_p = \dfrac{\alpha^2 P}{1 - \alpha^2} = 4 \) with P = 5.
2. \( 5\alpha^2 = 4 - 4\alpha^2 \), so \( \alpha^2 = 4/9 \).
3. \( \alpha = 2/3 \).

Teacher's Note:
1. Use \( K_p = \alpha^2 P/(1 - \alpha^2) \) for one mole giving two.
2. Solve for \( \alpha^2 \) first and then take the square root.

Answer: (b) 3.2

Explanation:
1. 20% is left, so \( [PCl_5] = 0.2 \) and 0.8 mol each of \( \ce{PCl3} \) and \( \ce{Cl2} \) are formed in 1 L.
2. \( K_c = \dfrac{0.8 \times 0.8}{0.2} = 3.2 \).

Teacher's Note:
1. First find the amount that has dissociated from the amount left.
2. Volume is 1 L, so moles equal concentration.

Answer: (c) \( \dfrac{P}{3} \)

Explanation:
1. \( K_p = \dfrac{\alpha^2 P}{1 - \alpha^2} \) with \( \alpha = 0.5 \).
2. \( K_p = \dfrac{0.25P}{0.75} = \dfrac{P}{3} \).

Teacher's Note:
1. Use the standard \( \alpha^2 P/(1 - \alpha^2) \) for \( A \rightleftharpoons B + C \).
2. Substitute 0.5 for \( \alpha \).

Answer: (d) \( \dfrac{1}{49} \)

Explanation:
1. HI left = 1 - 2/9 = 7/9 mol, and H2 = I2 = 1/9 mol (2 mol HI give 1 mol H2).
2. \( K_c = \dfrac{(1/9)(1/9)}{(7/9)^2} = \dfrac{1}{49} \).

Teacher's Note:
1. Two moles of HI give one mole each of H2 and I2, so the dissociated amount is halved.
2. Volume cancels because \( \Delta n = 0 \).

Answer: (a) 7

Explanation:
1. Let [HI] = 2y and \( [H_2] = [I_2] = z \). Then \( K_c = (2y)^2/z^2 = 49 \).
2. Taking the square root, \( [HI]/[I_2] = 7 \).

Teacher's Note:
1. Since H2 and I2 are equal, \( K_c \) is a perfect square of [HI]/[I2].
2. Taking the square root gives the ratio directly.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. Both statements are true.
2. K depends only on temperature, so adding an inert gas cannot change it.
3. The Reason therefore explains the Assertion.

Teacher's Note:
1. K is a function of temperature alone.
2. Concentration, pressure and inert gases do not change it.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. For this reaction \( \Delta n_g = 0 \), so \( K_p = K_c \). The Assertion is true.
2. In general \( K_p = K_c (RT)^{\Delta n_g} \), which equals \( K_c \) only when \( \Delta n_g = 0 \). The Reason is false.

Teacher's Note:
1. \( K_p = K_c \) only when the gas moles do not change.
2. Check \( \Delta n_g \) before comparing them.

Answer: (d) Assertion is false but the Reason is true.

Explanation:
1. K depends only on temperature and does not change with concentration. The Assertion is false.
2. Adding reactants shifts the equilibrium forward by Le Chatelier's principle. The Reason is true.

Teacher's Note:
1. A shift in position is not a change in the value of K.
2. Only temperature changes K.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. The Assertion is true: temperature dependence of K is given by the van't Hoff equation.
2. The correct form is \( \log\dfrac{K_2}{K_1} = \dfrac{\Delta H}{2.303R}\left[\dfrac{T_2 - T_1}{T_1 T_2}\right] \). The equation in the Reason has \( K_1/K_2 \) on the left, so it has the wrong sign and the Reason is false.

Teacher's Note:
1. Remember the form \( \log(K_2/K_1) \) with \( (T_2 - T_1)/(T_1T_2) \).
2. Inverting the K ratio reverses the sign.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. Opening the bottle lowers the pressure above the liquid, so the equilibrium \( \ce{CO2(g) <=> CO2(aq)} \) shifts to the gas side.
2. Both statements are true and the Reason explains the Assertion.

Teacher's Note:
1. Lower pressure reduces gas solubility.
2. The system moves to a new equilibrium at the lower pressure.

Answer: (b) Both Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.

Explanation:
1. Both statements are true.
2. Higher pressure favours the side with less gas volume (dissolved gas), which explains the pressure effect.
3. Being exothermic explains the effect of temperature, not of pressure, so the Reason does not explain the Assertion.

Teacher's Note:
1. Pressure affects solubility through the volume change.
2. The heat of solution explains the temperature effect instead.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. For an exothermic process, raising the temperature shifts the equilibrium towards the undissolved gas, so solubility decreases.
2. Both statements are true and the Reason explains the Assertion.

Teacher's Note:
1. Heating favours the endothermic direction, here the escape of gas.
2. That is why warm water holds less gas.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. A catalyst speeds up the forward and backward reactions equally, so the equilibrium composition and K stay the same. The Assertion is true.
2. The Reason is false, since the catalyst does not move the equilibrium to a new state.

Teacher's Note:
1. A catalyst only shortens the time to reach equilibrium.
2. The composition at equilibrium and K stay the same.

Answer: (b) Both Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.

Explanation:
1. \( K = \dfrac{[B][C]}{[A]} = \dfrac{(3 \times 10^{-4})(3 \times 10^{-4})}{3 \times 10^{-4}} = 3 \times 10^{-4} \), so the Assertion is true.
2. Heat appears on the product side, so the reaction is exothermic and the Reason is true.
3. The heat change has no role in the calculation of K from concentrations, so the Reason does not explain the Assertion.

Teacher's Note:
1. Substitute the concentrations directly into the expression of K.
2. The sign of the heat change only tells how K varies with temperature.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. At constant volume the concentrations, and so the partial pressures, of A, B and AB do not change when an inert gas is added. The equilibrium does not shift and the Assertion is true.
2. The Reason is false because the partial pressures stay the same.

Teacher's Note:
1. At constant volume an inert gas does not change the concentrations of the reacting gases.
2. It matters only at constant pressure, where the volume expands.