JEE Chemistry Equilibrium MCQs Set 15

Multiple Choice Questions (MCQs) for JEE Chemistry: Equilibrium

Access targeted multiple-choice questions for Equilibrium designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Equilibrium MCQs for JEE Chemistry

Access the complete set of multiple-choice questions for Equilibrium below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: In a closed container at constant volume the equilibrium \( \ce{SO2Cl2(g) <=> SO2(g) + Cl2(g)} \) is attained. If helium is introduced at constant volume, which statement is correct?
(a) More chlorine is formed
(b) \( \ce{SO2} \) concentration is reduced
(c) \( \ce{SO2Cl2} \) formation increases
(d) concentrations of \( \ce{SO2Cl2} \), \( \ce{SO2} \) and \( \ce{Cl2} \) do not change
Show Answer & Explanation

Answer: (d) concentrations of \( \ce{SO2Cl2} \), \( \ce{SO2} \) and \( \ce{Cl2} \) do not change

Explanation:
1. At constant volume, adding an inert gas does not change the partial pressures or concentrations of the reacting gases.
2. So the equilibrium does not shift.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 2: Which of the following equilibria will shift to the right side on increasing the temperature?
(a) \( \ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)} \)
(b) \( \ce{2SO2(g) + O2(g) <=> 2SO3(g)} \)
(c) \( \ce{H2O(g) <=> H2(g) + 1/2 O2(g)} \)
(d) \( \ce{4HCl(g) + O2(g) <=> 2H2O(g) + 2Cl2(g)} \)
Show Answer & Explanation

Answer: (c) \( \ce{H2O(g) <=> H2(g) + 1/2 O2(g)} \)

Explanation:
1. Heating favours the endothermic direction.
2. The decomposition of water vapour into its elements is endothermic, so it shifts right.
3. The other three forward reactions are exothermic.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 3: Sodium sulphate dissolves in water with evolution of heat. If the temperature of a saturated solution is raised, then according to Le Chatelier's principle
(a) more solid will dissolve
(b) some solid will precipitate out of the solution
(c) the solution will become supersaturated
(d) the solution concentration will remain unchanged
Show Answer & Explanation

Answer: (b) some solid will precipitate out of the solution

Explanation:
1. Dissolution is exothermic, so on heating the equilibrium shifts in the endothermic direction, i.e. towards undissolved solid.
2. Some solid precipitates out.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 4: Consider the equilibrium \( \ce{N2(g) + 3H2(g) <=> 2NH3(g)} \), \( \Delta H=-93.6 \) kJ. The maximum yield of ammonia is obtained by
(a) decrease of temperature and increase of pressure
(b) increase of temperature and decrease of pressure
(c) decrease of both temperature and pressure
(d) increase of both temperature and pressure
Show Answer & Explanation

Answer: (a) decrease of temperature and increase of pressure

Explanation:
1. The forward reaction is exothermic, so low temperature favours it.
2. It reduces the moles of gas (4 to 2), so high pressure favours it.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 5: Le Chatelier's principle is applicable only to a
(a) system in equilibrium
(b) irreversible reaction
(c) homogeneous reaction
(d) heterogeneous reaction
Show Answer & Explanation

Answer: (a) system in equilibrium

Explanation:
1. The principle describes how a system already at equilibrium responds to a stress.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 6: In a vessel containing \( \ce{SO3} \), \( \ce{SO2} \) and \( \ce{O2} \) at equilibrium, some helium gas is introduced so that the total pressure increases while temperature and volume remain constant. According to Le Chatelier's principle, the dissociation of \( \ce{SO3} \)
(a) increases
(b) decreases
(c) remains unaltered
(d) changes unpredictably
Show Answer & Explanation

Answer: (c) remains unaltered

Explanation:
1. At constant volume the partial pressures of the three gases are unchanged by the added helium, so the equilibrium position does not change.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 7: In which of the following systems is the rate of the backward reaction favoured by increase of pressure?
(a) \( \ce{PCl5 <=> PCl3 + Cl2} \)
(b) \( \ce{2SO2 + O2 <=> 2SO3} \)
(c) \( \ce{N2 + 3H2 <=> 2NH3} \)
(d) \( \ce{N2 + O2 <=> 2NO} \)
Show Answer & Explanation

Answer: (a) \( \ce{PCl5 <=> PCl3 + Cl2} \)

Explanation:
1. Increase of pressure shifts equilibrium towards fewer gas moles.
2. For \( \ce{PCl5 <=> PCl3 + Cl2} \) the forward reaction increases the moles (1 to 2), so the backward reaction is favoured.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 8: Which of the following equilibria is not shifted by an increase in the pressure?
(a) \( \ce{H2(g) + I2(g) <=> 2HI(g)} \)
(b) \( \ce{N2(g) + 3H2(g) <=> 2NH3(g)} \)
(c) \( \ce{2CO(g) + O2(g) <=> 2CO2(g)} \)
(d) \( \ce{2C(s) + O2(g) <=> 2CO(g)} \)
Show Answer & Explanation

Answer: (a) \( \ce{H2(g) + I2(g) <=> 2HI(g)} \)

Explanation:
1. Pressure has no effect when the moles of gas are equal on both sides.
2. In \( \ce{H2 + I2 <=> 2HI} \), \( \Delta n_g=0 \).

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 9: According to Le Chatelier's principle, adding heat to a solid and liquid in equilibrium (an endothermic process) will cause the
(a) temperature to rise
(b) temperature to fall
(c) amount of solid to decrease
(d) amount of liquid to decrease
Show Answer & Explanation

Answer: (c) amount of solid to decrease

Explanation:
1. Melting is endothermic, so added heat favours the forward process.
2. The solid melts and its amount decreases.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 10: On addition of an inert gas at constant volume to the equilibrium \( \ce{N2 + 3H2 <=> 2NH3} \), the
(a) reaction remains unaffected
(b) forward reaction is favoured
(c) reaction halts
(d) backward reaction is favoured
Show Answer & Explanation

Answer: (a) reaction remains unaffected

Explanation:
1. At constant volume the concentrations are unchanged by an inert gas, so the equilibrium is undisturbed.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 11: For the reaction \( \ce{A + B + Q <=> C + D} \), if the temperature is increased, the concentration of the products will
(a) increase
(b) decrease
(c) remain the same
(d) become zero
Show Answer & Explanation

Answer: (a) increase

Explanation:
1. The forward reaction absorbs heat (endothermic).
2. Raising the temperature favours it, so the products increase.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 12: For \( \ce{H2(g) + I2(g) <=> 2HI(g)} \), when the pressure is increased the reaction direction
(a) does not change
(b) forward
(c) backward
(d) decreases
Show Answer & Explanation

Answer: (a) does not change

Explanation:
1. \( \Delta n_g=0 \), so pressure does not shift the equilibrium.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 13: In the equilibrium \( \ce{N2 + 3H2 <=> 2NH3} \) + 22 kcal, the formation of ammonia is favoured by
(a) increasing the pressure
(b) increasing the temperature
(c) decreasing the pressure
(d) adding ammonia
Show Answer & Explanation

Answer: (a) increasing the pressure

Explanation:
1. The forward reaction reduces gas moles (4 to 2), so high pressure favours ammonia.
2. High temperature and added product would oppose it.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 14: The reaction \( \ce{2SO2 + O2 <=> 2SO3} \), \( \Delta H \) negative, is favoured by
(a) low temperature, low pressure
(b) low temperature, high pressure
(c) high temperature, high pressure
(d) high temperature, low pressure
Show Answer & Explanation

Answer: (b) low temperature, high pressure

Explanation:
1. Exothermic, so low temperature favours it.
2. Moles fall from 3 to 2, so high pressure favours it.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 15: Which of the following will favour the reverse reaction in a chemical equilibrium?
(a) Increasing the concentration of the reactants
(b) Removal of at least one of the products at regular intervals
(c) Increasing the concentration of one or more of the products
(d) Increasing the pressure
Show Answer & Explanation

Answer: (c) Increasing the concentration of one or more of the products

Explanation:
1. Adding more product raises Q above K, so the reaction proceeds in the backward direction.
2. Increasing reactant or removing product favours the forward reaction.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 16: Under what conditions of temperature and pressure will the formation of atomic hydrogen from molecular hydrogen be favoured most?
(a) High temperature and high pressure
(b) Low temperature and low pressure
(c) High temperature and low pressure
(d) Low temperature and high pressure
Show Answer & Explanation

Answer: (c) High temperature and low pressure

Explanation:
1. \( \ce{H2 <=> 2H} \) absorbs heat (bond breaking), so high temperature favours it.
2. The number of moles increases, so low pressure favours it.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 17: For \( \ce{BaO2(s) <=> BaO(s) + O2(g)} \) with \( \Delta H>0 \), at equilibrium the pressure of oxygen depends upon
(a) the mass of BaO2
(b) the mass of BaO
(c) the temperature
(d) the masses of BaO2 and BaO both
Show Answer & Explanation

Answer: (c) the temperature

Explanation:
1. Only the gas appears in the expression \( K_p=p_{O_2} \) (solids are omitted).
2. So the pressure is fixed by the temperature alone.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 18: The yield of product in the reaction \( \ce{A2(g) + 2B(g) <=> C(g)} \) + Q kJ would be high at
(a) high temperature and high pressure
(b) high temperature and low pressure
(c) low temperature and high pressure
(d) low temperature and low pressure
Show Answer & Explanation

Answer: (c) low temperature and high pressure

Explanation:
1. Exothermic (Q released): low temperature favours the forward reaction.
2. 3 moles of gas become 1, so high pressure favours it.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 19: Which reaction is not affected by change in pressure?
(a) \( \ce{H2 + I2 <=> 2HI} \)
(b) \( \ce{2C + O2 <=> 2CO} \)
(c) \( \ce{N2 + 3H2 <=> 2NH3} \)
(d) \( \ce{PCl5 <=> PCl3 + Cl2} \)
Show Answer & Explanation

Answer: (a) \( \ce{H2 + I2 <=> 2HI} \)

Explanation:
1. Equilibrium shifts with pressure only when the number of gas moles changes.
2. In \( \ce{H2 + I2 <=> 2HI} \) they are equal.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 20: The gaseous reaction \( \ce{A + B <=> 2C + D} \) + Q is most favoured at
(a) low temperature and high pressure
(b) high temperature and high pressure
(c) high temperature and low pressure
(d) low temperature and low pressure
Show Answer & Explanation

Answer: (d) low temperature and low pressure

Explanation:
1. Exothermic, so low temperature favours products.
2. Gas moles increase from 2 to 3, so low pressure favours products.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 21: The reaction \( \ce{2SO2(g) + O2(g) <=> 2SO3(g)} \), \( \Delta H^\circ=-198 \) kJ. On the basis of Le Chatelier's principle, the conditions favourable for the forward reaction are
(a) lowering of temperature as well as pressure
(b) increasing temperature as well as pressure
(c) lowering the temperature and increasing the pressure
(d) any value of temperature and pressure
Show Answer & Explanation

Answer: (c) lowering the temperature and increasing the pressure

Explanation:
1. Heat is released, so low temperature favours the forward reaction.
2. Moles of gas decrease, so high pressure favours it.

Teacher's Note:
1. Decide the effect of concentration, pressure (change in moles of gas) and temperature (sign of the heat term) separately.
2. Remember that a catalyst or an inert gas at constant volume does not shift the equilibrium.

Question 22: For the reaction \( \ce{2NO2(g) <=> 2NO(g) + O2(g)} \), \( K_c=1.8\times10^{-6} \) at 185 C. At 185 C, the value of \( K_c \) for the reaction \( \ce{NO(g) + 1/2 O2(g) <=> NO2(g)} \) is
(a) \( 0.9\times10^6 \)
(b) \( 7.5\times10^2 \)
(c) \( 1.95\times10^{-3} \)
(d) \( 1.95\times10^3 \)
Show Answer & Explanation

Answer: (b) \( 7.5\times10^2 \)

Explanation:
1. Reverse the reaction and halve the coefficients.
2. New \( K=1/\sqrt{1.8\times10^{-6}}=1/(1.342\times10^{-3})=745\approx7.5\times10^2 \).

Teacher's Note:
1. Reversing a reaction inverts K; multiplying coefficients by n raises K to the power n.

Question 23: One mole of AB reacts with one mole of CD: \( \ce{AB + CD <=> AD + CB} \). When equilibrium is established, 3/4 mole each of AB and CD has been converted to AD and CB. There is no change in volume. The equilibrium constant is
(a) \( 9/16 \)
(b) \( 1/9 \)
(c) \( 16/9 \)
(d) 9
Show Answer & Explanation

Answer: (d) 9

Explanation:
1. At equilibrium AB and CD are 1/4 mol each; AD and CB are 3/4 mol each.
2. K = (3/4 \( \times \) 3/4)/(1/4 \( \times \) 1/4) = 9.

Teacher's Note:
1. Volume cancels because the number of moles is equal on both sides.

Question 24: For the reaction \( \ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)} \) at a given temperature, the equilibrium amount of \( \ce{CO2(g)} \) can be increased by
(a) adding a suitable catalyst
(b) adding an inert gas
(c) decreasing the volume of the container
(d) increasing the amount of CO(g)
Show Answer & Explanation

Answer: (d) increasing the amount of CO(g)

Explanation:
1. Adding more CO increases the reactant concentration and shifts the equilibrium forward, making more \( \ce{CO2} \).
2. A catalyst, an inert gas or a change in volume (equal moles of gas on both sides) do not change the equilibrium amount.

Teacher's Note:
1. Equal gas moles on both sides means pressure has no effect.

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