JEE Chemistry Equilibrium MCQs Set 12

Multiple Choice Questions (MCQs) for JEE Chemistry: Equilibrium

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Practice Equilibrium MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: For \( \ce{2X(g) + Y(g) <=> 2Z(g)} \) + 80 kcal, which combination of pressure and temperature gives the highest yield of Z at equilibrium?
(a) 1000 atm and 500 C
(b) 500 atm and 500 C
(c) 1000 atm and 100 C
(d) 500 atm and 100 C
Show Answer & Explanation

Answer: (c) 1000 atm and 100 C

Explanation:
1. The forward reaction is exothermic, so a low temperature (100 C) favours Z.
2. Three moles of gas become two, so a high pressure (1000 atm) favours Z.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 2: In the reversible reaction \( \ce{2SO2 + O2 <=> 2SO3} \) + Q cal, the most suitable conditions for a higher production of \( \ce{SO3} \) are
(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. The reaction is exothermic, so low temperature favours the products.
2. Moles of gas decrease (3 to 2), so high pressure favours the products.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 3: In the reaction \( \ce{A(g) + 2B(g) <=> C(g)} \) + Q kJ, the forward reaction is favoured by
(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 reaction: low temperature favours the forward direction.
2. Three moles of gas become one: high pressure favours the forward direction.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 4: The gaseous reaction \( \ce{C2H4 + H2 <=> C2H6} \) is exothermic (\( \Delta H=-32.7 \) kcal). The equilibrium concentration of \( \ce{C2H6} \) can be increased by
(a) increasing the temperature
(b) reducing the temperature
(c) removing some \( \ce{H2} \)
(d) adding some \( \ce{C2H6} \)
Show Answer & Explanation

Answer: (b) reducing the temperature

Explanation:
1. Lowering the temperature shifts an exothermic equilibrium to the product side.
2. Removing \( \ce{H2} \) or adding \( \ce{C2H6} \) would shift it to the left.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 5: For the equilibrium \( \ce{2NO2(g) <=> N2O4(g)} \) + 14.6 kcal, an increase in temperature would
(a) favour the formation of \( \ce{N2O4} \)
(b) favour the decomposition of \( \ce{N2O4} \)
(c) not alter the equilibrium
(d) stop the reaction
Show Answer & Explanation

Answer: (b) favour the decomposition of \( \ce{N2O4} \)

Explanation:
1. The forward reaction releases heat.
2. Heating shifts the equilibrium in the endothermic direction, i.e. towards decomposition of \( \ce{N2O4} \).

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 6: Which of the following factors will favour the reverse reaction in a chemical equilibrium?
(a) Increase in the concentration of one of the reactants
(b) Removal of at least one of the products at regular intervals
(c) Increase in the concentration of one or more products
(d) None of these
Show Answer & Explanation

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

Explanation:
1. Adding products raises the reaction quotient above K, so the system shifts backwards.
2. Adding reactants or removing products shifts it forwards.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 7: The reaction \( \ce{A + B <=> C + D} \) + heat has reached equilibrium. The reaction may be made to proceed forward by
(a) adding more C
(b) adding more D
(c) decreasing the temperature
(d) increasing the temperature
Show Answer & Explanation

Answer: (c) decreasing the temperature

Explanation:
1. Adding C or D favours the reverse reaction.
2. Since heat is a product, lowering the temperature favours the forward reaction.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 8: For the equilibrium \( \ce{HCN(aq) <=> H+(aq) + CN-(aq)} \), the addition of \( \ce{CN-} \) ions will
(a) reduce the \( \ce{HCN} \) concentration
(b) decrease the \( \ce{H+} \) ion concentration
(c) increase the equilibrium constant
(d) decrease the equilibrium constant
Show Answer & Explanation

Answer: (b) decrease the \( \ce{H+} \) ion concentration

Explanation:
1. Extra \( \ce{CN-} \) shifts the equilibrium to the left, so [HCN] increases and \( \ce{H+} \) decreases.
2. K does not change at constant temperature.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 9: The yield of product will increase with an increase in pressure in the reaction
(a) \( \ce{H2(g) + I2(g) <=> 2HI(g)} \)
(b) \( \ce{H2O(g) + CO(g) <=> CO2(g) + H2(g)} \)
(c) \( \ce{H2O(g) + C(s) <=> CO(g) + H2(g)} \)
(d) \( \ce{CO(g) + 3H2(g) <=> CH4(g) + H2O(g)} \)
Show Answer & Explanation

Answer: (d) \( \ce{CO(g) + 3H2(g) <=> CH4(g) + H2O(g)} \)

Explanation:
1. Higher pressure favours the side with fewer moles of gas.
2. Only the methanation reaction has fewer moles of gas on the right (4 to 2).

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 10: In the reaction \( \ce{N2(g) + 3H2(g) <=> 2NH3(g)} \) (\( \Delta H=-93.6 \) kJ), the yield of ammonia does not increase when
(a) pressure is increased
(b) temperature is lowered
(c) pressure is lowered
(d) the volume of the reaction vessel is decreased
Show Answer & Explanation

Answer: (c) pressure is lowered

Explanation:
1. Increasing pressure (or decreasing volume) and lowering the temperature all favour ammonia.
2. Lowering the pressure shifts the equilibrium to the side with more moles, so the yield falls.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 11: In the Haber process \( \ce{N2 + 3H2 <=> 2NH3} \) + Q kcal, the conditions that give the maximum yield are
(a) high temperature, high pressure and high concentrations of the reactants
(b) high temperature, low pressure and low concentrations of the reactants
(c) low temperature and high pressure
(d) low temperature, low pressure and low concentration of \( \ce{H2} \)
Show Answer & Explanation

Answer: (c) low temperature and high pressure

Explanation:
1. The reaction is exothermic: low temperature gives a higher equilibrium yield.
2. Moles of gas decrease: high pressure favours ammonia.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 12: In a closed vessel the equilibrium \( \ce{PCl5(s) <=> PCl3(s) + Cl2(g)} \) is established. The effect on the equilibrium concentration of \( \ce{Cl2} \) of adding more \( \ce{PCl5} \) at constant temperature is
(a) it decreases
(b) it increases
(c) it is unaffected
(d) it cannot be described without the value of \( K_p \)
Show Answer & Explanation

Answer: (c) it is unaffected

Explanation:
1. Concentration of a pure solid is constant and does not appear in K.
2. Here \( K_p=p_{Cl_2} \) is fixed, so adding solid has no effect.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 13: In which of the following equilibria would the equilibrium shift to the right if the total pressure is increased?
(a) \( \ce{N2 + 3H2 <=> 2NH3} \)
(b) \( \ce{H2 + I2 <=> 2HI} \)
(c) \( \ce{H2 + Cl2 <=> 2HCl} \)
(d) \( \ce{N2O4 <=> 2NO2} \)
Show Answer & Explanation

Answer: (a) \( \ce{N2 + 3H2 <=> 2NH3} \)

Explanation:
1. Increase in pressure favours the side with fewer gas molecules.
2. Only the ammonia synthesis has fewer moles on the right (4 to 2).

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 14: In which of the following gaseous equilibria will an increase in pressure increase the yield of the products?
(a) \( \ce{2HI <=> H2 + I2} \)
(b) \( \ce{2SO2 + O2 <=> 2SO3} \)
(c) \( \ce{H2 + Br2 <=> 2HBr} \)
(d) \( \ce{H2O + CO <=> H2 + CO2} \)
Show Answer & Explanation

Answer: (b) \( \ce{2SO2 + O2 <=> 2SO3} \)

Explanation:
1. Only \( \ce{2SO2 + O2 <=> 2SO3} \) has fewer moles of gas on the right (3 to 2).
2. The others have the same number of moles on both sides.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 15: In the reaction \( \ce{A(g) + B(g) <=> C(g)} \), the backward reaction is favoured by
(a) a decrease of pressure
(b) an increase of pressure
(c) either of the two
(d) none of the two
Show Answer & Explanation

Answer: (a) a decrease of pressure

Explanation:
1. Two moles of gas on the left and one on the right.
2. A decrease in pressure favours the side with more moles, i.e. the backward reaction.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 16: If the pressure is increased on the equilibrium \( \ce{C(s) + H2O(g) <=> CO(g) + H2(g)} \), the reaction proceeds in the
(a) forward direction
(b) backward direction
(c) neither direction (no effect)
(d) none of these
Show Answer & Explanation

Answer: (b) backward direction

Explanation:
1. Gas moles: 1 on the left, 2 on the right.
2. An increase in pressure favours the side with fewer moles, so the backward direction.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 17: The formation of \( \ce{ClF3} \) is exothermic: \( \ce{Cl2(g) + 3F2(g) <=> 2ClF3(g)} \), \( \Delta H=-329 \) kJ. Which of the following will increase the quantity of \( \ce{ClF3} \) in the equilibrium mixture?
(a) Increasing the temperature
(b) Removing \( \ce{Cl2} \)
(c) Increasing the volume of the container
(d) Adding \( \ce{F2} \)
Show Answer & Explanation

Answer: (d) Adding \( \ce{F2} \)

Explanation:
1. Adding the reactant \( \ce{F2} \) shifts the equilibrium to the right.
2. Raising the temperature, removing \( \ce{Cl2} \) or increasing the volume (4 to 2 moles) all shift it to the left.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 18: Which of the following reactions proceeds at low pressure?
(a) \( \ce{N2 + 3H2 <=> 2NH3} \)
(b) \( \ce{H2 + I2 <=> 2HI} \)
(c) \( \ce{PCl5 <=> PCl3 + Cl2} \)
(d) \( \ce{N2 + O2 <=> 2NO} \)
Show Answer & Explanation

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

Explanation:
1. A reaction that increases the number of gas moles is favoured at low pressure.
2. Only \( \ce{PCl5} \) decomposition does (1 to 2 moles).

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 19: Le Chatelier's principle is applicable to
(a) both physical and chemical equilibria
(b) only chemical equilibrium
(c) only physical equilibrium
(d) neither physical nor chemical equilibrium
Show Answer & Explanation

Answer: (a) both physical and chemical equilibria

Explanation:
1. The principle applies to any system in equilibrium, e.g. ice-water, solubility and chemical reactions.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 20: When pressure is applied to the system ice \( \rightleftharpoons \) water, what will happen?
(a) more water will form
(b) more ice will form
(c) there will be no effect
(d) water will decompose into \( \ce{H2} \) and \( \ce{O2} \)
Show Answer & Explanation

Answer: (a) more water will form

Explanation:
1. Water has a smaller volume than ice.
2. Pressure favours the smaller-volume phase, so more water forms (ice melts).

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 21: Select the correct statement about catalysts.
(a) The equilibrium constant changes with the addition of a catalyst
(b) A catalyst increases only the rate of the forward reaction
(c) The ratio of mixture at equilibrium is not changed by a catalyst
(d) Catalysts are active only in solution
Show Answer & Explanation

Answer: (c) The ratio of mixture at equilibrium is not changed by a catalyst

Explanation:
1. A catalyst speeds up the forward and backward reactions equally.
2. It lowers the time to reach equilibrium but does not change K or the equilibrium composition.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 22: In the reaction \( \ce{A2(g) + 4B2(g) <=> 2AB4(g)} \), \( \Delta H<0 \), the formation of \( \ce{AB4} \) is favoured at
(a) low temperature and high pressure
(b) high temperature and low pressure
(c) low temperature and low pressure
(d) high temperature and high pressure
Show Answer & Explanation

Answer: (a) low temperature and high pressure

Explanation:
1. Exothermic: low temperature favours products.
2. Moles of gas decrease (5 to 2): high pressure favours products.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 23: For a reaction at equilibrium, the standard free energy change \( \Delta G^\circ=0 \). The equilibrium constant K is
(a) 0
(b) 1
(c) 2
(d) 10
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. \( \Delta G^\circ=-RT\ln K \); if it is zero then ln K = 0 and K = 1.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

Question 24: A reaction attains equilibrium when the free energy change accompanying it is
(a) positive and large
(b) zero
(c) negative and large
(d) negative and small
Show Answer & Explanation

Answer: (b) zero

Explanation:
1. At equilibrium the system has minimum free energy, so \( \Delta G=0 \) at constant temperature and pressure.

Teacher's Note:
1. Decide whether the change affects the concentration, pressure (moles of gas) or temperature, then apply Le Chatelier's principle.
2. Remember that only a change in temperature changes the value of K; a catalyst does not shift the equilibrium.

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