Chemical Equilibrium: Le Chatelier's Principle - Free MCQ Practice Test

• The equilibrium that is not affected by a change in volume at constant temperature is the one where the number of moles of gas on the reactant and product side are the same.
• For the reaction: H₂(g) + I₂(g) ⇌ 2HI(g), the number of moles on both sides is equal (2 moles of gases on each side).
• This means that changes in volume do not affect the position of equilibrium for this reaction.

The effect of pressure on equilibrium yield depends on the change in number of gas moles (Δn_g) between reactants and products.

The general relationship is: yield ∝ (1 / pressure)^Δn_g

• For (a) N₂O₄ ⇌ 2NO₂: Δn_g = 2 - 1 = +1 Increasing pressure 4 times decreases product yield by (1/4)⁺¹ = 1/4 times (yield decreases).
• For (b) C + H₂O ⇌ CO + H₂: Δn_g = (1 + 1) - 1 = 1 (Note: C is solid, so only gases count) Increasing pressure 4 times decreases product yield by (1/4)⁺¹ = 1/4 times (yield decreases).
• For (c) N₂ + 3H₂ ⇌ 2NH₃: Δn_g = 2 - (1 + 3) = 2 - 4 = -2 Increasing pressure 4 times increases product yield by (4)² = 16 times (yield greatly increases).
• For (d) 2Fe + 3H₂O ⇌ Fe₂O₃ + 3H₂: Δn_g = 3 - 3 = 0 No change in number of gas moles, so pressure change has no effect on yield.

Thus, the equilibrium where Δn_g is negative and largest in magnitude (case c) shows the maximum increase in product yield with increased pressure.

During exercise muscle need energy hence consume more oxygen and produce more CO₂ and lactic acid gets accumulated as well. According to bohr's effect oxygen binding affinity is inversely proportional to CO₂ concentration and acidity. Thus favours the dissociation​ or release of O₂.

Pressure increase solubility also increase in accordance with the Henry's law and option B is the closest to this. Hence B is correct.

The correct answer is Option A.
When ammonia is added after equilibrium is established, the partial pressure of ammonia will increase.When the temperature of an endothermic reaction is increased, the equilibrium will shift in the forward direction so that the heat is absorbed which will nullify the effect of increased temperature. Hence, the partial pressure of ammonia will increase.When the volume of the flask is increased, the pressure will decrease.

The correct answer is option C
Let initially
AB = 1M
AB = A⁺  + B^-
⇒1 - 0.1

K should remain same
AB ⇌ A⁺ + B^-
1-x

According to the given information, the solution turns pink when cooled in freezing mixture. So, the equilibrium has shifted in backward direction. So we can say that the reaction is endothermic and ∆H for the reaction is greater than zero. This is because, endothermic reactions have ∆H > 0 and they proceed in reverse direction. 

We know that K_p depends only on Temperature. As the temp. remains constant , K_p also remains constant.
α depends on the concentration of the reactant
So as the volume of the reaction container is halved, α also changes
So option d is correct.

Heat of reaction is dependent on temperature (Kirchhoff's equation) in heterogeneous system; equilibrium constant is independent on the molar concentration of solid species. Heat of reaction is not affected by catalyst. It lowers activation energy.

The correct answers are option A & D
As the reaction is exothermic decreasing the temperature will move it in forward direction in order to maintain equilibrium . increasing the amount of CO would increase the reactant concentration so in order to balance it out the reaction moves forward . hence the amount of CO₂ increases. so options A and D , both are correct.

c) partial pressure of NH₃ as well as that of N₂ increases when equilibrium is reached - Correct

• More N₂ is added → forward reaction is favored → more NH₃ formed.

• So, NH₃ partial pressure increases.

• Since some N₂ remains unreacted, its partial pressure is also higher than before.

d) the equilibrium constant Kp remains constant - Correct

• True. As long as temperature is unchanged, Kp does not change, regardless of changes in pressure, concentration, etc.

The correct answers are Options B and D. 
 
As we know that reaction is exothermic it means heat is released in the reaction so, if we apply pressure then reaction will proceed in backward direction but if there is gas phase equilibrium the reaction will shift in that direction in which less number of moles are present. If pressure increases then the ice will melt and ice gets more energy at low temp. To melt ,so it’s melting point decreases.
 

When ammonia is added, solubility of AgCl increases due to formation of complex salt which decreases the concentration of radicals in the product side and thus drives the reaction in forward direction.
When we add KCl common ion effect is applied in presence of common ion solubility decreases and reaction goes in backward direction.

For option a; with decrease in volume or with increase in pressure, reaction shifts towards less no. of moles. So, here combustion of CO will increase.
   For option b; adding argon at constant volume doesn’t make any effect on equilibrium. Also, if we add argon at constant pressure, the reaction will shift towards more no. of moles and so the combustion of CO will decrease.
   For option c; adding O₂ will shift the reaction in forward direction (according to Le Chatelier principle), so combustion of CO will increase.
   For option d; with decrease in pressure or increases in volume, reaction shifts towards more no. of moles and so combustion of CO will decrease.
 

The correct answers are Options A and D. 
As in this equilibrium, liquid converts into gas so increase in the pressure will favour reverse reaction; so formation of more H2O(l). Also, as we increase pressure boiling point of water increases.

The correct answers are option A,B,C
Because of silver salt ie . silver nitrate with powerful germicidal activity
Ag⁺ + SCN⁻ → AgSCN↓
Hg²⁺ + SCN− → Hg(SCN)²↓
4Fe³⁺ + 3(COO)²⁻₂ → 2Fe2(COO)₃↓

Volume decreased

i.e p increased

no.of moles reactant more