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Le Chatelier's Principles on Equilibrium - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

Syllabus

Quick Facts

Le Chatelier’s principle, Le Chatelier’s principle is considered one of the most asked concept.
31 Questions around this concept.

Solve by difficulty

MEDIUM

Which one of the following conditions will favor maximum formation of the product in the reaction,

$A_{2}(g) + B_{2}(g) \rightleftharpoons X_{2}(g) \Delta _{r}H = -X kJ$ ?

High temperature and high pressure

Low temperature and low pressure

Low temperature and high pressure

High temperature and low pressure

View Solution

The effect of addition of helium gas to the following reaction in equilibrium state, is :
$\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$

helium will deactivate $\mathrm{PCl}_{5}$ and reaction will stop.

the equilibrium will shift in the forward direction and more of $\mathrm{Cl}_{2}$ and $\mathrm{PCl}_{3}$ gases will be produced.

the equilibrium will go backward due to suppression of dissociation of $\mathrm{PCl}_{5}$ .

addition of helium will not affect the equilibrium.

View Solution

Concepts Covered - 3

Reaction coefficient/quotient

It is defined as the ratio of concentration of products to the concentration of the reacting species raised to their stoichiometric coefficient at any point of time other than the equilibrium stage. It has the exact same expression as that of the Equilibrium constant except that the concentration values are at any instant. Mathematically, it can be determined as follows:

If we consider a reaction
$\mathrm{m A +n B \rightleftharpoons pC +q D}$

$\mathrm{Q\: =\: \frac{[C]^{p}[D]^{q}}{[A]^{m}[B]^{n}}}$
Q can be denoted as Q_c or Q_p if we use concentration in terms of mole per litre or partial pressure respectively.

The value of Q is useful to determine the direction in which the equlibrium will shift at any instant for a particular set of activity of the species involved.

When Q = K, the reaction is at equilibrium and rate of forward and backward reactions are equal.
When Q > K, the reaction will proceed or favour backward direction. This means products convert into reactant to attain equilibrium.
When Q < K, the reaction will proceed or favour forward direction. This means that the reactants convert into product to attain equilibrium.

Relation between K, q and $\mathrm{\Delta G}$

$\mathrm{\Delta G = \Delta G^0 + RTlnQ}\ \ \ \ \rightarrow (1)$

where

$\mathrm{\Delta G} = \text{Change in Gibbs Free energy}$

$\mathrm{\Delta G^0} = \text{Change in Gibbs Free energy under Standard Conditions}$

$\text{Q = Reaction quotient}$

Now, we know that

$\mathrm{\Delta G^0= -RTlnK_{eq}}$

Putting this value in equation (1)

$\mathrm{\Delta G = -RT lnK_{eq} + RTlnQ}$

which can be simplified to

$\mathrm{\Delta G = RT ln\left ( \frac{Q}{K_{eq}}\right )}\ \ \ \ \rightarrow (2)$

From Equation (2) it is clear that

$\mathrm{When\ Q = K_{eq},\ \Delta G = 0} \text{ and the reaction is at equilibrium}$
$\mathrm{When\ Q < K_{eq},\ \Delta G < 0} \text{ the reaction will move in the forward direction}$
$\mathrm{When\ Q > K_{eq},\ \Delta G > 0} \text{ the reaction will move in the backward direction}$

Le Chatelier’s principle

It describes the effect of change in concentration, pressure and temperature on the reversible system.
According to it, "If the system at equilibrium is subjected to a change of concentration or temperature or pressure, the system adjusts itself in such a way as to nul the effect of that change i.e., the effect of these changes can be neglected or minimized."

Effect of Concentration

An increase in the concentration of any substance favours the reaction in which it is used up i.e, in the opposite direction.
An increase of concentration of reactant favours the formation of more product i.e., forward reaction. Increase in concentration of product favours.
Increase in concentration of product favours backward reaction. I.e., its continuous removal is essential for more formation of it.

Effect of Pressure

High pressure is favourable for the reaction in which there is a decrease in volume or n_r > n_p.
Low pressure is favourable for the reaction in which there is an increase in volume or n_r < n_p.
Pressure is kept constant when the volume is constant or n_r = n_p.

Here n_r = moles of gaseous reactant

n_p = moles of gaseous product

Le Chatelier’s principle

Effect of change in temperature

On increasing the temperature, equilibrium shifts to that direction which proceeds with the absorption of heat.

$\mathrm{A\: +\: B\:+\: \Delta \: \rightleftharpoons \: C\: +\: D}$
This is an endothermic reaction. Thus, on increasing the temperature, equilibrium shifts in the forward direction.
$\mathrm{A\: +\: B\:: \rightleftharpoons \: C\: +\: D\: +\: \Delta }$
This is an exothermic reaction. Thus on increasing the temperature, equilibrium shifts in backward direction.

For example:

$\mathrm{H_{2}\: +\: I_{2}\: \rightleftharpoons \: 2HI+\: 3000\, calories }$
In this reaction, the product formed is HI and the release of 3000 calories of energy. Thus if temperature is increased then equilibrium will shift backward and form the reactants.

Effect of Adding Inert Gas on Equilibrium

When n_p is equal to n_r there is no effect of adding an inert gas either at constant volume or pressure.
When n_p ≠n_r there is no effect of adding an inert gas at constant volume.
When n_p ≠n_r at constant pressure on adding inert gas equilibrium will shift towards more volume side. e.g., dissociation of ammonia will be more at constant pressure by adding inert gas like argon (Ar).