Le Chatelier's Principles on Equilibrium

This principle was discovered by the Le Chatelier a French chemist Henri Louis Le Chatelier in 1884, this principle tells us that the change of equilibrium conditions by changing any physical parameter such as temperature, pressure, and the concentration of the reaction. In this principle, we get to know the shift of equilibrium and attaining the equilibrium at the different positions if the standard condition of the reaction is disturbed. Le Chatelier introduced this principle to predict the changes in any chemical equilibrium conditions, which is very important in determining the optimum chemical reactions, and in the lab it also needed for maintaining the equilibrium

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This Story also Contains

1. Le Chatelier Principle On Equilibrium
2. Some Solved Examples
3. SUMMARY

Le Chatelier Principle On Equilibrium

It is defined as the ratio of the 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 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$\begin{aligned} & \mathrm{mA}+\mathrm{nB} \rightleftharpoons \mathrm{pC}+\mathrm{qD} \\ & \mathrm{Q}=\frac{[\mathrm{C}]^{\mathrm{p}}[\mathrm{D}]^{\mathrm{q}}}{[\mathrm{A}]^{\mathrm{m}}[\mathrm{B}]^{\mathrm{n}}}\end{aligned}$
Q can be denoted as Q_c or Q_p if we use concentration in terms of a mole per liter or partial pressure respectively.

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

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

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

$
\Delta \mathrm{G}=\Delta \mathrm{G}^0+\mathrm{RT} \ln \mathrm{Q} \quad \rightarrow(1)
$
where
$\Delta \mathrm{G}=$ Change in Gibbs Free energy
$\Delta \mathrm{G}^0=$ Change in Gibbs Free energy under Standard Conditions
$
\mathrm{Q}=\text { Reaction quotient }
$

Now, we know that
$
\Delta \mathrm{G}^0=-\mathrm{RT} \ln \mathrm{K}_{\mathrm{eq}}
$

Putting this value in equation (1)
$
\Delta \mathrm{G}=-\mathrm{RT} \ln \mathrm{K}_{\text {eq }}+\mathrm{RT} \ln \mathrm{O}
$
which can be simplified to
$
\Delta \mathrm{G}=\mathrm{RT} \ln \left(\frac{\mathrm{Q}}{\mathrm{K}_{\mathrm{eq}}}\right) \rightarrow(2)
$

From Equation (2) it is clear that
- When $\mathrm{Q}=\mathrm{K}_{\text {eq }}, \Delta \mathrm{G}=0$ and the reaction is at equilibrium
. When $\mathrm{Q}<\mathrm{K}_{\text {eq }}, \Delta \mathrm{G}<0$ the reaction will move in the forward direction
- When $\mathrm{Q}>\mathrm{K}_{\text {eq }}, \Delta \mathrm{G}>0$ the reaction will move in the backward direction

Recommended topic video on (Le Chatelier's Principles on Equilibrium)

Some Solved Examples

1. For the reaction, $A+B \rightleftharpoons 3 C$ at 25^o C, a 3-litre vessel contains 1,2,4 mole of A, B and C respectively. Find the Reaction quotient.

1)5.66

2)64

3) (correct)10.66

4)8.66

Solution

Reaction:-

$\begin{gathered}A+B \rightleftharpoons 3 C \\ \text { Before reaction, }[A]_0=1 / 3 \\ \qquad[B]_0=2 / 3 \\ {[C]_0=4 / 3} \\ \text { Thus, } Q=\frac{[C]_0^3}{[A]_0[B]_0} \\ Q=\frac{4^3 \times 3 \times 3}{3^3 \times 1 \times 2} \\ Q=\frac{64}{6}=10.66\end{gathered}$

Hence, the answer is the option (3).

2. The standard Gibbs energy change at 300 K for the reaction $2 A \rightleftharpoons B+C$ is 2494.2 J. At a given time, the composition of the reaction mixture is [A] = 1/2, [B] = 2 and [C] = 1/2. The reaction proceeds in the:

1)forward direction because $Q>K_C$
2) $\left(\right.$ correct) ${ }^{\text {reverse }}$ direction because $Q>K_C$
3) forward direction because $Q<K_C$
4) reverse direction because $Q<K_C$

Solution

The standard Gibbs energy change at 300 K for the reaction $2 A \rightleftharpoons B+C$
So,
$$
\begin{aligned}
& Q=\frac{[B][C]}{[A]^2} \\
& Q=\frac{2 \times \frac{1}{2}}{\left(\frac{1}{2}\right)^2}=4 \\
& \Delta G^{\circ}=-2.303 R T \log K_C \\
& 2494.2=-2.303 \times 8.314 \times 300 \log K_C \\
& \therefore K_C=0.36
\end{aligned}
$$
$\because Q>K_C$ thus the reaction moves in the backward direction.

Hence, the answer is the option (2).

3.The $K_{s p}$ for the following dissociation is $1.6 \times 10^{-5}$.
$
\mathrm{PbCl}_2(s)=\mathrm{Pb}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)
$

Which of the following choices is correct for a mixture of 300m, 0.134 M ${ }_4 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_2$ and 100 mL 0.4 M NaCl?

1) $Q<K_{s p}$
2) (correct) $Q>K_{\text {sp }}$
3) $Q=K_{s p}$
4)Not enough data is provided

Solution
The $K_{s p}$ for the following dissociation is $1.6 \times 10^{-5}$.
$
\mathrm{PbCl}_2(s)=\mathrm{Pb}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)
$

So,
$
\begin{aligned}
Q & =\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^2\right]^2 \\
Q & =\frac{300 \times 0.134}{400} \times\left[\frac{100 \times 0.4}{400}\right]^2 \\
Q & =\frac{3 \times 0.134}{4} \times(0.1)^2 \\
Q & =0.105 \times 10^{-2} \\
Q & =1.005 \times 10^{-3}
\end{aligned}
$

So, Q > Ksp

Hence, the answer is the option (2).

4. Which of the following reaction proceeds nearly to completion?

${ }_{1)} \mathrm{CO}_2 \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3 \quad K_c=1.7 \times 10^{-2}$
2) $\mathrm{H}_2 \mathrm{CO}_3 \rightleftharpoons \mathrm{HCO}_3+\mathrm{H}^{+} \quad K_c=8.4 \times 10^{-7}$
3) $\mathrm{CaCO}_3 \rightleftharpoons \mathrm{Ca}^{2+}+\mathrm{CO}_3^{-} \quad K_c=8.4 \times 10^{-9}$
4) (correct) $\mathrm{H}_2+\mathrm{Br}_2 \rightleftharpoons 2 \mathrm{HBr} \quad k_c=5.4 \times 10^{18}$

Solution

As we have learned

The composition of the equilibrium mixture when Kc > 1000 -

If K_c > 10³, products predominate over reactants i.e. if K_c is very large, the reaction proceeds nearly to completion.

e.g.
$
\begin{aligned}
& \mathrm{H}_2+\mathrm{Cl}_2 \rightleftharpoons 2 \mathrm{HCl} \\
& K_c=4 \times 10^{31}
\end{aligned}
$
$
\begin{aligned}
& \mathrm{H}_2+\mathrm{Br}_2 \rightleftharpoons 2 \mathrm{HBr} \\
& K_c=5.4 \times 10^{18}
\end{aligned}
$

Reactions a,b, and c have a K_c value less than 10³, but reaction d has a value greater than 10³. We can say this reaction proceeds to completion.

Hence, the answer is the option (4).

5. In which one of the following reactants predominates over products?

${ }_{1)} \mathrm{H}_2+\mathrm{Al}_2 \rightleftharpoons 2 \mathrm{HCl} \quad k_c=4 \times 10^{31}$
2) $\mathrm{H}_2+\mathrm{Br}_2 \rightleftharpoons 2 \mathrm{HBr} \quad k_c=5.4 \times 10^{18}$
3) $\left(\right.$ correct) $N_2+O_2 \rightleftharpoons 2 N O \quad k_c=4.8 \times 10^{31}$
4)both a and b

Solution

As we have learned

Composition of equilibrium when Kc < 0.001 -

Reactants predominate over products i.e. If K_c is very small, the reaction proceeds rarely.

- wherein

$\begin{aligned} & \text { e.g. } N_2+O_2 \rightleftharpoons 2 N O \\ & K_c=4.8 \times 10^{-31}\end{aligned}$In Kc is very small then reaction proceeds rarely or say, reactant predominates over product.

Hence, the answer is the option (3).

SUMMARY

Le Chatelier's principle are that principle whose invention tells us the equilibrium shift of any reaction by change the concentration of reactant and the product. If the temperature is decreased equilibrium shifts to the forward direction in any exothermic reactions. If you increase the pressure on a gaseous system, the equilibrium will shift toward the side with fewer gas molecules. It helps us to predict the reaction shift and how changes in conditions will affect the position of equilibrium in chemical reactions. And also optimizing reactions: will assist in adjusting reaction conditions to maximize yields in industrial processes.