Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

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11. Disturbance in equilibrium: The Le-Chatelier’s Principle

This principle, which is based on the fundamentals of a stable equilibrium, states that “When a chemical reaction at equilibrium is subjected to any stress, then the equilibrium shifts in that direction in which the effect of the stress is reduced”.

Confused with “stress” Well by stress here what we mean is any change of reaction conditions e.g. in temperature, pressure, concentration etc.

This statement will be explained by the following example.

Let us consider the reaction:   Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Let the moles o f N2, H2 and NH3 at equilibrium be a, b and c moles respectively. Since the reaction is at equilibrium,

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Where, 

X terms denote respective mole fractions and PT is the total pressure of the system.

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Here,

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Now, let us examine the effect of change in certain parameters such as number of moles. Pressure, temperature etc.

If we increase a or b, the left hand side expressio n becomes QP (as it is disturbed from equilibrium) and we can see that Q> KP

 The reaction therefore moves backward to make QP = KP

 If we increase c, QP < KP and the reaction has to move forward to revert back to equilibrium.

If we increase the volume of the container (which amounts to decreasing the pressure), QP < KP and the reaction moves forward to attain equilibrium.

If we increase the pressure of the reaction, then equilibrium shifts towards backward direction since in reactant side we have got 2 moles and on product side we have got 4 moles. So pressure is reduced in backward direction.

However from the expression if we increase the temperature of the reaction, the left hand side increase (QP) and therefore does it mean that the reaction goes backward (since QP > KP

Does this also mean that if the number of moles of reactant and product gases are equal no change in the reaction is observed on the changing temperature (as T would not exist on the left hand side)?. The answer to these quest ions is No. This is because KP also change with temperature. Therefore we need to know the effect of temperature on both QP and KP to decide the course of the reaction.


1. Effect of Concentration on Equilibrium:

If the concentration of a component is increased, reaction shifts in a direction which tends to decrease its concentration

For exmaple: Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

At equilibrium: Moles        1        1        1           1          KC = 1

If added 1 mole of D at equilibrium:

                                               1       1          1         1        QC = 2

QC> KC react ion implies reaction proceeds in backward direct ion.

At new equilibrium: 1 +x          1+ x           1-x           2-x

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Note: Here KC will remain constant even after addition of D Because KC depends on temperature only. 

2. Effect of change of pressure / volume on equilibrium

Three cases arises

· Δng = 0 

· Δng > 0 

· Δng < 0

Case I:Δng = 0 

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

At equilibrium 1 atm 1 atm 2 atm       1 

When we double the pressure or half the volume at equilibrium at t

                           2 atm 2 atm 4 atm      1

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Here, K= QP

 Hence reaction will not move in any direction. It resembles the reaction is still at equilibrium. 

Case II: Δng> 0

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

At equilibrium  1 atm          1 atm          1 atm             3       1 

When we double the pressure or half the volume 

At t                     2 atm           2 atm          2 atm            6      2

At this time, Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

QP = 2

QP>KP; hence reaction will move in backward direction.

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

At new equation 2+P          2-P          2-P

Case III:Δng < 0

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

At t = teq           1 atm        1 atm          1 atm       3           1

When we double the pressure or half the volume at equilibrium

At t                     2 atm        2 atm       2 atm    6         2

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Since, QP< KP React ion will move in forward direction

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

New equilibrium      2–P        2–P         2+P

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Table to calculate

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

3. Effect of Temperature on Equilibrium:

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Now we will derive the dependence of KP on temperature.

Starting with Arrhenius equation of rate constant

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Where, k= rate constant for forward reaction, A= Arrhenius constant of forward reaction, 

Eaf = Energy of activation of forward reaction

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Dividing (i) by (ii) we get,

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

We know that Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev (equilibrium constant)

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

At temperature T1

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

At temperature T2

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Dividing (iv) by (iii) we get 

The enthalpy of a reaction is defined in terms of activation energies as Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

For and exothermic reaction, ΔH would be negative. If we increase the temperature of the system (T2 > T1), the right hand side of the equation (V) becomes negative.

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev 

that is, the equilibrium constant at the higher temperature would be less than that at the lower temperature.

Now let us analyse our question. Will the reaction go forward or backward?

Before answering this, we must first encounter problem. If temperature is increased. The new KP would either increase or decrease or remain unchanged. If Kp increase and QP decreases. 

Then Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev therefore the reaction moves forward. If Kincrease and QP remains same then Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev Again, the reaction moves forward. What, if KP increase and QP also increase?

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev This can be answered by simply looking at the dependence of QP and KP on temperature. You can see from the equation that KP depends on temperature exponentially. While Q’s dependence on T would be either to the power g, l, t………..Therefore the variat ion in KP would still be greater QP and the reaction moves forward again. Therefore, to see it decreases, reaction moves backward and if it remains fixed, then no change at all.

Therefore, two cases arises:

(1) ΔH is +ve (Endothermic Reactions)

(2) ΔH is -ve (Exothermic Reactions)


Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Endothermic Reaction:

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

As we increase temperature, it is clearly understood by equation (1). That Keq also increase. Hence, with increase in temperature Keq increases in endothermic reactions.

Reactions will shift in forward direction.

Exothermic Reaction (ΔH = -ve)

As temperature is increased, the magnitude o f Keq decreases. Hence, increase in temperature leads to decrease in Keq. React ion will shift in backward direction.

4. Effect of addition of inert gases to a reaction at equilibrium

1. Addition at constant pressure

Let us take a general reaction

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

We know,  

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Where,

nC , nD , n A , n B denotes the no. of moles of respective components and PT is the total pressure and ∑n = total no. of moles of reactants and products.

Now, rearranging, Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Where Δn = (c + d) - (a + b)

Now, Δn can be = 0, < 0 or > 0

Let us take each case separately.

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

Addition of inert gas increases the Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev is decreased and so is Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

So products have to increase and reactants have to decrease to maintain constancy of KP. So the equilibrium moves forward.

Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev

In this case Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev decreases but Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev increases. So products have to decrease and reactants have to increase to maintain constancy of KP. So the equilibrium moves backward.

2. Addition at Constant Volume:

Since at constant volume, the pressure increases with addition of inert gas and at the same time ∑n also increases, they almost counter balance each other. So Chemical Equilibrium Introduction and Irreversible and Reversible Reactions (Part-2) Chemistry Notes | EduRev can be safely approximated as constant. Thus addition of inert gas has no effect at constant volume.

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