Le Chatelier’s Principle

Introduction to Le Chatelier’s Principle

“When an equilibrium is subjected to either a change in concentration, temperature or, in external pressure, the equilibrium will shift in that direction where the effects caused by these changes are nullified”.

Refer to the following Video for Le Chatelier’s Principle

This can be understood by the following example. Overall we can also predict the direction of equilibrium by keeping in mind following theoretical assumption.

PCl₅  ——> PCl₃ + Cl₂

Let us assume that we have this reaction at equilibrium and the moles of Cl₂, PCl₃ and PCl₅ at equilibrium are a, b and c respectively, and the total pressure be P_T.

 

Since

P_T  =  (a+b+c) RT / V  

K_P = abRT / cV  

Now if d moles of PCl₃ is added to the system, the value of Q would be, a(b+d)RT / cV 

We can see that this is more than KP. So the system would move reverse to attain equilibrium.

If we increase the volume of the system, the Q becomes  abRT / cV' where V' > V. Q becomes less, and the system would move forward to attain equilibrium.

If we add a noble gas at constant pressure, it amounts to increasing the volume of the system and therefore the reaction moves forward.

If we add the noble gas at constant volume, the expression of Q remains as Q = abRT / cV and the system continues to be in equilibrium. Nothing happens.

Therefore for using Le-Chatlier’s principle, convert the expression of K_P and K_C into basic terms and then see the effect of various changes.

Effect of Change of Concentration on Equilibrium

Let us have a general reaction,

 aA + bB  cC + dD

at a given temperature, the equilibrium constant,

K_c = [C]^c[D]^d / [A]^a[B]^b 

again if α, β, γ and δ are the number of mole of A, B, C and D are at equilibrium

then, K_c = [γ]^c[δ]^d / [α]^a[β]^b 

If any of product will be added, to keep the Kc constant, concentration of reactants will increase i.e. the reaction will move in reverse direction. Similarly if any change or disturbance in reactant side will be done, change in product’s concentration will take place to minimise the effect.
 

Effect of Chenage of Temperature on Equilibrium

The effect of change in temperature on an equilibrium cannot be immediately seen because on changing temperature the equilibrium constant itself changes. So first we must find out as to how the equilibrium constant changes with temperature.

For the forward reaction, according to the Arrhenius equation,

logK₂/K₁ = ΔH^o / 2.303R (1/T₁ – 1/T₂) it is clear that                                      

(b) If ΔH^o is -ve (exothermic), an increase in temperature (T₂ > T1), will make K₂ < K₁ i.e., the reaction goes in the reverse direction.

Increase in temperature will shift the reaction towards left in case of exothermic reactions and right in endothermic reactions. 

Increase of pressure (decrease in volume) will shift the reaction to the side having fewer moles of the gas; while decreases of pressure (increase in volume) will shift the reaction to the side having more moles of the gas.

Example . Under what conditions will the following reactions go in the forward direction?

(i)  N₂(g)+ 3H₂(g)  2NH₃(g) + 23 k cal.

(ii) 2SO₂(g) + O₂(g)    2SO₃(g) + 45 k cal.

(iii) N₂(g) + O₂(g)  2NO(g) - 43.2 k cal.

(iv) 2NO(g) + O₂(g)   2NO₂(g) + 27.8 k cal.

(v) C(s) + H₂O(g)   CO₂(g) + H₂(g) + X k cal.

(vi) PCl₅(g)  PCl₃(g) + Cl­₂(g)- X k cal.

(vii) N₂O₄(g)  2NO₂(g) - 14 k cal.

Solution:     

(i)  Low T, High P, excess of N₂ and H₂. 

(ii) Low T, High P, excess of SO₂ and O₂.

(iii) High T, any P, excess of N₂ and O₂         

(iv) Low T, High P, excess of NO and O₂

(v) Low T, Low P, excess of C and H₂O  

(vi) High T, Low P, excess of PCl₅

(vii) High T, Low P, excess of N₂O₄.

Example: The equilibrium constant K_P for the reaction N₂(g) + 3H₂(g)  2NH₃(g)  is 1.6 × 10^-4atm^-2 at 400^oC. What will be the equilibrium constant at 500^oC  if heat of the reaction in this temperature range is-25.14 k cal?

Solution: Equilibrium constants at different temperature and heat of the reaction are related by the equation,  

log KP₂ = –25140 / 2.303 × 2 [773 – 673 / 773 × 673]  + log 1.64 10^-4

log KP₂ = –4.835

KP₂ = 1.462 × 10–5 atm^–2
 

Effect of Change of Pressure on Equilibrium

The effect of change of pressure on chemical equilibrium can be done by the formula.

Kp = QP × P^(Δn)

Case A: When Δn = 0

K_p = Q_p

And equilibria is independent of pressure.

Case B: When Δn = – ve

K_p = Q_P × P^–Δn = Q_P / P^Δn

Here with increase in external pressure, will shift the equilibrium towards forward direction to maintain K_­p. Similarly, decrease in pressure will shift the equilibrium in the reverse direction.

Case C: When Δn = +ve

 K_p = Q_P × P^–Δn and, effect will be opposite to that of Case ‘B’
 

Effect of Addition of Inert Gases to a Reaction at Equilibrium

1.   Addition at constant pressure

Let us take a general reaction

 aA + bB  cC + dD

We know,

Where,

 n_C n_D, n_A, n_B denotes the no. of moles of respective components and P_T is the total pressure and n = total no. of moles of reactants and products.

Now, rearranging,

Where n = (c + d) – (a + b)

Now, n can be = 0, < 0 or > 0

Lets take each case separately.

a)   n = 0 : No effect

b)   n = ‘+ve’ : Addition of inert gas increases the  n i.e. is decreased and so is . So products have to increase and reactants have to decrease to maintain constancy of Kp. So the equilibrium moves forward.

c) n = ‘–ve’  In this case  decreases but 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 ån also increases, they almost counter balance each other. So  can be safely approximated as constant. Thus addition of inert gas has no effect at constant volume.
 

Effect of Catalyst on Equilibrium

Catalysts are the substances which  alter the rate of a reaction without themselves getting consumed in the reaction.  Since the catalyst is associated with forward as well as reverse direction reaction. So, at equilibrium, rate of forward reaction will be equal to rate of reverse reaction and hence catalyst effect will be same on both forward as well as reverse. Hence catalyst never effect the point of equilibrium but it reduces the time to attain the equilibrium. 

Example: What will be the effect on the equilibrium constant on increasing the temperature. 

Solution: Since the forward reaction is endothermic, so increasing the temperature, the forward reaction is favoured. Thereby the equilibrium constant will increase.

Sample problem Kp for the reaction 2BaO_2(s)  2BaO_(s) + O_2(g) is 1.6 × 10^–4 atm, at 400°C. Heat of reaction is – 25.14 kcal. What will be the no. of moles of O2 gas produced at 500°C temperature, if it is carried in 2 litre reaction vessel?

Solution:   We know that

log L_P2/K_P1 = ΔH / 2.303R [1/T₁ – 1/T₂] 

log K_P2/1.6 × 10^–4 = –25.14 / 2.330 × 2 × 10^–3 [773 – 673 / 773 × 673] 

=> K_P2/1.46 × 10^–5 atm 

K_P2 = p₀₂ = 1.46 × 10^–5 atm 

Since, PV = nRT

Related Resources

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Reference books of Chemistry

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