Factors Affecting Equilibria

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".

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 $\rightleftharpoons$ 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.

$\therefore$ $K_P=\frac{\left(\frac{a}{a+b+c}\times P_T\right)\left(\frac{b}{a+b+c}\times P_T\right)}{\left(\frac{c}{a+b+c}\times P_T\right)}$= $\frac{ab{P}_T}{c(a+b+c)}$

Since P_T = $\frac{(a+b+c)RT}{V}$

$\therefore$ K_P = $\frac{\mathbf{abRT}}{\mathbf{cV}}$

Now if d moles of PCl₃ is added to the system, the value of Q would be, $\frac{a(b+d)RT}{cV}$

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

(i) If we increase the volume of the system, the Q becomes $\frac{abRT}{c{V}^{\prime }}$ where V¢ > V.

$\therefore$ Q becomes less, and the system would move forward to attain equilibrium.

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

(iii) If we add the noble gas at constant volume, the expression of Q remains as Q = $\frac{abRT}{cV}$. and the system continues to be in equilibrium.

$\therefore$ nothing happens.

(iv) 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 Concentration

Let us have a general reaction,

aA + bB $\rightleftharpoons$ cC + dD

at a given temperature, the equilibrium constant,

$K_c=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$

again if $\alpha ,\beta ,\gamma$and $\delta$ are the number of mole of A, B, C and D are at equilibrium

then, $K_c=\frac{{[\gamma ]}^c{[\delta ]}^d}{{[\alpha ]}^a{[\beta ]}^b}$

If any of product will be added, to keep the K_c 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.

Temperature Effect

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,

$k_f=A_f{e}^{-\frac{E_{af}}{RT}}$

And for the reverse reaction,

$k_r=A_r{e}^{-\frac{E_{ar}}{RT}}$

$\therefore$ $\frac{k_f}{k_r}=K=\frac{A_f}{A_r}{e}^{\frac{(E_{ar}-E_{af})}{RT}}$

$\therefore$ $K_{T_1}=\frac{A_f}{A_r}{e}^{\frac{\left(E_{ar}-E_{af}\right)}{R{T}_1}}$

$K_{T_2}=\frac{A_f}{A_r}{e}^{\frac{\left(E_{ar}-E_{af}\right)}{R{T}_2}}$

It can be seen that

$\frac{K_{T_2}}{K_{T_1}}=e^{\frac{\left(E_{ar}-E_{af}\right)}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}$

or in $\frac{K_{T_2}}{K_{T_1}}=\frac{\left(E_{ar}-E_{af}\right)}{R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]$

For any reaction, $\Delta$H = E_af - E_ar

$\ln \frac{K_{T_2}}{K_{T_1}}$=$\frac{-\Delta H}{R}$$\left[\frac{1}{T_2}-\frac{1}{T_1}\right]$or in$\frac{K_{T_2}}{K_{T_1}}$=$\frac{\Delta H}{R}$$\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$

From the equation,

$\log \frac{K_2}{K_1}=\frac{\Delta H^o}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$it is clear that

(a) If $\Delta$H^o is +ve (endothermic), an increase in temperature (T₂ > T₁) will make K₂ > K₁, i.e., the reaction goes more towards the forward direction and vice-versa.

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

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

(ii) 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.

(iii) If no gases are involved in the reaction higher pressure favours the reaction to shift towards higher density solid or liquid.

Illustration 1. Under what conditions will the following reactions go in the forward direction?

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

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

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

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

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

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

(vii) N₂O₄(g) $\rightleftharpoons$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.

Effect of change of Pressure

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

$\rightleftharpoons$$K_p=Q_p\times P^{(\Delta n)}$

Case A: When $\Delta$n = 0

K_p = Q_p

And equilibria is independent of pressure.

Case B: When $\Delta$n = – ve

$K_p=Q_p\cdot P^{-\Delta n}=\frac{Q_p}{P^{\Delta n}}$

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 $\Delta$n = +ve

$K_p=Q_p\cdot P^{\Delta n}$and, effect will be opposite to that of Case 'B'

EFFECT OF CATALYST ON EQUILIBRIUM

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.

Effect on equilibrium due to the addition of inert gases

Following are the cases, where this effect in different manner.

Case A: When $\Delta$n = 0 and total volume change at equilibrium remain constant.

So, addition of inert gases will increase the number of moles in the mixture but partial pressure of each component remains constant. Hence equilibrium remains unaffected.

Case B: When $\Delta$n = +ve and $\Delta$V_T $\ne$ 0

The total volume will increase with addition of inert gases. This will shift the equilibrium in forward direction.

Case C: When $\Delta$n = –ve and $\Delta$V_T $\ne$ 0

In this case, the addition of inert gases will increase the total volume and the reaction will shift in reverse direction.

Dependence of K_P or K_c on Temperature

With the increase of temperature, equilibrium favours forward reaction in case of an endothermic ($\Delta$H > 0) reaction while it favours backward reaction in the case of an exothermic ($\Delta$H < 0) reaction.

Illustration 2. For the equilibrium

NH₄I_(g) $\rightleftharpoons$ NH_3(g) + HI_(g) $\Delta H=+ve$

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.