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 I 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:

2 NH_{3} (g) N_{2} (g) + 3H_{2} (g)

Let the moles of N_{2 }, H_{2} and NH_{3} at equilibrium be a , b and c moles respectively.

Since the reaction is at equilibrium,

$\large \frac{P_{N_2} \times (P_{H_2})^3}{(P_{N H_3 })^2} = K_P = \frac{(X_{N_2} P_T )\times (X_{H_2} P_T)^3}{(X_{N H_3} P_T)^2} $

Where, X terms denote respective mole fractions and P_{T} is the total pressure of the system.

$\Large \frac{(\frac{a}{a+b+c}\times P_T)(\frac{b}{a+b+c}\times P_T)^3}{(\frac{c}{a+b+c} \times P_T)^2} = K_P$

Here, $\large \frac{a}{a+b+c}$ = mole fraction of N_{2}

$\large \frac{b}{a+b+c}$ = mole fraction of H_{2}

$\large \frac{c}{a+b+c}$ = mole fraction NH_{3}

⇒$\large \frac{a b^3}{c^2} \times \frac{(P_T)^2}{(a+b+c)^2} = K_P$

Since $\large P_T = \frac{(a+b+c)RT}{V}$ ( assuming all gases to be ideal)

∴ $\large \frac{a b^3}{c^2} \times (\frac{R T}{V})^2 = K_P$ ……….(1)

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 expression becomes Q_{P} ( as it is disturbed from equilibrium) and we can see that Q_{P} > K_{P}. The reaction therefore moves backward to make Q_{P} = K_{P}

If we increase c , Q_{P} < K_{P} 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), Q_{P} < K_{P} 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.

If temperature is increased the equilibrium will shift in forward direction since the forward reaction is endothermic and temperature is reduced in this direction.

However from the expression if we increase the temperature of the reaction, the left hand side increases (Q_{P}) and therefore does it mean that the reaction goes backward (since Q_{P} > K_{P} )?.

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 questions is No. This is because K_{P} also changes with temperature. Therefore, we need to know the effect of temperature on both Q_{P} and K_{P} to decide the course of the reaction.

### Dependence of KP or Kc on Temperature

###### 1. Addition at constant pressure

Let us take a general reaction

a A_{(g)} + b B_{(g)} ↔ c C_{(g)} + d D_{(g)}

We know,

$\Large K_P = \frac{(\frac{n_C}{\Sigma n}\times P_T)^c (\frac{n_D}{\Sigma n}\times P_T)^d}{(\frac{n_A}{\Sigma n}\times P_T)^a (\frac{n_B}{\Sigma n}\times P_T)^b}$

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 Σn = total no. of moles of reactants and products.

Now, rearranging,

$\large K_P = \frac{n_C^c \times n_D^d}{n_A^a \times n_B^b}\times [\frac{P_T}{\Sigma n}]^{\Delta n} $

Where Δn = (c + d) – (a + b)

Now, Δn can be = 0 , < 0 or > 0

Lets take each case separately.

a)Δn = 0 : No effect

b) Δn = ‘+ve’ :

Addition of inert gas increases the Σn i.e. $\large (\frac{P_T}{\Sigma n})$ is decreased and so is . $\large (\frac{P_T}{\Sigma n})^{\Delta n}$

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

(c) Δn = ‘-ve ‘ : In this case $\large (\frac{P_T}{\Sigma n})$decreases but $\large (\frac{P_T}{\Sigma n})^{\Delta n}$ 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 Σn also increases, they almost counter balance each other.

So $\large (\frac{P_T}{\Sigma n})^{\Delta n}$ can be safely approximated as constant. Thus addition of inert gas has no effect at constant volume.

Now we will derive the dependence of K_{P} on temperature.

Starting with Arrhenius equation of rate constant

$\large k_f = A_f e^{-E_{af}/RT}$ …(i)

Where, k_{f} = rate constant for forward reaction, A_{f} = Arrhenius constant of forward reaction

E_{af} = Energy of activation of forward reaction

$\large k_r = A_r e^{-E_{ar}/RT}$ …(ii)

Dividing (i) by (ii) we get,

$\large \frac{k_f}{k_r} = \frac{A_f}{A_r} e^{\frac{E_{ar} – E_{af}}{RT}}$

We know that

$\large \frac{k_f}{k_r} = K$ (equilibrium Constant)

$\large K = \frac{A_f}{A_r} e^{\frac{E_{ar} – E_{af}}{RT}}$

At temperature T_{1}

$\large K_{T_1} = \frac{A_f}{A_r} e^{\frac{E_{ar} – E_{af}}{R T_1}}$ …(iii)

At temperature T_{2}

$\large K_{T_2} = \frac{A_f}{A_r} e^{\frac{E_{ar} – E_{af}}{R T_2}}$ …(iv)

Dividing (iv) by (iii) we get

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

$\large log\frac{K_{T_2}}{K_{T_1}} = \frac{E_{ar} – E_{af}}{2.303 R} (\frac{1}{T_2}-\frac{1}{T_1})$

The enthalpy of a reaction is defined in terms of activation energies as

E_{af} – E_{ar} = ΔH

$\large log\frac{K_{T_2}}{K_{T_1}} = \frac{-\Delta H}{2.303 R} (\frac{1}{T_2}-\frac{1}{T_1})$

$\large log\frac{K_{T_2}}{K_{T_1}} = \frac{\Delta H}{2.303 R} (\frac{1}{T_1}-\frac{1}{T_2})$

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

K_{T2} < K_{T1} , 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?

Now, Q_{P} can also increase, decrease or remain unchanged. If K_{P} increases and Q_{P} decreases, then Q_{PT2} < K_{PT2} , therefore the reaction moves forward.

If K_{P} increase and Q_{P} remains same, then Q_{PT2}(Q_{PT1}) < K_{PT2} . Again, the reaction moves forward. What, if K_{P} increase and Q_{P} also increases?

Will Q_{PT2} = K_{PT2} or Q_{PT2} < K_{PT2} or Q_{PT2} > K_{PT2} ?

This can be answered by simply looking at the dependence of Q_{P} and K_{P} on temperature. As from above equation ,

the KP depends on temperature exponentially. While Q ‘s dependence on T would be either to the power g , l , t ….. Therefore the variation in KP due to T would be more than in Q_{P} due to T.

∴ K_{P} would still be greater than Q_{P} and the reaction moves forward again. Therefore, to see the temperature effect, we need to look at KP only. If it increases the reaction moves forward, if it decreases, reaction moves backward and if it remains fixed, then, no change at all.

**Illustration :** For the equilibrium

NH4I(g) ↔ NH3(g) + HI(g)

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.