**Let us consider a reaction of the type,**

$ A(g) + B(g) \rightleftharpoons C(g) + D(g) $

The double headed arrow signifies that the reaction occurs in both the directions in measurable extent.

What do you do at the start ? You start with pure A & B which are the reactants, in a closed system. Initially A and B reacts very fast to give C and D.

Now as soon as C and D are formed they also started giving back A and B. But A and B being present in greater quantity forces the reaction to occur in forward direction much faster than the backward one.

So what is the net result ? It’s nothing but the net formation of C & D.

In this way as the time passes, A & B decreases and hence their strength, on the other hand C & D increases and so is their strength. So ultimately a time comes when the forward reaction is balanced by backward reaction. **This state is the equilibrium** .

Now, before deriving the different related expression let us know what is Law of Mass Action. It states that,

**” The rate of a reaction is directly proportional to the product of concentrations of reactants with the stoichiometric co-efficient being raised to the power “**

e.g. for the reaction.

$ a A(g) + b B(g) \rightleftharpoons c C(g) + d D(g) $

for the forward reaction,

R_{f} ∝ [A]^{a} [B]^{b}

Where R_{f} denotes rate of forward reaction.

[A] denotes concentration of A

‘ a ‘ , ‘ b ‘ are the powers which are the coefficients of A & B respectively.

If k_{f} be the rate constant for forward reaction the expression can be rewritten as

R_{f} = k_{f}[A]^{a} [B]^{b} …..(1)

Similarly, for the backward reaction:

R_{b} = k_{b} [C]^{c} [D]^{d} ….(2)

**At equilibrium:**

Rate of forward reaction = Rate of backward reaction

From equation (1) and (2)

R_{f} = R_{b}

or k_{f} [A]^{a} [B]^{b} = k_{b} [C]^{c} [D]^{d}

Rearranging we get,

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

Now, the expression on the L.H.S. is a constant. So it means the expression on the R.H.S. has to be a constant. In fact it is so for a particular reaction.

But obviously at a given temperature. Effect of temperature we will study later. This constant is named as equilibrium constant and is denoted by K.

This implies that no matter what we start with (i.e., A & B or C&D or A + B + C or A + B + D or A + C + D or B + C + D or A + B + C + D ) and how much of these we start with, the ratio of $\large \frac{[C]^c [D]^d}{[A]^a [B]^b}$

is a fixed quantity at a given temperature when the reaction reaches equilibrium.

That is , if we assume that this reaction has K = 4 then no matter what we take initially and irrespective of how much we take, once equilibrium is reached the ratio of $\large \frac{[C]^c [D]^d}{[A]^a [B]^b}$ will always be equal to 4.

Now let us consider the following reaction

$ A(s) + B(g) \rightleftharpoons C(s) + D(g) $

Its equilibrium constant K would be ,

$\large K = \frac{[C][D]}{[A][B]}$

Concentration of C is the number of moles of C per unit volume of solution. Concentration of D is the number of moles of D per unit volume of the container (we can assume that the gas occupies the entire container).

The concentration of A is the number of moles of A per unit volume of A. The concentration of all solids and pure liquids is a constant. This is because if initially we take w gm of A , then the moles of A are w/M .

The volume of A is w/d where d is the density of A.

Therefore, the initial concentration of A is

$\large \frac{w/M}{w/d} = \frac{d}{M}$

We can see that at equilibrium also the concentration of A remains as d/M ( d and M are constants).

In fact even if A were a pure liquid, its concentration would have remained constant

Therefore, we bring all the constant terms on one side and we get

$\large K [A] = \frac{[C][D]}{[B]}$

This ratio which is a constant and which involves only those concentration terms which are variables is called K_{c }, the equilibrium constant in terms of concentration.

K_{c} = [C][D]/[B] = K [A]

Now, let us consider the reaction,

$ A(g) + 2B(g) \rightleftharpoons 3C(g) + 4D(g) $

$\large K_c = \frac{[C]^3[D]^4}{[A][B]^2}$

We know that concentration of a gas can be expressed as P/RT as shown below:

P V = n R T

P/RT = n/V = number of moles per litre = concentration

Hence ,

$\large [C] = \frac{P_C}{R T}$

$\large [D] = \frac{P_D}{R T}$

$\large [A] = \frac{P_A}{R T}$

$\large [B] = \frac{P_B}{R T}$

Where,

P_{C} = partial pressure of C

P_{D}= partial pressure of D

P_{A} = partial pressure of A

P_{B} = partial pressure of B

$\large K_C = \frac{(\frac{P_C}{R T})^3(\frac{P_D}{R T})^4}{(\frac{P_A}{R T})(\frac{P_B}{R T})^2}$

$\large K_C (R T)^{(3+4)-(1+2)} = \frac{P_C^3 P_D^4}{P_A P_B^2}$

Since K_{C} is a constant and RT is also a constant, so the right hand side of the above expression is also a constant. This is called K_{p }, the equilibrium constant in terms of the partial pressure.

$\large K_P = \frac{P_C^3 P_D^4}{P_A P_B^2}$

Therefore, for this particular equilibrium, the ratio of partial pressures is also a constant.

### Relation between KP & KC

**In general, the relation between K _{P} and K_{C} is **

K_{P} = K_{C} (RT)^{Δn}

Where Δn = number of moles of gaseous products − number of moles of gaseous reactants.

Now Δn can have three possibilities.

Δn < 0 , Δn = 0 , Δn > 0

Accordingly we can predict, out of Kp and Kc which one will be higher or lower.

At Δn = 0 both K_{p} & K_{c} are the same.

Now let us assume that A is a solid or pure liquid. The changes now would be that K_{C} would look like this,

$\large K_C = \frac{[C]^3[D]^4}{[B]^2}$

and following the above given sequence of derivation, K_{p} would like this

$\large K_P = \frac{P_C^3 P_D^4}{P_B^2} $

Next we assume that A was a solute present in a solution then K_{c} would remain the same

i.e,$\large K_C = \frac{[C]^3[D]^4}{[B]^2}$

Now if we try to express the concentrations in terms of partial pressures, we would fail to do that for A.

It is not possible to express the concentration of a solution in terms of its pressure or vapour pressure and constants.

Therefore [A] remains as such

$\large K_C = \frac{(P_C/RT)^3(P_D/RT)^4}{[A](P_B/RT)^2}$

$\large K_C (R T)^{(3+4)-2} = \frac{P_C^3 P_D^4}{[A]P_B^2}$

The R.H.S. of the above expression is a constant which implies that the L.H.S is also a constant.

This new expression cannot be called as either KC or KP since it contains both concentration terms and pressure terms. We call it KPC.

We can also see that if we take [A] to the R.H.S. the L.H.S. contains only pressure terms, but then it is not a constant since [A] is not a constant.

Therefore, we can conclude that for K_{P} to exist for a reaction it must fulfill two conditions:–

**(i) it must have at least one gas either in the reactants or in the products and **

**(ii) it must not have any component in solution phase.**

**Exercise : **In a gaseous reaction of the type A(g) + 2B(g) 2C(g) + D(g), the initial concentration of B was 1.5 times that of A. At equilibrium the equilibrium concentration of A and D were equal. Calculate the equilibrium constant ?

### Important relationships involving Equilibrium Constant

If we reverse an equation, K_{c} or K_{p} is inverted i.e.

If $ A(g) + B(g) \rightleftharpoons C(g) + D(g) \; K_C = 10 $

Then for $ C(g) + D(g) \rightleftharpoons A(g) + B(g) \; K’_C = 10^{-1}$

*If we multiply each of the coefficient in a balanced equation by a factor n , then equilibrium constant is raised to the same factor.*

If (1/2) N_{2} + (1/2) O_{2} ↔ NO K_{c} = 5

Then for N_{2} + O_{2} ↔ 2NO K_{c}‘ = 5^{2} = 25

*If we divide each of the coefficients in a balanced equation by the factor n, then new equilibrium constant is nth root of the previous value.*

If 2SO_{2} + O_{2} ↔ 2SO_{3} K_{c} = 25

Then for SO_{2} + (1/2)O_{2} ↔ SO_{3} K_{c}‘ = √25 = 5

When we combine individual equation, we have to multiply their equilibrium constants for net reaction.

If K_{1} , K_{2} and K_{3} are step wise equilibrium constant for A <–> B , B <–> C , C ↔ D .

Then for A ↔ D , equilibrium constant is K = K_{1} K_{2} K_{3}

### Significance of the Magnitude of Equilibrium Constant

∎ A very large value of K_{C} or K_{P} signifies that the forward reaction goes to completion or very nearly so.

∎ A very small value of K_{C} or K_{P} signifies that the forward reaction does not occur to any significant extent/.

∎ A reaction is most likely to reach a state of equilibrium in which both reactants and products are present if the numerical value of K_{c} or K_{P} is neither very large nor very small.

**Illustration : **Calculate Kc for the reaction ,

HI (g) ↔ (1/2) H_{2} (g) + I_{2} (g)

Given that the reaction H_{2} (g) + I_{2} ↔ 2HI (g), had been started with 1 mole of H_{2} (s) and 3 mole of I_{2} (g).

At the equilibrium the moles of HI formed is 1.5 moles. The volume of the reaction vessel is 2 litre.

**Solution:**

H_{2}(g) + I_{2} ↔ 2HI(g)

Initially moles 2 3 0

Moles at eqm. 2 – x 3 – x 2x

Conc. of eqm.

From question, 2x = 1.5

x = 0.75

$\large [H_2] = \frac{2-x}{2} = \frac{1.25}{2}$

$\large [I_2] = \frac{3-x}{2} = \frac{2.25}{2}$

$\large [H I] = \frac{1.5}{2}$

∴ K_{c} for the reaction $\large = \frac{[HI]^2}{[H_2][I_2]}$

On putting the above values ,

K_{c} = 0.8

Let the equilibrium constant K’_{c} for the reaction

HI(g) <–> H_{2}(g) + I_{2}(g)

$\large K_C’ = \frac{1}{\sqrt{K_C}} = \frac{1}{\sqrt{0.8}}$

= 1.124

**Exercise :** At a certain temperature, equilibrium constant (KC) is 16 for the reaction:

SO_{2}(g) + NO_{2} (g) ↔ SO_{3}(g) + NO(g)

If we take one mole of each of the four gases in one litre container , what would be the equilibrium concentration of NO and NO_{2} ?

### The Reaction Quotient ‘ Q ‘

Consider the equilibrium

PCl_{5} (g) ↔ PCl_{3}(g) + Cl_{2} (g)

At equilibrium $\large K_C = \frac{[Cl_2][PCl_3]}{[PCl_5]}$

**When the reaction is not at equilibrium this ratio is called ‘ Q _{C} ‘ i.e. , Q_{C} is the general term used for the above given ratio at any instant of time. And at equilibrium Q_{C} becomes K_{C} .**

Similarly , $\large \frac{P_{Cl_2} P_{Cl_3}}{P_{Cl_5}}$ is called Q_{P} and at equilibrium it becomes K_{P}.

∎ If the reaction is at equilibrium , Q = K_{c}

∎ A net reaction proceeds from left to right (forward direction) if Q < K_{C}.

∎ A net reaction proceeds from right to left (the reverse direction) if Q > K_{c}

**Illustration : **For the reaction,

A(g) + B(g) ↔ 2C(g) at 25°C , at a 2 litre vessel contains 1 , 2 , 3 moles of respectively.

Predict the direction of the reaction of

(a) K_{c} for the reaction is 3

(b) K_{d} for the reaction 6

(c) K_{c} for the reaction

**Solution:** A(g) + B(g) ↔ 2C(g)

Reaction quotient $\large Q = \frac{[C]^2}{[A][B]}$

$\large Q = \frac{(3/2)^2}{(1/2)(2/2)} $

= 9/2 = 4.5

(a) Q < K_{C}

therefore backward reaction will be followed

(b) Q > K_{C}

The forward reaction is followed

(c) Q = K_{C}

The reaction is at equilibrium

**Exercise : **The equilibrium constant is 0.403 at 1000K for the process

FeO (s) + CO (g) ↔ Fe (s) + CO_{2} (g)

A steam of pure CO is passed over powdered FeO at 1000K until equilibrium is reached. What is the mole fraction of CO in the gas stream leaving the reaction zone?