Relation between Kp and Kc (derivation)

Relationship between Kc and Kp

In a reaction,

aA + bB = cC + dD

we can write

Assuming the gaseous components to behave ideally,

P_iV_i = n_iRT

or

where (i) is the molar concentration of the species 'i'.

n = (Number of moles of gaseous products)-(Number of moles of gaseous reactants).

Thus, n is equal to the difference in the number of gaseous moles of products and the number of gaseous moles of reactants. The above equation can be rewritten as

K_p = K_c (RT)ⁿ

Problems

2. Write the relationship between K_p and K_c for the following reactions.

Solution

(a) Here, n = 1+ 1 - 1 = 1

K_p = K_c (RT)¹= K_cRT

(b) Here, n = 2- (3+1) = -2

K_p = K_c (RT)^-2

(c) Here, n = 2- (1+1) = 0

K_p = K_c (RT)⁰ = K_c

(d) Here, n = 2+1-2 =1

K_p = K_c (RT)¹= K_cRT

(d) Here, n = 2- (2+1)= -1

K_p = K_c (RT)^-1= K_c / RT

3. At 700 K, the equilibrium constant, K_p for the reaction

is 1.80 x 10^-3 kPa. What is the numerical value in moles per litre of K_c for this reaction at the same temperature?

Solution

For the reaction,

K_p = 1.80 x 10^-3 kPa, K_c = ?

We know that, K_p = K_c (RT)ⁿ

where n = n_g(products) - n_g(reactants)

For the given reaction,

n = 2 + 1 - 2 = 1

So, K_p = K_c (RT)¹= K_c x RT

R = 8.31 L kPa K^-1 mol^-1

4. Two litres of a solution of acetic acid contains 15 g of acetic acid. What is the active mass?

Solution

Molecular weight of acetic acid = 60

Weight of acetic acid = 15 gms.

Volume of the solution = 2 litres.

Active mass = number of moles/litre

5.

Solution

Here the products are X and Y and reactants are A and B. The

powers of the concentrations of A, B, X and Y correspond to the

respective number of moles in the equation.

6. Write down the expression for the equilibrium constants

What is the relation between K_p and K_c for this reaction?

Solution

n = number of moles of products - number of moles of reactants

= 3 - 5 = -2

K_p = K_c (RT)ⁿ = K_c (RT)^-2

K_p = K_c(RT) ^-2 for this reaction.