### Relationship between Kc and Kp

In a reaction,

aA + bB = cC + dD

we can write

Assuming the gaseous components to behave ideally,

PiVi = niRT

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

Kp = Kc (RT)n

### Problems

2. Write the relationship between Kp and Kc for the following reactions.

### Solution

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

Kp = Kc (RT)1= KcRT

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

Kp = Kc (RT)-2

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

Kp = Kc (RT)0 = Kc

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

Kp = Kc (RT)1= KcRT

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

Kp = Kc (RT)-1= Kc / RT

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

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

### Solution

For the reaction,

Kp = 1.80 x 10-3 kPa, Kc = ?

We know that, Kp = Kc (RT)n

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

For the given reaction,

n = 2 + 1 - 2 = 1

So, Kp = Kc (RT)1= Kc 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 Kp and Kc for this reaction?

### Solution

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

= 3 - 5 = -2

Kp = Kc (RT)n = Kc (RT)-2

Kp = Kc(RT) -2 for this reaction.

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