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,

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

Relationship between Kc and Kp

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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