Question

Ina reaction A+B=C+Dthe initial concentration of A and B were 0.9mol dm-3each.at equilibrium the concentration of D was found to be 0.6 mole dm-3. What is the valu of equilibrium constant for the reaction? ​

Answer 1

Given:

In the reaction A + B = C + D , the initial concentration of A and B were 0.9 mole/dm³. At equilibrium , the concentration of D is found to be 0.6 mole/dm³.

To find:

Value of Equilibrium Constant

Calculation:

$Reaction : \: \: \: \: \: A \: \: \: \: + \: \: \: \: B \: = \: \: \: C \: + \: \: \: \: D$

$\sf{at \: t = 0 \: sec} : 0.9 \: \: \: \: \: \: \: \: \: \: 0.9 \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: 0$

$\sf{at \: t = t \: sec} : \small{(0.9 - x) \: \: (0.9 - x)\: \: x \: \: \:\:\:\:\: \:\:x}$

At t = t sec , equilibrium is reached , so we can say that

$\boxed{ \sf{x = 0.6 \: \frac{mol}{ {dm}^{3} } }}$

Hence concentration of C and D can be represented as :

[C] = [D] = 0.6 mol/dm³

So, concentration of A and B will be :

[A] = [B] = (0.9 - 0.6) = 0.3 mol/dm³

As per Law Of Mass Action , let Equilibrium Constant be k :

$\therefore \: k = \dfrac{ \{A \} \{B \}}{ \{C \} \{D \}}$

$= > \: k = \dfrac{ \{0.3 \} \times \{0.3\}} { \{0.6 \} \times \{0.6 \}}$

$= > \: k = \dfrac{0.09} {0.36}$

$= > k = 0.25$

So final answer:

Equilibrium constant is 0.25