Question

pls solve the above question .​

Answer 1

★ Concept :-

Here the concept of Equilibrium Constant has been used. We see that we are given a equation which is simply word equation. So here we no need to balance it. Also we are given the initial concentrations of A and B and even the concentrations at equilibrium of C and D . Firstly we can substitute the the initial concentration of C and  as 0 . Then from the equilibrium concentration we can find the value of equilibrium concentration of A and B. And finally by using the formula of Equilibrium Constant we can find the answer.

Let's do it  !!

________________________________________________________

★ Formula Used :-

$\;\boxed{\sf{\pink{K_{c}\;=\;\bf{\dfrac{[C][D]}{[A][B]}}}}}$

_________________________________________________________

★ Solution :-

Given,

The reaction is ::

⇒ A + B ⇌ C + D

⇒ Initial concentration of A = 1 M

⇒ Initial concentration of B = 1 M

⇒ Equilibrium Concentration of C = 0.8 M

⇒ Equilibrium Concentration of D = 0.8 M

This reaction is in equilibrium. This means sum of reactants will be equal to sum of products. Since we are not given the initial concentrations of A and B , thus

• Initial Concentration of C = 0 M

• Initial Concentration of D = 0 M

________________________________________________________

~ For equilibrium concentration of A and B ::

Since, A + B and C + D are in equilibrium . So the equilibrium concentration of A and B ill be the difference of their initial concentration and equilibrium concentration of C and D.

→ Equilibrium Concentration of A = 1 - 0.8 = 0.2 M

→ Equilibrium Concentration of B = 1 - 0.8 = 0.2 M

________________________________________________________

~ For the Equilibrium Concentration of reaction :-

From formula, we know that

$\;\sf{\rightarrow\;\;K_{c}\;=\;\bf{\dfrac{[C][D]}{[A][B]}}}$

$\;\sf{\rightarrow\;\;K_{c}\;=\;\bf{\dfrac{[0.8][0.8]}{[0.2][0.2]}}}$

$\;\sf{\rightarrow\;\;K_{c}\;=\;\bf{\dfrac{0.8\:\times\:0.8}{0.2\:\times\;0.2}}}$

$\;\sf{\rightarrow\;\;K_{c}\;=\;\bf{4\:\times\:4}}$

$\;\bf{\green{\rightarrow\;\;K_{c}\;=\;\bf{16}}}$

This is required answer.

So, option D) 16 is correct.

Hence, required equilibrium constant = 16

$\;\underline{\boxed{\tt{Required\;\: Equilibrium\;\: Constant\;=\;\bf{\purple{16}}}}}$

Answer 2

In the reversible reaction A + B ⇔ C + D, the concentration of each C and D at equilobrium was 0.8 mole/litre, then the equilibrium constant Kc will be. ... Suppose 1 mole of A and B is each taken, then 0.8 mole/litre of C and D each formed remaining concentration of A and B will be (1-0.8)=0.2 mole/litre.

Hope it will help you..