Question

When 3.00 mol of a and 4.00 mol of b are placed in a container and allowed to come to equilibrium, the resulting mixture is found to contain 0.80 mol of d. what is the value of k at equilibrium? a(g) + 2b (g) c (g) + d(g)

Answer 1

Correct question :- When 3.00 mol of A and 4.00 mol of B are placed in a container and allowed to come to equilibrium, the resulting mixture is found to contain 0.80 mol of D. what is the value of k at equilibrium?

The reaction is. :- A(g) + 2B (g) $\rightarrow$ C (g) + D (g)

Soln:-

Given

The reaction :- A(g) + 2B (g) $\rightarrow$ C (g) + D (g)

To Find

The equilibtium constant of the reaction.

Concept :-

For a reaction :- xA(g) + yB (g) $\rightarrow$ mC (g) + nD (g) ,if [A] ,[B] ,[C] and [D] are equilibtium concentration of the species A,B,C and D respectively. Then rate constant of the reaction is given as :

$\boxed{\green{K\:=\: \frac{[C]^m\:[D]^n}{[A]^x\:[B]^y}}}$

Solutions

Initially 3 moles of A and 4 moles of B were present in the container. At equilibrium, 0.8 mole of D is formed.

• Let x moles of A is dissociated.

See the attachment now

We get , [D] = x

Hence, x = 0.8 moles.

• 0.8 moles of [A] is dissociated at equilibrium.

• At equilibrium

[A] = 3 - x = 3 - 0.8 = 2.4

[B] = 4 -2x = 4 - 1.6 = 2.4

[C] = [D] = 0.8

Now, for the reaction :- A(g) + 2B (g) $\rightarrow$ C (g) + D (g)

$K\:=\: \frac{[C]\:[D]}{[A]\:[B]^2}$

=> $K\:=\: \frac{0.8\:\times\:0.8}{2.4\:\times{2.4}^2 }$

=> $K\:=\: \frac{0.8\:\times\:0.8}{2.4\:\times{2.4}^2 }$

=> K = 0.046 .

Hence, equilibrium constant of the reaction is 0.046 .