Question

"In a reaction, $A+2B\to 2C$, in starting the $3.0$ mole of 'B' and $2.0$ of mole 'C' are placed in a $2.0$L flask and the equilibrium concentration of 'C' is $0.5$ mol/L. The equilibrium constant (K) for the reaction is

• A. $0.073$
• B. $0.147$
• C. $0.05$
• D. $0.026$

Answer 1

The correct option is C $0.05$

$A+2B\to 2C$

Let t = 0 (Initially) 2 3 2

t = t 2 - x 3 - 2x 2x + 2

x = -0.5

(Initially Now, [C] = 0.5 mol/L = (2x + 2)/2 mol/L

Put x = - 0.5 at time t

Thus, A = 2.5

B = 4

C = 1

$K=\left(C{)}^2/\right(A\left)\right(B{)}^2$

$K=\left(1/2{)}^2/\right(2.5/2).(4/2{)}^2$

[Dividing moles by 2 {volume given} as we express here in terms of concentration

K = 1/20

K = 0.05