# In a reaction,A+2B-------->2C,2 moles of A 3 mole of B and 2 mole of C are placed in a 2 L flask and the equilibrium concentration of C is 0.5 mol/L.The equilibrium constant K for the reaction isans.-0.05please tell how?

Vikas TU
14149 Points
10 years ago

A          +          2B        -------->             2C

Let t = 0 (Initially)                2                         3                                        2

t = t                                  2 - x                    3 - 2x                               2x + 2 (Initially it was 2)

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

x  = - 0.5

Put x = - 0.5  at time t

Thus, A = 2.5

B = 4

C = 1

K = (C)2 / (A) (B)2

K = (1/2)2/ (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

Approve plz!

Rishi Sharma
3 years ago
Dear Student,

A + 2B → 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=(C)^2/(A)(B)^2
K=(1/2)2/(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

Thanks and Regards