Question

Solve using both ICE table and approx method

A2B ⇌ 2A +B

Kc=.00145

assume all to be gas

A2B to be .326 Moles

Answer 1

A2B <-------> 2A + B

Kc= [ A]^2 [ B ]/[ A2B]​ = 0.00145

Initial concentration

[ A2B ] = 0.326M

[ A ] = 0

[ B ] = 0

Change in Concentration

[ A2B ] = - X

[ A ] = +2X

[ B ] = X

Concentration at equilibrium

[ A2B ] = 0.326 - X

[ A ] = 2X

[ B] = X

Kc = (2X)^2 .X/ (0.326 - X )

0.00145 = 4X^3/(0.326 - X)

If we assume 0.326 - X = 0.326

4X^3 = 4.73×10^-4

X^3 = 1.18 × 10^-4

X = 4.90×10^-2

Therefore,

[ A2B ] = 0.326 - 0.0490 = 0.277M

[ A ] = 2×0.0490 = 0.098M

[ B ] = 0.0490M

moles of A2B = 0.277

moles of A = 0.098

moles of B = 0.049