Answer to Question #155664 in Physical Chemistry for Anand

Question

For the equilibrium 

 2SO₂(g)+O₂(g) 2SO₃(g) K_c=245 (at 100 K) 

 The equilibrium concentrations are [SO₂]=0.204 M,[O₂]=0.0264 M and 

[SO_3]=0.368 M. Suppose that the concentration of SO₂ is suddenly halved calculate 

Q_c.

Answer 1

Q_c = [so3]² \ ([o2]x[so2]²) ; [so2] = 0.204\2 = 0.102 M (new concentration)

= (0.368²)\(0.0264 x 0.102²)

= (0.1350)\(0.0264 x 0.0104)

= 491.5 (ans)

as here conc of reactant so2 is reduced so Qc become larger than Kc so that the reaction move backward to restore Kc and produce reactant from product.