Answer to Question #176125 in Physical Chemistry for Vikram

Question

For the equilibrium

2SO2(g)+O2(g)=2SO3(g) K_c=245(at100k)

The equilibrium concentrations are [SO₂]=0.204M,[O₂]=0.0264M and [SO₃]=0.368M. Suppose that the concentration of SO₂ is suddenly halved calculate Qc and use it to show that the forward reaction takes place to reach a new equilibrium.

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.