Question #176125
For the equilibrium
2SO2(g)+O2(g)=2SO3(g) Kc=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.
Expert's answer
Qc = [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.
