Question

For the equilibrium

2SO_2(g)+O_2(g)->2SO_3(g) K_c=245(at 100K)

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

$2SO_{2(g)} +O_{2(g)} \to 2SO_{3(g)}$

[SO_2] = 0.204 \\ [O_2] = 0.0264 \\ [SO_3] = 0.368

$Q = \dfrac{SO_3]^2}{[O_2][SO_2]^2} = \dfrac{0.368}{0.102×0.0264} = 136.7$

Q < K therefore, a forward reaction takes place to reach a new equilibrium.