# At a certain temperature, 0.620 mol of SO3 is placed in a 3.00 L containter. `2SO3 (g) harr 2SO2 (g) + O2 (g)` At equilibrium, 0.130 mol of O2 is present. Calculate Kc.

*print*Print*list*Cite

Expert Answers

jeew-m | Certified Educator

`2SO_3 harr 2SO_2+O_2`

Initial `[SO_3] = 0.62/3M = 0.2067M`

Let us say xM of `SO_3` has dissociated.

Mole ratio

`SO_3:SO_2 = 2:2 = 1:1`

`SO_3:O_2 = 2:1`

So xM of SO_2 and x/2M of O_2 will be produced.

Final concentrations

`[SO_3] = 0.2067-x`

`[SO_2] = x`

`[O_2] = x/2`

`K_C = ([SO_2]^2[O_2])/[SO_3]^2`

According to the given data x = 0.13

` K_C = ([0.13]^2[0.13/2])/[0.0767]^2`

`K_c = 0.1867`

** So `K_c` for the reaction is 0.1867**.

Further Reading: