hi

Question

i have difficulty in this question.any help would be much appreciated.

1) a mixture of 0.0080 mol of So2(g) and 0.0056mol of O2(g) is placed in one liter container at 100K. when equilibrium is established, 0.0040 mol of So3(g)is present

2SO2(g)+O2(g)->2So3(g)
<-
a) what are the equilibrium concentration of SO2 and O2?b)what is the value of K for the equilibrium at 100k?

Answer 1

(a) To form 0.0040 mol of SO3, you must have used up 0.0040 moles of SO2 and 0.0020 moles of O2. That is because of the coefficients in the reaction itself.

(b) You end up with the following amounts: 0.0040 mol SO2, 0.036 mol O2 and 0.0040 mol SO3. They are still contained in the one liter container, so those are also the molar concentrations in mole/liter.

The value of Kc is [SO3]^2/[SO2]^2[O2].
Use the final values.