Question

The numerical value of the equilibrium constant, Kc, for the following gas phase reaction is 0.50 at a certain temperature. When a certain reaction mixture reaches equilibrium, the concentration of O2 is found to be 2.0 M, while the concentration of SO3 is found to be 10 M. What is the equilibrium concentration of SO2 in this mixture

Answer 1

The equilibrium concentration of
`SO_(2)` is 10 mol/L or 10 M

Step-by-step explanation:

Firstly, the balanced equation for this reaction is:

`2SO_(2)(g)` +
`O_(2)(g)` ⇆
`2SO_(3) (g)`

We're given:

Equilibrium constant, Kc = 0.50

Equil conc. of product, [SO3] = 10M

Equil conc. of O2, [O2] = 2.0M

But
`K_(c)` =
`([product]^(n) )/([reactants]^(n) )` =
`([SO_(3)]^(2) )/([SO_(2)]^(2)[O_(2) ] )` (where n = no. of moles)

⇒ 0.5 =
`(10^(2)M )/([SO_(2)]^(2). 2M)`

{SO2] =
`√(100)`

= 10 M

Answer 2

Answer: 10 M

Step-by-step explanation:

The Balanced Reaction is;

2SO2 + O2 <----> 2SO3

As Equilibrium Concentration of O2 is given to be;

[O2] = 2.0 M

And the Equilibrium Concentration of SO3 is given to be

[SO3] = 10 M

With the Equilibrium Constant as;

Kc = 0.50

Then we can solve for the concentration of SO2 as,

Kc = [SO3]2 / ( [SO2]2 × [O2] )

0.50 = (10)2 / ( [SO2]2 × 2.0 )

0.50( [SO2]2 × 2.0 )

[SO2]2 = 100

Solving for SO2 we get,

[SO2] = (100)1/2 = 10 M

Therefore, the Equilibrium Concentration of SO2 in the mixture is 10 M