Question

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2 in a 8.00 liter container forms an equilibrium mixture containing 0.309 mole of H2O and corresponding amounts of CO, H2, and CH4.

Answer 1

Kc = 3.90

Step-by-step explanation:

CO reacts with
`H_2` to form
`CH_4` and
`H_2O`. balanced reaction is:

`CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)`

No. of moles of CO = 0.800 mol

No. of moles of
`H_2` = 2.40 mol

Volume = 8.00 L

Concentration =
`(Moles)/(Volume\ in\ L)`

Concentration of CO =
`(0.800)/(8.00) = 0.100\ mol/L`

Concentration of
`H_2` =
`(2.40)/(8.00) = 0.300\ mol/L`

`CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)`

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium
`H_2O (x)` = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium
`H_2` = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium
`CH_4` = 0.0386 M

`Kc=([H_2O][CH_4])/([CO][H_2]^3)`

`Kc=(0.0386 * 0.0386)/((0.184)^3 * 0.0614) =3.90`