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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.

Respuesta :

Answer:

Kc = 3.90

Explanation:

CO reacts with [tex]H_2[/tex] to form [tex]CH_4[/tex] and [tex]H_2O[/tex]. balanced reaction is:

[tex]CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)[/tex]

No. of moles of CO = 0.800 mol

No. of moles of [tex]H_2[/tex] = 2.40 mol

Volume = 8.00 L

Concentration = [tex]\frac{Moles}{Volume\ in\ L}[/tex]

Concentration of CO = [tex]\frac{0.800}{8.00} = 0.100\ mol/L[/tex]

Concentration of [tex]H_2[/tex] = [tex]\frac{2.40}{8.00} = 0.300\ mol/L[/tex]

                 [tex]CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)[/tex]

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

It is given that,

at equilibrium [tex]H_2O (x)[/tex] = 0.309/8.00 = 0.0386 M

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

At equilibrium [tex]H_2[/tex] = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium [tex]CH_4[/tex] = 0.0386 M

[tex]Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}[/tex]

[tex]Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90[/tex]

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