Question

CH4 and H2O were mixed in a 0.64 L reactor at 1800 K. Steam reforming took place according to: CH4 (g) + H2O (9) CO (g) + 3 H2 (9) The equilibrium constant for this reaction is K+0.28. At equilibrium, the reactor contained 0.36 mol of CO, 0.081 mol of Ha and 0.051 mol of CH. What is the concentration of H20 at equilibrium?

Answer 1

0.05110 mol/L is the concentration of water at equilibrium.

Step-by-step explanation:

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

An equilibrium constant of the given reaction =
`K_c=0.28`

Volume of the container = V = 0.64 L

Concentration =
`(Moles)/(Volume)`

Concentration of CO=
`[CO]=(0.36 mol)/(0.64 L)`

Concentration of
`H_2=[H_2]=(0.081 mol)/(0.64 L)`

Concentration of
`CH_4=[CH_4]=(0.051 mol)/(0.64 L)`

Concentration of
`H_2O=[H_2O]=?`

An expression of an equilibrium constant is given as:

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

`0.28=((0.36 mol)/(0.64 L)* ((0.081 mol)/(0.64 L))^3)/((0.051 mol)/(0.64 L)* [H_2O])`

`[H_2O]=0.05110 mol/L`

0.05110 mol/L is the concentration of water at equilibrium.