Question

In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous CH4 and H2O in a 0.32-L flask at 1200 K. At equilibrium the flask contains 0.26 mol of CO, 0.091 mol of H2, and 0.041 mol of CH4. What is the [H2O] at equilibrium

Answer 1

Answer: The
`[H_(2)O]` at equilibrium is 0.561 M.

Step-by-step explanation:

The given data is as follows.

Volume of flask = 0.32 L,

No. of moles of
`CH_(4)` = 0.041 mol,

No, of moles of CO = 0.26 mol, No. of moles of
`H_(2)` = 0.091 mol

Equilibrium constant, K = 0.26

Balanced chemical equation for this reaction is as follows.

`CH_(4)(g) + H_(2)O(g) \rightleftharpoons CO(g) + 3H_(2)(g)`

Hence,

K =
`([CO][H_(2)]^(3))/([CH_(4)][H_(2)O])` ...... (1)

First, we will calculate the molarity or concentration of given species as follows.

`CH_(4) = (0.041)/(0.32)` = 0.128 M,

`CO = (0.26)/(0.32)` = 0.8125 M,

`H_(2) = (0.091)/(0.32)` = 0.284 M

Therefore, using expression (1) we will calculate the
`[H_(2)O]` as follows.

K =
`([CO][H_(2)]^(3))/([CH_(4)][H_(2)O])`

or,
`[H_(2)O] = ([CO][H_(2)]^(3))/([CH_(4)] * K)`

=
`(0.8125 * (0.284)^(3))/(0.128 * 0.26)`

=
`(0.0187)/(0.0333)`

= 0.561 M

Thus, we can conclude that the
`[H_(2)O]` at equilibrium is 0.561 M.