Question

A mixture of water and graphite is heated to 600 K. When the system comes to equilibrium, it contains 0.13 mol of H2, 0.13 mol of CO, 0.43 mol of H2O, and some graphite. Some O2 is added to the system, and a spark is applied so that the H2 reacts com- pletely with the O2. Find the amount of CO in the flask when the system returns to equilibrium.

Answer 1

Step-by-step explanation:

Expression to calculate the value of for the given reaction is as follows.

And, it is given that

[CO] =
`[H_(2)]` = 0.13 mol

= 0.43 mol

Putting the given values into the above formula as follows.

`K_(c) = ([CO][H_(2)])/([H_(2)O])`

=
`(0.13 * 0.13)/(0.43)`

= 0.04

When additional amount of is added then all of has reacted.

So, new = 0 mols

new = 0.43 + 0.13 = 0.56 mols

The reaction equation is as follows.

`C + H_(2)O \rightleftharpoons CO + H_(2)`

Initial: - 0.56 0.13 0

Change: - -x +x +x

Equilibm.: - 0.56 - x 0.13 + x x

So,

0.04 =
`((0.13 + x)(x))/((0.56-x))`

`0.0224 - 0.04x = x^(2) + 0.13x`

`x^(2) + 0.17x - 0.0224` = 0

x = 0.087 mols

Therefore, the amount of [CO] at equilibrium is as follows.

0.13 + 0.087

= 0.217 mols

thus, we can conclude that the amount of CO in the flask when the system returns to equilibrium is 0.217 moles.