Question

For the reaction given below at 700°c, kc = 0.534. h2(g + co2(g h2o(g + co(g calculate the number of moles of h2 that are present at equilibrium if a mixture of 0.229 moles of co and 0.229 moles of h2o is heated to 700°c in a 11.0-l container.

Answer 1

Answer : The concentration of
`H_2` at equilibrium is 0.012 M

Solution :

First we have to calculate the concentration of
`CO` and
`H_2O`.

Concentration of
`CO` =
`(Moles)/(Volume)=(0.229)/(11)=0.0208M`

Concentration of
`H_2O` =
`(Moles)/(Volume)=(0.229)/(11)=0.0208M`

The given equilibrium reaction is,

`H_2O(g)+CO(g)\rightleftharpoons H_2(g)+CO_2(g)`

Initially conc. 0.0208 0.0208 0 0

At equilibrium. (0.0208-x) (0.0208-x) x x

The expression of
`K_c` will be,

`K_c=([H_2][CO_2])/([H_2O][CO])`

For given reaction the value of
`K_c` will be,
`(1)/(0.534)` (for reverse reaction).

`(1)/(0.534)=((x)(x))/((0.0208-x)* (0.0208-x))`

`(1)/(0.534)=((x)^2)/((0.0208-x)^2)`

`(1)/(0.534)=(((x))/((0.0208-x)))^2`

`\sqrt{(1)/(0.534)}=((x))/((0.0208-x))`

`1.368=((x))/((0.0208-x))`

By solving the term x, we get

`x=0.012`

Thus, the concentration of
`H_2` at equilibrium = x = 0.012 M

Answer 2

The balanced chemical reaction is written as:

H2 + CO2 <---> H2O + CO

To determine the moles of H2 at equilibrium, we use the ICE table as follows:

Initial concentration = 0.229/11.0 = 0.021 M

H2O CO H2 CO2
I 0.021 0.021 0 0
C -x -x x x
--------------------------------------------------------
E 0.021- x 0.021-x x x

Kc = 0.534 = (H2)(CO2) / (H2O)(CO)
0.534 = (x)(x) / (0.021-x) (0.021-x)
x = 8.87x10^-3 M

mole H2 = 8.87x10^-3 M (11.0 L) = 0.098 mol H2