Question

A mixture of water and graphite is heated to 600 k in a 1 l container. when the system comes to equilibrium it contains 0.15 mol of h2, 0.15 mol of co, 0.57 mol of h2o, and some graphite. some o2 is added to the system and a spark is applied so that the h2 reacts completely with the o2.

Answer 1

(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???

when the reaction equation is:

C(s) + H2O(g)  ↔ H2(g) + CO(g) 

∴ Kc = [H2] [CO] / [H2O]

and we have Kc = 0.0393 (given missing in the question)

when the O2 is added so, the reaction will be:

2H2(g) + O2(g) → 2H2O(g)

that means that 0.15 mol H2 gives 0.15 mol of H2O

∴ by using ICE table:

            [H2O]          [H2]        [CO]

initial 0.57 + 0.15      0               0.15

change  -X                +X              +X

Equ   (0.72-X)             X            (0.15+X)

by substitution:

0.0393 = X (0.15+X) / (0.72-X)  by solving for X

∴ X = 0.098 

∴[H2] = X = 0.098 M

∴[CO] = 0.15 + X
 
           = 0.15 + 0.098 = 0.248 M

∴[H2O] = 0.72 - X

             = 0.72 - 0.098

             = 0.622 M

Answer 2

The concentration of $\rm H_2$ is 0.098 mol, CO is 0.248 mol, and $\rm H_2O$ is 0.622 mol after equilibrium.

The complete question is to find the concentration of CO, $\rm H_2$ and $\rm H_2O$. Given, $\rm k_c$ = 0.0393.

The reaction will be:

$\rm H_2O\;+\;C\;\rightarrow\;H_2\;+CO$

By the addition of excess $\rm O_2$,

$\rm H_2\;+O_2\;\rightarrow\;H_2O$

At equillibrium,

$\rm H_2$ = 0.15 mol

CO= 0.15 mol

$\rm H_2O$ = 0.57 mol

Initially, $\rm H_2O$ = 0.57 + 0.15

CO = 0.15

After equillibrium,

$\rm H_2O$ = 0.57 + 0.15 - X

$\rm CO$ = 0.15 + X

$\rm H_2$ = +X

0.0393 = $\rm \frac{0.15\;+\;X}{0.72\;-X}$

X = 0.098

Concentration after equillibrium :

$\rm H_2$ = 0.098 mol

CO= 0.15 + 0.098 = 0.248 mol

$\rm H_2O$ = 0.72 - 0.098 = 0.622 mol

For more information about equilibrium, refer the link:

https://brainly.com/question/1389022?referrer=searchResults