Respuesta :
(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)Â
∴ 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 + 0.15Â Â Â 0Â Â Â Â Â Â Â Â 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)Â by solving for X
∴ X = 0.098Â
∴[H2] = X = 0.098 M
∴[CO] = 0.15 + X
Â
      = 0.15 + 0.098 = 0.248 M
∴[H2O] = 0.72 - X
       = 0.72 - 0.098
       = 0.622 M
when the reaction equation is:
C(s) + H2O(g)  ↔ H2(g) + CO(g)Â
∴ 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 + 0.15Â Â Â 0Â Â Â Â Â Â Â Â 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)Â by solving for X
∴ X = 0.098Â
∴[H2] = X = 0.098 M
∴[CO] = 0.15 + X
Â
      = 0.15 + 0.098 = 0.248 M
∴[H2O] = 0.72 - X
       = 0.72 - 0.098
       = 0.622 M
The concentration of [tex]\rm H_2[/tex] is 0.098 mol, CO is 0.248 mol, and [tex]\rm H_2O[/tex] is 0.622 mol after equilibrium.
The complete question is to find the concentration of CO, [tex]\rm H_2[/tex] and [tex]\rm H_2O[/tex]. Given, [tex]\rm k_c[/tex] = 0.0393.
The reaction will be:
[tex]\rm H_2O\;+\;C\;\rightarrow\;H_2\;+CO[/tex]
By the addition of excess [tex]\rm O_2[/tex],
[tex]\rm H_2\;+O_2\;\rightarrow\;H_2O[/tex]
At equillibrium,
[tex]\rm H_2[/tex] = 0.15 mol
CO= 0.15 mol
[tex]\rm H_2O[/tex] = 0.57 mol
Initially, [tex]\rm H_2O[/tex] = 0.57 + 0.15
CO = 0.15
After equillibrium,
[tex]\rm H_2O[/tex] = 0.57 + 0.15 - X
[tex]\rm CO[/tex] = 0.15 + X
[tex]\rm H_2[/tex] = +X
0.0393 = [tex]\rm \frac{0.15\;+\;X}{0.72\;-X}[/tex]
X = 0.098
Concentration after equillibrium :
[tex]\rm H_2[/tex] = 0.098 mol
CO= 0.15 + 0.098 = 0.248 mol
[tex]\rm H_2O[/tex] = 0.72 - 0.098 = 0.622 mol
For more information about equilibrium, refer the link:
https://brainly.com/question/1389022?referrer=searchResults