Respuesta :
Answer:
[tex][H_2]_{eq}=0.0173M[/tex]
Explanation:
Hello,
In this case, for the given reaction, the law of mass action turns out:
[tex]Kc=\frac{[H_2O][CO]}{[H_2][CO_2]}[/tex]
In such a way, the reactants initial concentrations are:
[tex][CO_2]_0=[H_2]_0=\frac{0.300mol}{10.0L} =0.030M[/tex]
And considering the change [tex]x[/tex] due to the equilibrium, the law of mass action takes the following form:
[tex]Kc=\frac{(x)(x)}{(0.030M-x)(0.030M-x)} =\frac{x^2}{0.0009-0.06x+x^2} \\Kc(0.0009-0.06x+x^2)=x^2\\0.0004806-0.03204x+0.534x^2-x^2=0\\0.466x^2+0.03204x-0.0004806=0\\x_1=-0.0814M\\x_2=0.0127M[/tex]
The feasible answer is [tex]x=0.0127M[/tex]
Therefore, the hydrogen moles at equilibrium are:
[tex][H_2]_{eq}=0.030M-0.0127M=0.0173M[/tex]
Best regards.
The number of moles of H2 present at equilibrium is 0.04moles
Data;
- Kc = 0.534
- CO = 0.3mol
- H2O = 0.30mol
- V = 10L
- T = 700ºC
Equilibrium Concentration
let's find the concentration of the species in the reaction.
The concentration of CO is
[tex][CO] = \frac{0.3}{10} = 0.03M[/tex]
The concentration of H2O is
[tex][H_2O] = \frac{0.3}{10} 0.03M[/tex]
The equation of this reaction is
H2(g) + CO2(g) ⇌ H2O(g) + CO(g)
final -x -x 0.03 + x 0.03 + x
The Kc i.e the equilibrium constant of this reaction is
[tex]K_c = \frac{[product]}{[reactant]}\\[/tex]
let's substitute the values and solve.
[tex]K_c = \frac{product}{reactant} \\0.534 = \frac{(0.03+x)(0.03+x)}{x * x} \\x= 0.04M[/tex]
The number of moles of H2 present at equilibrium can be calculated as
[tex]0.04 = \frac{0.04}{1}[/tex]
The number of moles of H2 present at equilibrium is 0.04moles
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