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Sagot :
For the reaction,
[tex]CO(g) + H_2 O(g) = CO_2(g) + H_2(g)[/tex]
Initial concentration:
0.1M , 0.1M , 0 , 0
Let 'x' mole per litre of each of theproduct be formed.
At equilibrium:
(0.1 – x)M , (0.1 – x)M , xM , xM
where x is the amount of Carbon dioxide and Hydrogen, at equilibrium.
Hence, equilibrium constant can be written as,
[tex]K_c = \frac{x²}{(0.1 – x)²} = 4.24[/tex]
→ x² = 4.24 (0.01 + x² – 0.2x)
→ x² = 0.0424 + 4.24 x² - 0.848x
→ 3.24x² - 0.848x + 0.0424 = 0
a = 3.24, b = -0.848, c = 0.0424
(for quadratic equation ax² + bx+c=0)
[tex]x = \frac{( - b \: ± \: \sqrt{ {b}^{2} - 4ac) } }{2a} [/tex]
[tex] = > x = \frac{ - ( - 0.848 \: ± \: \sqrt{( - 0.848)^{2} - 4(3.24)(0.0424) } }{2 \times 3.24} [/tex]
[tex] = > x = \frac{ - 0.848±0.4118}{6.48} [/tex]
[tex]x_1 = \frac{0.848 - 0.4118}{6.48} = 0.067[/tex]
[tex]x_2 = \frac{0.848 + 0.4118}{6.48} = 0.194[/tex]
Here, the value 0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration.
∴ The equilibrium concentrations are :-
[tex][CO_2] [H_2] = x = 0.067M[/tex]
[tex][CO] [H_2 O] = 0.1 - 0.067 = 0.033M[/tex]
The equilibrium concentrations of each of the species can be obtained by the use of the ICE table.
What is Kc?
Kc refers to the value of the equilibrium constant obtained using the concentrations of the reactants.
Now we have to set up the ICE table as shown below;
CO + H2O ⇄ CO2 + H2
I 0.1 0.1 0 0
C - x - x +x +x
E 0.1 - x 0.1 - x x x
Kc = [CO2] [H2]/[CO] [H2O]
4.24 = x^2/(0.1 - x)^2
4.24(0.1 - x)^2 = x^2
0.042 - 0.848x + 4.24x^2 = x^2
x^2 - 4.24x^2 + 0.848x - 0.042 = 0
-3.24x^2 + 0.848x - 0.042 = 0
since x can not be greater then the initial concentration, x=0.066M
Final concentration of each specie;
CO = 0.1 M - 0.066M = 0.034 M
H2O = 0.1 M - 0.066M = 0.034 M
CO2 = 0.066M
H2 = 0.066M
Learn more about concentrations:https://brainly.com/question/15289741?
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