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Thae value of [tex]\sf K_c[/tex] is 4.24 at 800K for the reaction ,
[tex]\underline{\bf CO_(g)+H_2O_(g)\leftrightharpoons CO_2_(g)+H_2_(g)}[/tex]

Calculate equilibrium concentrations of CO_2 ,H_2 and H_2O at 800K,if only H_2O and CO are present initially at concentrations of 0.1M each .

[tex]\boxed{\bf Note:-}}[/tex]



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Respuesta :

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]

pstnonsonjoku

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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