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The following reaction was performed in a sealed vessel at 791 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.10M and [I2]=2.50M . The equilibrium concentration of I2 is 0.0800 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Respuesta :

ajeigbeibraheem

Answer:

4.31 × 10²

Explanation:

Equation of the reaction;

[tex]H_{2(g)} + I_{2(g)}[/tex]     ⇌     [tex]2HI_{(g)[/tex]

The ICE Table is shown as follows:

                            [tex]H_{2(g)}[/tex]         [tex]+[/tex]         [tex]I_{2(g)}[/tex]        ⇌     [tex]2HI_{(g)[/tex]

Initial                    3.10                     2.50                  0      

Change                 - x                       -x                     + 2x      

Equilibrium        (3.10 - x)                0.0800              2x

From [tex]I_{2(g)}[/tex]   ;

We can see that 2.50 - x = 0.0800

So; we can solve for x;

x = 2.50 - 0.0800

x = 2.42

[tex]H_{2(g)}[/tex]  which = (3.10 -x) will be :

= 3.10 - 2.42

= 0.68

[tex]2HI_{(g)[/tex] = 2x

= 2 (2.42)

= 4.84

[tex]K_c = \frac{[HI]^2}{[H_2][I_2]}[/tex]

[tex]K_c = \frac{(4.84)^2}{(0.68)(0.0800)}[/tex]

[tex]K_c =\frac{23.4256}{0.0544}[/tex]

[tex]K_c =[/tex] 430.62

[tex]K_c[/tex] ≅ 431

[tex]K_c[/tex] = 4.31 × 10²

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