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The equilibrium constant, Kc, for the following reaction is 55.6 at 698 K:
H2(g) + I2(g) 2HI(g)
Calculate the equilibrium concentrations of reactants and product when 0.234 moles of H2 and 0.234 moles of I2 are introduced into a 1.00 L vessel at 698 K.
[H2] = M
[I2] = M
[HI] = M

Respuesta :

sebassandin

Answer:

[tex][I_2]=[H_2]=0.369M[/tex]

[tex][HI]=0.0495M[/tex]

Explanation:

Hello,

In this case, considering the described reaction at equilibrium:

[tex]H_2(g) + I_2(g)\rightleftharpoons 2HI(g)[/tex]

We can write the law of mass action for the equilibrium concetrations:

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

That in terms of the change [tex]x[/tex] (considering the ICE procedure) is written as:

[tex]55.6=\frac{(2x)^2}{([I_2]_0-x)([H_2]_0-x)}[/tex]

Whereas the initial concentration of iodine and hydrogen are both 0.234 M (0.234 moles in 1 liter), therefore, we have:

[tex]55.6=\frac{(2x)^2}{(0.234-x)^2}[/tex]

That can be solved as shown below:

[tex]\sqrt{55.6} =\sqrt{\frac{(2x)^2}{(0.234-x)^2}} \\\\7.457=\frac{2x}{0.234-x}\\\\7.457(0.234-x)=2x\\\\1.745-7.457x=2x\\\\1.745=x(7.457+2)\\\\x=\frac{1.745}{9.457}\\ \\x=0.185M[/tex]

Thereby, equilibrium concentrations are:

[tex][I_2]=[H_2]=2*0.185M=0.369M[/tex]

[tex][HI]=0.234-0.185=0.0495M[/tex]

Regards.

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