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The reaction of bromine gas with chlorine gas, shown here, has a Keq value of 7.20 at 200°C. If a closed vessel was charged with the two reactants, each at an initial concentration of 0.200 M, but with no initial concentration of BrCl, what would be the equilibrium concentration of Br2, Cl2 and BrCl(g)?
Br2(g) + Cl2(g) ↔ 2BrCl(g) K = 7.20

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Answer:

[tex][Cl_2]=[Br_2]=0.856M[/tex]

[tex][BrCl]=0.229M[/tex]

Explanation:

Hello there!

In this case, for this chemical equilibrium problem, it turns out possible for us to solve for the equilibrium concentrations by firstly setting up the equilibrium expression:

[tex]Keq=\frac{[BrCl]^2}{[Cl_2][Br_2]}[/tex]

Thus, by plugging in an ICE chart, in terms of x (reaction extent), we can write:

[tex]7.20=\frac{(2x)^2}{(0.200-x)^2}[/tex]

And could be solved for x as follows:

[tex]\sqrt{7.20} =\sqrt{\frac{(2x)^2}{(0.200-x)^2} } \\\\2.68=\frac{2x}{0.200-x} \\\\x=0.1146M[/tex]

Therefore, the equilibrium concentrations turn out to be:

[tex][Cl_2]=[Br_2]=0.200M-0.1146M=0.856M[/tex]

[tex][BrCl]=2*0.1146M=0.229M[/tex]

Regards!

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