Question
Asked Oct 21, 2019
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The following reaction has Kc = 0.145 at 298 K in carbon tetrachloride solution:
2BrCl(soln)⇌Br2(soln)+Cl2(soln)
If the concentrations are 

[BrCl]=4.3×10−2M, 

[Br2]=3.8×10−2M

[Cl2]=3.8×10−2M,

Determine the equilibrium concentrations of Br2, BrCl, and Cl2.

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Expert Answer

Step 1
The equilibrium constant is given as 0.145. The equation for the given reaction and
concentrations are given below:
2BrCl Br Cl
BrCl 4.3x10 M
[Br3.8x10 M
[CI,] 3.8x10 M
help_outline

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The equilibrium constant is given as 0.145. The equation for the given reaction and concentrations are given below: 2BrCl Br Cl BrCl 4.3x10 M [Br3.8x10 M [CI,] 3.8x10 M

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Step 2
The ICE table for the reaction is given below
2BIC1 Br+ Cl2
2 BrCl
Br
Cl
Initial
0.043
0.038
0.038
Change
2x
-X
-X
Equilibirium
0.0432x
0.038-x
0.038-X
+
help_outline

Image Transcriptionclose

The ICE table for the reaction is given below 2BIC1 Br+ Cl2 2 BrCl Br Cl Initial 0.043 0.038 0.038 Change 2x -X -X Equilibirium 0.0432x 0.038-x 0.038-X +

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Step 3
Substituting the values of equilibrium concentrations in the equilibrium constant
equation
Br,C
К.
BICI
(0.038-х) (0.038- х)
(0.043+2x)
0.145
x =0.012
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Substituting the values of equilibrium concentrations in the equilibrium constant equation Br,C К. BICI (0.038-х) (0.038- х) (0.043+2x) 0.145 x =0.012

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Equilibrium Concepts

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