At 350 k, kc = 0.142 for the reaction 2 brcl (g) ⇀↽ br2 (g) + cl2 (g) an equilibrium mixture at this temperature contains equal concentrations of bromine and chlorine, 0.043 mol/l. what is the equilibrium concentration of brcl? answer in units of m.

Question

Chemistry

Draven Grant

29 January, 16:00

At 350 k, kc = 0.142 for the reaction 2 brcl (g) ⇀↽ br2 (g) + cl2 (g) an equilibrium mixture at this temperature contains equal concentrations of bromine and chlorine, 0.043 mol/l. what is the equilibrium concentration of brcl? answer in units of m.

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Angie Chambers · Answer 1

0.114 mol/l

The equilibrium equation will be:

Kc = ([Br2][Cl2]) / [BrCl]^2

The square factor for BrCl is due to the 2 coefficient on that side of the equation.

Now solve for BrCl, substitute the known values and calculate.

Kc = ([Br2][Cl2]) / [BrCl]^2

[BrCl]^2 * Kc = ([Br2][Cl2])

[BrCl]^2 = ([Br2][Cl2]) / Kc

[BrCl] = sqrt (([Br2][Cl2]) / Kc)

[BrCl] = sqrt (0.043 mol/l * 0.043 mol/l / 0.142)

[BrCl] = sqrt (0.001849 mol^2/l^2 / 0.142)

[BrCl] = sqrt (0.013021127 mol^2/l^2)

[BrCl] = 0.114110152 mol/l

Rounding to 3 significant figures gives 0.114 mol/l