Question

At 1323°C the equilibrium constant for the reaction :
Br2(g) ------> 2Br(g); Kc=10^(-3)
A 0.2L vessel containing an equilibrium mixture of gases has 0.24g Br2(g) in it. What is the mass of Br(g)
in the vessel?

Answer 1

moles Br2 = 0.245 g/ 159.808 g/mol=0.00153

[Br2]= 0.00153 / 0.190 L=0.00807 M

1.04 x 10^-3 = [Br]^2/ [Br2] = [Br]^2 / 0.00807

[Br]= 8.93 x 10^-6 M

moles Br = 8.93 x 10^-6 x 0.190=1.59 x 10^-6

mass Br = 1.59 x 10^-6 x 79.904 g/mol=0.000127 g

Answer 2

At 1323°C the equilibrium constant for the reaction :

Br2(g) ------> 2Br(g); Kc=10^(-3)

A 0.2L vessel containing an equilibrium mixture of gases has 0.24g Br2(g) in it. The mass of Br(g) in the vessel would be 0.0439 grams.

Given:

T= 1323 C = 1596 K

Br2(g) ------> 2Br(g); Kc = 10^(-3)

Volume of vessel = 0.2L

mass of gas = 0.24 grams.

To find:

mass of Br(g) in the vessel

Solution:

First, let us find the number of moles of Br2 in the vessel.

That is, nBr2 = mBr2/MBr2

                     = 0.24/159

                    = 0.00151 moles

now, the Concentration of Br2 in the vessel

i.e., [Br2] = nBr2/V

               = 0.00151 moles/ 0.2 L

               = 0.00755 M

To find the concentration of Br

kc = [Br]^2/[Br2]

now, [Br]^2 = kc * [Br2]

                   = 10^(-3) * 0.00755

        [Br] ^2= 0.00000755

        [Br] = 0.00275 M

Now, the mass of Br in the vessel would be

n Br = [Br] * V

       = 0.00275 M * 0.2

       = 0.000549 moles

m Br = n Br * M Br

        = 0.000549 moles * 79.9

        = 0.0439 grams

The mass of Br(g) in the vessel would be 0.0439 grams.

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