Question

At 298 K, Kc = 1.45 for the following reaction 2 BrCl (g) Br2(g) + Cl2(g) A reaction mixture was prepared with the following initial concentrations. [BrCl] = 0.0400 M, [Br2] = 0.0300 M and [Cl2] = 0.0300 M Calculate their equilibrium concentrations.

Answer 1

Answer:

[BrCl]=0.02934M

[Br2]=[Cl2]=0.03533M

Explanation:

First, consider the Kc definition:

$1.5=K_{c}=\frac{[Br_{2}][Cl_{2}]}{[BrCl]^{2}}$

It is necessary to define a new variable, 'x', as the amount of moles of Br2 that are produced. If 'x' moles of Br2 are produced, the moles of the compounds will be calculated as;

$n_{Br_{2}}=n_{{Br_{2}}^{0}} +x\\n_{Cl_{2}}=n_{{Cl_{2}}^{0}} +x\\n_{BrCl}=n_{{BrCl}}^{0}} -2x$

Where the zero superscript means the initial moles.

Dividing the last equations by the volume, which is constant because the reaction does not change the total moles number (2 moles of BrCl produce 2 moles, one of Cl2 and another of Br2), we have the molarity equations for all species:

$M_{Br_{2}}=M_{{Br_{2}}^{0}} +x\\M_{Cl_{2}}=\\M_{BrCl}=M_{{BrCl}}^{0}} -2x$

And now 'x' is a change in molarity.

Replacing these in the Kc equation we have:

$1.45=\frac{({M_{{Br_{2}}}^{0}}+x)({M_{{Cl_{2}}}^{0}}+x)}{({M_{{BrCl}}^{0}}-2x)^{2}}$

Where the only unknown is 'x'. So, let's solve the equation:

$1.45=\frac{(0.03+x)(0.03+x)}{(0.04-2x)^{2} } \\1.45(0.04-2x)^2=(0.03+x)^2\\1.45(0.0016-0.16x+4x^2)=0.0009+0.06x+x^2\\4.8x^2-0.292x+0.00142=0\\x_{1}=0.0555\\x_{2}=0.00533$

The result $x_{1}=0.0555$ lacks of sense because it will give a negative concentration for BrCl, so the result is $x_{2}=0.00533$.

Applying the result, the concentrations at equilibrium are:

$M_{Br_{2}}=M_{Cl_{2}}=0.03+0.00533=0.03533M\\M_{BrCl}=0.04-2*0.00533=0.02934M$

If you calculate Kc with this concentrations it will give 1.45 as a result.

Greets, I will be happy to solve any doubt you have.