Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

2BrCl(g) ⇌ Br2(g)   +   Cl2(g)

For which Kc= 32 at 500 K.

If initially pure BrCl is present at a concentration of $$3.3 \times 10^{-3}molL^{-1}$$, what is its molar concentration in the mixture at equilibrium?

Asked by Abhisek | 11 months ago |  63

##### Solution :-

Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

2BrCl(g)    Br2(g)  +  Cl2(g)

Initial conc. $$3.3\times 10^{-3}$$    0       0

At equilibrium  $$3.3\times 10^{-3}-2x$$

Now, we can write,

So,,at equilibrium

Answered by Pragya Singh | 11 months ago

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