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

Related Questions

The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M.

The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2 . in which of these solutions precipitation will take place?

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K?

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 × 10–6).

What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide

What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10–18).