Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium : 2BrCl_((g)) hArr Br_(2(g)) + Cl_(2(g)) for which K_c = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 xx10^(-3) "mol L"^(-1), what is its molar concentration in the mixture at equilibrium ?

Bromine monochloride, BrCl decomposes into bromine and chlorine and re

1.	Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium : 2BrCl_((g)) hArr Br_(2(g)) + Cl_(2(g)) for which K_c = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 xx10^(-3) "mol L"^(-1), what is its molar concentration in the mixture at equilibrium ?
Answer» Solution :In starting BRCL = `3.3xx10^(-3) "MOL L"^(-1)` Suppose x mol `L^(-1)` of decompose. So, EQUILIBRIUM `[Br_2]=[Cl_2]= x/2 "mol L"^(-1)` Because , according to stoichiometry, 2 mol BrCl `to` 1 mol `Br_2` + 1 mol `Cl_2` `therefore` x mol Cl `to x/2` mol `Br_2 + x/2` mol `Cl_2` `{:("Reaction :",2BrCl_((g)) hArr , Br_(2(g)) +,Cl_(2(g))),("Initial mol L"^(-1):, 3.3xx10^(-3), 0 ,0):}` `{:("Change in reaction mol L"^(-1):, -x, +x//2,x//2),("At equilibrium mol L"^(-1) :, (3.3xx10^(-3)-x),x//2,x//2):}` According to CHEMICAL equilibrium `K_c=([Br_2][Cl_2])/([BrCl]^2)` `therefore 32.0=((x/2)(x/2))/(3.3xx10^(-3)-x)^2=x^2/(4(0.0033-x)^2)` `therefore 128=(x/"0.003-x")^2` `therefore sqrt128=x/(0.0033-x)` `therefore` 11.31 (0.0033-x)=x `therefore` 0.03732 - 11.31x=x `therefore` 12.31x =0.03732 `therefore x=0.03732/12.31`=0.0030 At equilibrium [BrCl]=`3.3xx10^(-3)-0.003` `=3.3xx10^(-3)-3.0xx10^(-3)` `=0.3xx10^(-3)=3.0xx10^(-4) "mol L"^(-1)`