A sample of pure PCl_5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl_5 was found to be 0.5 xx 10^(-1) "mol L"^(-1).If value of K_c is 8.3 xx 10^(-3), what are the concentrations of PCl_3and Cl_2 at equilibrium ? PCl_(5(g)) hArr PCl_(3(g)) +Cl_(2(g))

A sample of pure PCl_5 was introduced into an evacuated vessel at 473

1.	A sample of pure PCl_5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl_5 was found to be 0.5 xx 10^(-1) "mol L"^(-1).If value of K_c is 8.3 xx 10^(-3), what are the concentrations of PCl_3and Cl_2 at equilibrium ? PCl_(5(g)) hArr PCl_(3(g)) +Cl_(2(g))
Answer» SOLUTION :At equilibrium `PCl_5(0.5xx10^(-1))=0.05 "mol L"^(-1)` Product `[Cl_2]`= Product `[PCl_3]`= x M so, `{:("BALANCE reaction:",PCl_(5(g)) hArr , PCl_(3(g)) + ,Cl_(2(g))),("Equilibrium CONCENTRATION :",0.5xx10^(-1),x ,x):}``K_c=([PCl_3][Cl_2])/([PCl_5])` `therefore 8.3xx10^(-3) =((x)(x))/(0.5xx10^(-1))` `therefore x^2=8.3xx10^(-3)xx0.5xx10^(-1)= 4.15xx10^(-4)` `therefore x=sqrt(4.15xx10^(-4))=2.037xx10^(-2)`=0.02037 M `therefore x approx 0.02 M = [PCl_3]=[Cl_2]`