The equilibrium composition for the reaction is
`{:(PCl_(3),+,Cl_(2),hArr,PCl_(5)),(0.20,,0.10,,0.40 mol L^(-1)):}`
What will be the equilibrium concentration of `PCl_(5)` on adding `0.10 mol` of `Cl_(2)` at the same temperature?

Answer

`0*45 "mol "L^(-1)`

Explanation

`K_(c) = (0*40)/(0*20 xx 0*10)=20`
`" New initial conc. of " Cl_(2) = 0*10+0*10 = 0*20 " mol "L^(-1)`
`" New initial conc. of " PCl_(3) and PCl_(5) " remain the same " `
Supposing x mole of `PCl_(3)` reacts , the new equilibrium concs. will be
`[PCl_(3)] = 0*20 - x , [C_(2)]= 0*20 - x and [ PCl_(5)] = 0*40 + x`
Putting the values in `K_(c) = ([PCl_(5)])/([PCl_(3)] [ Cl_(2)])`.
` ((0*40 + x))/((0*20 - x)(0*20-x))=20 or (0*40 +x) = 20 (0*4 + x^(2)-0*40 x)or 20 x^(2) - 9 x + 0*40 = 0`
or ` x = (-b pm sqrt(b^(2) - 4 ac))/(2a)=(9pm sqrt(81-4 xx 20 xx 0*4))/(2 xx 20)=(9 pm 70)/40`
=` 0*4 or 0*05 (0*4 " is impossible because x cannot be greater than " 0*2) `
Hence, `[PCl_(5)] = 0*40 + 0*05 = 0*45 " mol " L^(-1)`