Sponsor Area

Equilibrium

Question
CBSEENCH11006432

3·00 mol of PCl5 kept in 1L closed reaction vessel was allowed to attain equilibrium at 380K. Calculate the composition of the mixture at equilibrium, Kc = 1·80.

Solution
Let x mol of PCl5 be dissociated.
The reaction is
                    PCl subscript 5 space space rightwards harpoon over leftwards harpoon space space space space space PCl subscript 3 space space plus Cl subscript 2
Initial conc.      3.0           0           0
At eqm.         3 - x           x           x
Applying the law of chemical equilibrium
                 straight K subscript straight c space equals space fraction numerator open square brackets PCl subscript 3 close square brackets space open square brackets Cl subscript 2 close square brackets over denominator open square brackets PCl subscript 5 close square brackets end fraction                ...(1)
Putting the values in expression (1), we have
               1.8 space equals space fraction numerator straight x squared over denominator 3 minus straight x end fraction
or space space straight x squared plus space 1.8 straight x space minus 5.4 space equals space 0
space space space space space space straight x space equals space fraction numerator open square brackets negative 1.8 space plus-or-minus square root of left parenthesis 1.8 right parenthesis squared minus 4 left parenthesis negative 5.4 right parenthesis end root close square brackets over denominator 2 end fraction
space space space space space straight x space equals space fraction numerator open square brackets negative 1.8 space plus-or-minus square root of 3.24 space plus-or-minus 21.6 end root close square brackets over denominator 2 end fraction
space space space space space straight x space equals space fraction numerator left square bracket 1.8 space plus-or-minus space 4.98 right square bracket over denominator 2 end fraction
space or space space straight x space equals space fraction numerator left square bracket negative 1.8 space plus 4.98 right square bracket over denominator 2 end fraction space equals 1.59
therefore space space space open square brackets PCl subscript 5 close square brackets space equals space 3.0 space space minus space straight x space equals space 3.0 space minus space 1.59 space equals space 1.41 space straight M
space space space space space space open square brackets PCl subscript 3 close square brackets space equals space open square brackets Cl subscript 2 close square brackets space equals space straight x space equals space 1.59 space straight M