-->

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