Question

how much pcl5 must be added to a one litre vessel at 250c in order to obtain a concentration of 0.1 mole of CL2?kc for PCL5<-->PCL3+CL2 is 0.0414 mol/litre

Answer 1

Answer : Moles of $PCl_{5}$ added = 0.3416 moles

Solution : Given,

Concentration of $Cl_{2}$ = 0.1 mole

$K_{C}$ = 0.0414 mole/L

The given reaction is,

                                     $PCl_{5}$      ⇄        $PCl_{3}+Cl_{2}$

initial concentration       C                            0              0

final concentration     C(1-x)                         Cx           Cx

where,

c = concentration

x = degree of dissociation

Concentration of $Cl_{2}$ = 0.1 = Cx         .................(1)

From the given reaction, the expression for $K_{C}$ is,

$K_{C}=\frac{[PCl_{3}]\times [Cl_{2}]}{[PCl_{5}]}$ = $\frac{Cx\times Cx}{C(1-x)}$ = $\frac{Cx^{2} }{(1-x)}$

now put all the given values in this relation, we get

0.0414 - 0.0414 x = 0.1 x

by rearranging the terms, we get the value of 'x'

x = 0.2927

Now put the value of 'x' in equation (1), we get value of concentration

Cx = 0.1

C × 0.2927 = 0.1

C = 0.3416 mol/L

So, moles of $PCl_{5}$ added = 0.3416 moles

Answer 2

Answer:check out pic

Explanation: