Question

Pcl5 => pcl3+cl2

The no of moles of cl2 produced will be ______? If one mole of pcl5 is heated 250c in vessel having a capacity of 10dm^3. At 250°c kc= 0.041

Answer 1

10dm^3. At 250°c kc= 0.041

Answer 2

Answer: 0.47 moles

Step-by-step explanation:

Initial moles of
`PCl_5` = 1 mole

Volume of container =
`10dm^3=10L`

Initial concentration of
`PCL_5=(moles)/(volume)=(1mole)/(10L)=0.1M`

The given balanced equilibrium reaction is,

`PCL_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)`

Initial conc. 0.1 M 0 0

At eqm. conc. (0.1-x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

`K_c=([PCl_3][Cl_2])/([PCl_5])`

`0.041=(x* x)/(0.1-x)`

Solving for x:

`x=0.047`

Thus concentration of
`Cl_2` produced is 0.047 M

`Molarity=(n)/(V_s)`

where,

n= moles of solute

`V_s` = volume of solution in L

`0.047M=(n)/(10L)`

`n=0.47moles`

Thus moles of
`Cl_2` produced is 0.47.