Question

4 moles of PCl5 dissociate at 760K in a 2 litre flask, PCl5 gives PCl3 and Cl2 at equilibrium. 0.8moles of Cl2 was present in the flask. Kc would be?

Answer 1

Answer: $K_c=0.533$

Explanation: $\text{Initial moles of }PCl_5=4$

Volume of Container = 2 liter

$Molarity=\frac{\text{Moles of solute}}{\text{volume of solution in liters}}$

Therefore, $Molarity=\frac{4}{2}=2 moles/litre$

                         $PCl_5 \rightarrow PCl_3 + Cl_2$

Initial molarity    2          0              0

Equilibrium      2-x          x              x

Given : Moles of $Cl_2$ at equilibrium = x = 0.8

$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

              =$\frac{(0.8)(0.8)}{{2-0.8}}$

              = 0.533

Answer 2

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2.0 mol of PCl5 were introduced in a vessel...

2.0 mol  of PCl5  were introduced in a vessel  of 5.0 L capacity  at a particular temperature . At   equilibrium PCl5 was found to be 35% dissociated into PCl3 and Cl2 . The value of Kc for the reaction is

2 years ago

Answers : (1)

                     PCl5     ---------->         PCl3         +        Cl2

t=0                 0.4                             0                        0

t=teq.           0.4-x                            x                         x

At eqm., (0.4 -x) = 0.65*0.4

Or

x = 0.4 – 0.4*0.65

   =0.4(1-0.65)

   =0.4*0.35

   = 0.14

 

Kc = 0.14*0.14/0.26  

     = 0.0753846