Question

4 moles of PCl5 dissociate at 760 k in a 2 litre flask at equilibrium .0.8 mole of Cl2 was present in the flask . The equilibrium constant would be.?

Answer 1

Answer:The equilibrium constant will be 0.1.

Explanation:

               $PCl_5\rightleftharpoons PCl_3+Cl_2$

at t=0        4 mol                  0           0

at eq'm    (4-0.8)mol         0.8 mol   0.8 mol

Volume of the container = 2 L

$[\text{concentration of the component}]=\frac{\text{moles of the component}}{\text{volume in liters}}$

$[PCl_5]=\frac{4-0.8 moles}{2}=1.6mol/L$

$[PCl_3]=\frac{0.8 moles}{2}=0.4mol/L$

$[Cl_2]=\frac{0.8 moles}{2}=0.4mol/L$

Expression of $K_c$ is written as:

$K_c=\frac{[PCl_3]\times [Cl_2]}{[PCl_5]}=\frac{0.4\times 0.4}{1.6}=0.1$

The equilibrium constant will be 0.1.

Answer 2

Explanation:

PCl_5\rightleftharpoons PCl_3+Cl_2PCl

5

⇌PCl

3

+Cl

2

at t=0 4 mol 0 0

at eq'm (4-0.8)mol 0.8 mol 0.8 mol

Volume of the container = 2 L

[\text{concentration of the component}]=\frac{\text{moles of the component}}{\text{volume in liters}}[concentration of the component]=

volume in liters

moles of the component

[PCl_5]=\frac{4-0.8 moles}{2}=1.6mol/L[PCl

5

]=

2

4−0.8moles

=1.6mol/L

[PCl_3]=\frac{0.8 moles}{2}=0.4mol/L[PCl

3

]=

2

0.8moles

=0.4mol/L

[Cl_2]=\frac{0.8 moles}{2}=0.4mol/L[Cl

2

]=

2

0.8moles

=0.4mol/L

Expression of K_cK

c

is written as:

K_c=\frac{[PCl_3]\times [Cl_2]}{[PCl_5]}=\frac{0.4\times 0.4}{1.6}=0.1K

c

=

[PCl

5

]

[PCl

3

]×[Cl

2

]

=

1.6

0.4×0.4

=0.1

The equilibrium constant will be 0.1.