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Answers
Answer:
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.
Answer:
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