Respuesta :
Answer : The equilibrium concentration of [tex]PCl_5[/tex] is, 0.16 M
Explanation :
First we have to calculate the concentration of [tex]PCl_5\text{ and }PCl_3[/tex]
[tex]\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution}}=\frac{0.20mol}{2.00L}=0.4M[/tex]
and,
[tex]\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}=\frac{1.0mol}{2.00L}=2.0M[/tex]
The given chemical reaction is:
[tex]PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)[/tex]
Initial conc. 0.4 2.0 0
At eqm. (0.4-x) (2.0+x) x
The expression for equilibrium constant is:
[tex]K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]
Now put all the given values in this expression, we get:
[tex]3.33=\frac{(2.0+x)\times (x)}{(0.4-x)}[/tex]
x = -5.57 and x = 0.24
We are neglecting the value of x = -5.57 because equilibrium concentration can not be more than initial concentration.
Thus, the value of x = 0.24
The equilibrium concentration of [tex]PCl_5[/tex] = (0.4-x) = (0.4-0.24) = 0.16 M
Therefore, the equilibrium concentration of [tex]PCl_5[/tex] is, 0.16 M
