Respuesta :
Answer:
The new concentration of [tex]PCl_5[/tex] will be 0.9 M.
Explanation:
[tex]PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)[/tex]
The equilibrium concentration of all species:
[tex][PCl_5]=0.40 M[/tex]
[tex][PCl_3]=[Cl_2]=0.20 M[/tex]
The equilibrium constant's expression can be written as:
[tex]K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]
[tex]K_c=\frac{0.20 M\times 0.20 M}{0.40 M}=0.1[/tex]
On halving the volume of the container, the concentration will get doubled;
[tex]PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)[/tex]
0.80 M 0.40M 0.40 M
[tex]Q_c=\frac{0.40 M\times 0.40 M}{0.80 M}=0.2[/tex]
[tex]Q_c>K_c[/tex]
Reaction will go backward.
[tex]PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)[/tex]
Initially
0.80 M 0.40M 0.40 M
After reestablishment of an equilibrium
(0.80+x) M (0.40-x)M (0.40-x) M
So, the equilibrium expression for above reaction can be written as :
[tex]K_c=\frac{(0.40-x) M\times (0.40-x) M}{(0.80+x) M}[/tex]
[tex]0.1=\frac{(0.40-x) M\times (0.40-x) M}{(0.80+x) M}[/tex]
x = 0.1 M
The new concentration of [tex]PCl_5[/tex] will be:
[tex][PCl_5]=(0.80+x) M=(0.80+0.1) M = 0.90[/tex]
