Answer:
The new partial pressures after equilibrium is reestablished:
[tex]PCl_3,p_1'=6.798 Torr[/tex]
[tex]Cl_2,p_2'=26.398 Torr[/tex]
[tex]PCl_5,p_3'=223.402 Torr[/tex]
Explanation:
[tex]PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g) [/tex]
At equilibrium before adding chlorine gas:
Partial pressure of the [tex]PCl_3=p_1=13.2 Torr[/tex]
Partial pressure of the [tex]Cl_2=p_2=13.2 Torr[/tex]
Partial pressure of the [tex]PCl_5=p_3=217.0 Torr[/tex]
The expression of an equilibrium constant is given by :
[tex]K_p=\frac{p_1}{p_1\times p_2}[/tex]
[tex]=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245[/tex]
At equilibrium after adding chlorine gas:
Partial pressure of the [tex]PCl_3=p_1'=13.2 Torr[/tex]
Partial pressure of the [tex]Cl_2=p_2'=?[/tex]
Partial pressure of the [tex]PCl_5=p_3'=217.0 Torr[/tex]
Total pressure of the system = P = 263.0 Torr
[tex]P=p_1'+p_2'+p_3'[/tex]
[tex]263.0Torr=13.2 Torr+p_2'+217.0 Torr[/tex]
[tex]p_2'=32.8 Torr[/tex]
[tex]PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g) [/tex]
At initail
(13.2) Torr (32.8) Torr (13.2) Torr
At equilbriumm
(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr
[tex]K_p=\frac{p_3'}{p_1'\times p_2'}[/tex]
[tex]1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}[/tex]
Solving for x;
x = 6.402 Torr
The new partial pressures after equilibrium is reestablished:
[tex]p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr[/tex]
[tex]p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr[/tex]
[tex]p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr[/tex]
