Answer:
[tex]p_{PCl_5}=223.407torr\\p_{PCl_3}=6.796torr\\p_{Cl_2}=26.396torr[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]PCl_3(g)+Cl_2(g)\leftrightarrow PCl_5(g)[/tex]
And the equilibrium constant at the reaction's temperature is:
[tex]Kp=\frac{p_{PCl_5}}{p_{PCl_3}p_{Cl_2}}=\frac{217.0torr}{(13.2torr)(13.2torr)} =1.2454[/tex]
Now, even when chlorine is added, such pressure equilibrium constant does not change, therefore, since the initial total pressure is:
[tex]p_{T0}=217.0torr+13.2torr+13.2torr=243.4torr[/tex]
The new pressures, due to the change [tex]x[/tex] owing to the chlorine's addition, turn out:
[tex]p_{PCl_5}=p_{PCl_5}^0+x\\p_{PCl_3}=p_{PCl_3}^0-x\\p_{Cl_2}=p_{Cl_2}^0+p_{Cl_2}^{added}-x[/tex]
Therefore, the added chlorine is:
[tex]p_{Cl_2}^{added}=263torr-243.4torr=19.6torr[/tex]
Thus, the new partial pressures are found via the law of mass action in terms of the change [tex]x[/tex] as follows:
[tex]1.2454=\frac{(217+x)}{(32.8-x)(13.2-x)}[/tex]
Solving for [tex]x[/tex] one obtains:
[tex]x=6.404torr[/tex]
Finally, the new partial pressures result:
[tex]p_{PCl_5}=217torr+6.404torr=223.407torr\\p_{PCl_3}=13.2torr-6.404torr=6.796torr\\p_{Cl_2}=32.8torr-6.404torr=26.396torr[/tex]
Best regards.
