Answer:
1.2 atm is the pressure of [tex]PCl_5[/tex] after the system has reached equilibrium.
Explanation:
Initial pressure of the [tex]PCl_5[/tex] =2.74 atm
The value of the equilibrium constant = [tex]K_p=1.80[/tex]
[tex]PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)[/tex]
Initially
2.74 atm 0 0
At equilibrium
(2.74-p) p p
The expression of an equilibrium constant can be given as:
[tex]K_p=\frac{p_{PCl_3}\times p_{Cl_2}}{p_{PCl_5}}[/tex]
[tex]1.80=\frac{p\times p}{(2.74-p)}[/tex]
Solving for p:
p = 1.50 atm
Pressure of the [tex]PCl_5[/tex] at equilibrium :
= (2.74-p) atm = 2.74 - 1.50 atm = 1.2 atm
1.2 atm is the pressure of [tex]PCl_5[/tex] after the system has reached equilibrium.
