Question

4.0 moles of argon and 5.0 moles of PCl₅ are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The K_p for the reaction is ______. [Given: R = 0.082 L atm K^-1 mol^-1]

Options

• 2.25

• 6.24

• 12.13

• 15.24

Answer 1

4.0 moles of argon and 5.0 moles of PCl₅ are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The K_p for the reaction is 2.25.

Explanation:

Given PCl₅ = 5.0 mole

`"n"_(PCl_5)` = 5, n_Ar = 4

V = 100 L, T = 610 K.

PV = nRT

`"P"_"Total" = ("n"_"Total" "RT")/"V"`

`"P"_"Total" = (9 xx 0.082 xx 610)/100`

= 4.5 atm

So, `"P"_(PCl_5) = ("n"_("PCl"_5) xx "P"_"Total")/("n"_"Total")`

`= 5/9 xx 4.5`

= 2.5 atm

P_Ar = `("n"_"Ar" xx "P"_"Total")/("n"_"Total")`

`= 4/9 xx 4.5`

= 2 atm

\[\ce{PCl5 <=> PCl3 + Cl2}\]

At t = 0	2.5	0	0
At t = teq	2.5 - x	x	x

P_Total = 2.5 - x + x + x + P_Ar

6 = 2.5 + x + 2

x = 1.5

So, at eq^m, `"P"_("PCl"_5)` = 2.5 - 1.5 = 1

`"P"_("PCl"_5)` = 1.5, `"p"_("Cl"_2)` = 1.5

`"K"_"p" = (["P"_("PCl"_3)]["P"_("Cl"_2)])/["P"_("PCl"_5)]`

`"K"_"p" = (1.5 xx 1.5)/1`

`"K"_"p"` = 2.25