Question

$PCl_5$ dissociates as
$PCl_5(g) ⇌ PCl_3(g)+ Cl_2(g)$
$5$ moles of $PCl_5$ are placed in a $200$ litre vessel which contains $2$ moles of $N_2$ and is maintained at $600\ K$. The equilibrium pressure is $2.46$ atm. The equilibrium constant Kp for the dissociation of $PCl_5$ is ______ $× 10^{–3}$. (nearest integer)
(Given: $R = 0.082$ L atm K–1 mol–1; Assume ideal gas behaviour)

Answer 1

$PCl_5(g) ⇌ PCl_3(g)+ Cl_2(g)$
t = 0 5 0 0
t = Equilibrium 5-x x x

Number of moles of $N_2= 2$
Equilibrium pressure = $2.46$ atm
$P_{eq} = \frac {(7+x)× 0.082×600)}{200}$
$P_{eq}= 2.46$
On solving, $x = 3$
Hence,
$K_p = \frac {(\frac {3P}{10})(\frac {3P}{10})}{\frac {2P}{10}}$

$K_p= \frac {9×2.46}{20}$
$K_p= 1107 × 10^{-3}$ atm

So, the correct answer is $1107$.