PCl5, PCl3 and Cl2 are at equilibrium at 500 K in a closed container and their concentrations are  and , respectively. The value of ${\mathrm{K}}_{\mathrm{c}}$ for the reaction

${\mathrm{PCl}}_{5}\left(\mathrm{g}\right)⇌{\mathrm{PCl}}_{3}\left(\mathrm{g}\right)+{\mathrm{Cl}}_{2}\left(\mathrm{g}\right)$ will be

(1)

(2) $1.8×{10}^{3}$

(3)

(4) $0.55×{10}^{4}$

Hint: ${\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{PCl}}_{3}\right]\left[{\mathrm{Cl}}_{2}\right]}{\left[{\mathrm{PCl}}_{5}\right]}$

Step 1:

For the reaction, ${\mathrm{PCl}}_{5}\left(\mathrm{g}\right)⇌{\mathrm{PCl}}_{3}\left(\mathrm{g}\right)+{\mathrm{Cl}}_{2}\left(\mathrm{g}\right)$
The given values are as follows:
At 500 K in a closed container,

Step 2:

${\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{PCl}}_{3}\right]\left[{\mathrm{Cl}}_{2}\right]}{\left[{\mathrm{PCl}}_{5}\right]}\phantom{\rule{0ex}{0ex}}=\frac{\left(1.2×{10}^{-3}\right)×\left(1.2×{10}^{-3}\right)}{\left(0.8×{10}^{-3}\right)}\phantom{\rule{0ex}{0ex}}=1.8×{10}^{-3}$