PCl _ 5 (g) → PCl _ 3 (g) + Cl _ 2 (g) In the above first order...

${PCl}_{5\left(g\right)}\dfrac{I\text{order}}{300K}{PCl}_{3\left(g\right)} + {Cl}_{2\left(g\right)}$

$t = 0, [ {PCl}_{5} ] = 50Mt = 120min, [ {PCl}_{5} ] = 10M$

$t = 120min10M$

$⇒ K = \dfrac{2 . 303}{t}log\dfrac{\left[{A}_{0}\right]}{\left[{A}_{t}\right]}$

$⇒ K = \dfrac{2 . 303}{120}log\dfrac{50}{10}$

$⇒ K = \dfrac{2 . 303}{120} \times 0 . 6989 = 0 . 013413\min^{- 1}$

$= 1 . 3413 \times {10}^{- 2}\min^{- 1}$

$1 . 34 ⇒$ Nearest integer $= 1$