PCl 5 g → PCl 3 g + Cl 2 g In the above first order reaction the concentration of PCl 5 reduces from initial concentration 50 mol L - 1 to 10 mol L - 1 in 120 minutes at 300 K . The rate constant for the reaction at 300 K is x × 10 - 2 min - 1 . The value of x is __________. [Given log 5 = 0 . 6989 ]

Step-by-step solution

PCl 5 g I  order  300   K PCl 3 g + Cl 2 g t = 0 ,   [ PCl 5 ] =    50 M t = 120   min ,   [ PCl 5 ] =    10 M t = 120   min 10 M ⇒ K = 2 . 303 t log A 0 A t ⇒ K = 2 . 303 120 log 50 10 ⇒ K = 2 . 303 120 × 0 . 6989 = 0 . 013413   min - 1 = 1 . 3413 × 10 - 2   min - 1 1 . 34 ⇒ Nearest integer = 1