`PCl _(5) (g) to 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)` m 120 minutes at 300K. The rate constant for the reaction at 300 K is `x xx 10 ^(-2) "min"^(-1).` The value of x is ______________[Given log `5 = 0.69989`]

In a first order reaction the concentration of the reactant is reduced from 0.6 "mol" L^(-1) to 0.2 "mol" L^(-1) in 5 minutes the rate constant of the reaction is

N_(2)O_(5(g)) rarr 2NO_(2(g)) + (1)/(2)O_(2(g)) . In the above first order reaction the initial concentration of N_(2)O_(5) is 2.40 xx 10^(-2) mol L^(-1) at 318 K. The concentration of N_(2)O_(5) after 1 hour was 1.60 xx 10^(-2) mol L^(-1) . The rate constant of the reaction at 318K is ______ xx 10^(-3) "min"^(-1) . (Nearest integer) [Given: log3= 0.477, log 5= 0.699]

The rate constant of a first order reaction is 2.0 xx 10^(-5)s^(-1) and the initial concentration is 0.10 mol L^(-1) . The initial rate is

The system PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) attains equilibrium. If the equilibrium concentration of PCl_(3)(g) is doubled, the concentration of Cl_(2)(g) would become

PCl_5 to PCl_3 + Cl_2 the above 1st order reaction has initial moles as 50 and final moles as 10 in 120seconds. Find the rate constant.

The initial rate of a second order reaction is 4xx10^(-4) "mol" L^(-1) s^(-1) and the initial concentration of the reaction substance is 0.20 "mol" L^(-1) the value of rate constant is

If t_(0.5) of a first order reaction, PCl_(5) to PCl_(3) + Cl_(2) is 10 minutes then the time in which the concentration of PCl_(5) reduces to 10% of the original is