The first order rate constant for the decomposition of `N_(2)O_(5)` is `6.2 xx 10^(-4)s^(-1)`. The half-life period for this decomposition is

The first order rate constant for the decomposition of N_(2)O_(5) is 6.2 xx 10^(-2)s^(-1) . The half-life period for this decomposition is

The first order rate constant for dissociation of N_(2)O_(5) is 6.2xx10^(-4)s^(-1) . The half-lite period (in s) of this dissociation will be

The first order rate constant for the decomposition of N_(2)O_(5) is 6.2 xx 10^(-4) sec^(-1) . The t_(1//2) of decomposition is

The first order rate constant for the decomposition of CaCO_3 , at 700 K is 6.36 xx 10^(-3)s^(-1) and activation energy is 209 "kJ mol"^(-1) . Its rate constant (in s^(-1) ) at 600 K is "x" xx 10^(-6) . The value of x is ______ .(Nearest integer) [Given R=8.31 J K^(-1) "mol"^(-1) , log 6.36 xx 10^(-3)= -2.19, 10^(-4.79)= 1.62 xx10^(-5) ]

Findingf concentration and time : The rate constant for the decomposition of gasous N_(2)O_(5) at 55^(@)C is 1.7xx10^(-3)s^(-1) . If the initial concentration of N_(2)O_(5) after five half-lives ? How long will it take for the N_(2)O_(5) concentration to fall to 12.5% of its initial values ? Strategy : Because the unit of rate constant is time^(-1) , the decomposition of N_(2)O_(5) is a first order reaqction. To find [N_(2)O_(5)] after n half lives, multiply its initial concentration by (1//2)^(n) since [N_(2)O_(5)] drops by a factor of 2 during each successive haslf-life.