The initial concentration of `N_(2)O_(5)` in the following first order reaction
`N_(2)O_(5(g))to2NO_(2(g))+(1)/(2)O_(2(g))` was `1.24xx10^(-2)`
Mol `L^(-1)` at 318 K.The concentration of `N_(2)O_(5)` after 60 minutes was `0.20xx10^(-2)` mol `L^(-1)`. calculate the rate constant of the reaction at 318 K.

Explanation

Given reaction is first order in it k=(?)
Initial t=0 time ,the concentration of `N_(2)O_(5)` is `1.24xx10^(-2) mol L^(-1)`
So.`t_(1)=0` and `[R]_(1)=1.24xx10^(-4)` mol `L^(-1)`
After 60 minutes the concentration of `N_(2)O_(5)=0.2xx10^(-2) mol L^(1)`
So, `t_(2)=60` minutes and `[R]_(2)=0.2xx10^(-2)` mol `L^(-1)`
For a first order reaction ,
`log([R]_(1))/([R]_(2))=(k(t_(2)-t_(2)))/(2.303)`
`therefore log (1.24xx10^(-2))/(0.2xx10^(-1))=k((60-0)"minute")/(2.303)`
`therefore log 6.2=k(60 "minute")/(2.303)`
`therefore 0.7924=k(60)/(2.303)`
`therefore k=(0.7924xx2.303)/(60 "minute") =0.0304 "minute"^(-1)`