The half life for the reaction, `N_2O_(5(g)) rarr 2NO_(2(g)) + O_(2(g))` is 2.4 hr at `30^@`C. (a) Starting with 10 g, what is the mass of `N_2O_5` left after 9.6 hr? (b) How much time is required to reduce `5.0xx10^(10)` molecules of `N_2O_5` to `1.0xx10^8` molecules?

Explanation

For a first order reaction
`k=(0.693)/(t_(1//2))=(0.693)/(2.4xx60xx60)=8.02xx10^(-5)s^(-1)`
(a) Now using the expression
`k=(2.303)/(t)log(a)/(a-x)`
a=10 g,t=9.6 hr =`9.6xx60xx60` s, k=`8.02xx10^(-5)s^(-1)`
`8.02xx10^(-5)=(2.303)/(9.6xx60xx60)log^(10)/(a-x)`
`log(10)/(a-x)=(8.02xx10^(-5)xx9.6xx60xx60)/(2.303)`
`log(10)/(a-x)=(2.77171)/(2.303),(10)/(a-x)=15.9`
`underset("amount left")(a-x)=(10)/(15.9)=0.628` g
(b) `t=(2.303)/(k)log(a)/(a-x)`
`t=(2.303)/(8.02xx10^(-5))log((5.0xx10^(10))/(1.0xx10^8))=(2.303xx2.698)/(8.02xx10^(-5))`
=77503 s =21.52 hr