The half life for the reaction `N_(2)O_(5) hArr 2NO_(2)+1/2 O_(2)` in `24 hr` at `30^(@)C`. Starting with `10 g` of `N_(2)O_(5)` how many grams of `N_(2)O_(5)` will remain after a period of `96` hours ?

Options

A. 1.25 g
B. 0.63 g
C. 1.77 g
D. 0.5 g

Explanation

`k= 0.93/(t_1//2) and k = 2.303/t log. (a)/(a-x)`
`:. 0.693/ 24 hr^(-1)- 2.303/tlog. (a)/(a-x)`
`:. 0.693/24 hr^(-1)= 2.303/96 log. (a)/(a-x)`
`or log. (10)/(a-x)=(0.693 xx 96)/(24 xx 2.303)= 1.2036`
`or 10/(a-x)= (0.693 xx 96)/(24 xx 2.303)= 1.2036`
`or 10/(a-x) = AL (1.2036) = 15.98 = 16`
`:. a -x =10/16= 0.625`
Atenatively, half lives `=n = t/(t_(1//2))= 96/24 = 4`
`[A]/[A]_0= 1/2^n=2^(1//4)= 1/16`
`:. [A] =[A]_0/16 = 10/16 = 0.625`