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

Answer

b

Explanation

`k = (0.063)/(24) hr^(-1) = (2.303)/(96)` log `(10)/(a -x)`
or log `(10)/(a-x) = 1.2036` or 1 - log ( a-x) = `1.2036`
or log (a-x) = `-0.2036 = 1.7964`
or (a-x) = antilog 1.7964
or (a-x) = antilog 1.7964 = 0.6258 gm .