The half-life of first order decomposition of `NH_(4)NO_(3)` is 2.10 hr at 288 K temperature
`NH_(4)NO_(3)(aq)toN_(2)O(g)+2H_(2)O(l)`. If 6.2 of `NH_(4)NO_(3)` is allowed to decompose, the required for `NH_(4)NO_(3)` to decompose 90 % is :

Options

A. 6.978 hr
B. 0.319
C. 0.319 hr
D. 0.6978 hrs

Answer

A

Explanation

`k=(0.693)/(2.1)=0.33 hr^(-1)`
Let t be the time for 90% decomposition, so x = 90 hence (a - x)=10
`k=(2.303)/(t)log((a)/(a-x))`
`implies t=(2.303)/(0.33)log((100)/(10))`, t = 6.978 hr