Half-life of a first order reaction is `5xx10^(4)`s. What percentage of the initial reactant will react in 2 hours? Calculate.

Explanation

`k=(0.693)/(t_(1//2)) and k=(2.303)/(t)"log"([R]_(0))/([R])`
From these two equations, we have
`(0.693)/(t_(1//2))=(2.303)/(t)"log"([R]_(0))/([R])`
Substituting the values, we have
`(0.693)/(5xx10^(4))=(2.303)/(7200)"log"([R]_(0))/([R])`
or `"log" ([R]_(0))/([R])=(0.693xx7200)/(5xx10^(4)xx2.303)=(4989.6)/(11.515xx10^(4))=0.0433`
or `([R]_(0))/([R])=1.105 or ([R])/([R]_(0))=(1)/(1.105)`
Percentage `=(1)/(1.105)xx100=90.49%`