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

Answer»

Solution :`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, wehave
`(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%`


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