1.

The half-life period of a 1^(st) order reaction is 60 minutes. What percentage will be left over after 240 minutes ?

Answer»

0.0625
0.0425
0.05
0.06

Solution : `:t_(1//2)=(0.693)/(K)Rightarrow(0.693)/(t_(1//2))=Krightarrow(0.693)/(60)=K`
`K=0.01155min^(-1)`
`K= (2.303)/(t)LOG ((a)/(a-x))`
LET the INITIAL amount (a) 100
`0.01155min^(-1)= (2.303)/(240min)log((100)/(a-x))`
`(0.01155min^(-1)xx40min)/(2.303)=log ((100)/(a-x))`
`1.204=log100-(a-x)`
`1.204=2-log(a-x)`
`log(a-x)=2-1.204`
log(a-x)=0.796
(a-x)=6.25%


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