1.

For a first order reaction, show that the time required for 99% completion of a first order reaction is twice the time required for the completion of 90%.

Answer»

SOLUTION :For FIRST ORDER reaction, `t=(2.303)/(k)log""(a)/(a-x)`
`99%` completion means that `x=99%" of a=0.99" a"`
`t_(99%)=(2.303)/(k)log""(a)/(a-0.99a)=(2.303)/(k)log10^(2)=2XX(2.303)/(k)`
`90%` completion means that `x=90%" of "a=0.90" a"`
`:.""t_(90%)=(2.303)/(k)log""(a)/(a-0.99a)=(2.303)/(k)log10=(2.303)/(k)`
`:.""(t_(99%))/(t_(90%))=((2xx2.303)/(k))//((2.303)/(k))=2" or "t_(99%)=2xxt_(90%).`


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