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A firstorderreactionis 20% complete in 10 min. How long it takes to complete 80% ?

Answer»

Solution :Applying first ORDER integral rate equation, equation, rate constant K is gives as,
`K=(2.303)/(t)"log"(a)/(a-x)RARR(2.303)/(10)"log"(100)/(100-20)=0.0223" MIN"^(-1)`
TIME required for `80%` completion of the first order reaction, `t_(0.8)`
`t_(0.8)=(2.303)/(k)"log"(100)/(!00-80)=(2.303)/(0.0223)"log"(100)/(20)=72.2` min


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