1.

A first order reaction is `20%` complete in `10 min`. Calculate (a) the specific rate constant of the reaction and (b) the time taken for the reaction to reach `75%` completion.

Answer» Correct Answer - (a) `k = 0.0223 min^(-1)`
(b) `62.16 min`
(a) `k = (2.303)/(t) log.(a)/(a-x)`
For `20%` completion,
`k = (2.303)/(10)log.(100)/(80) = 0.0223 min^(-1)`
(b) Half life `= (0.693)/(k) = (0.693)/(0.0023) = 31.08 min`
For `75%` completion of the reaction, two half life lives are required, which is equal to `31.08 xx 2 = 62.16 min`.Correct Answer - (a) `k = 0.0223 min^(-1)`
(b) `62.16 min`
(a) `k = (2.303)/(t) log.(a)/(a-x)`
For `20%` completion,
`k = (2.303)/(10)log.(100)/(80) = 0.0223 min^(-1)`
(b) Half life `= (0.693)/(k) = (0.693)/(0.0023) = 31.08 min`
For `75%` completion of the reaction, two half life lives are required, which is equal to `31.08 xx 2 = 62.16 min`.


Discussion

No Comment Found

Related InterviewSolutions