1.

A first order reaction is `40%` completed in 50 minutes. What is the time required for `90%` of the reaction to complete?

Answer» For the first order reaction,
`t=(2.303)/t log a/(a-x)`
Ist case: `a=100%, x=40%, (a-x)=100-40=60%, t=50` min
`t_(40%)=(2.303)/k log(100%)/(60%)`
`50"min" = 2.303/k log (100)/(60)=2.303/k log 1.667 = (2.303 xx 0.2219)/k`…………(i)
IInd case: `a=100%, x=90%, (a-x)=100-90=10%`
`t_(90%) = 2.03/k log (100%)/(10%)=2.303/k log 10=2.303/k`.........(ii)
Divide eqn. (ii) by eqn. (i),
`t_(90%)/(50"min") = 2.303/(2.303 xx 0.2219) = 4.506 t_(90%)=4.506 xx (50"min") = 2.25 xx 10^(2)`min.


Discussion

No Comment Found

Related InterviewSolutions