1.

The rate constant for a first order reaction is 2.3xx10^(-4)s^(-1). If the initial concentration of the reactant is 0.01m. What concentration will remain after 1 hour ?

Answer»

Solution :`k=2.3xx10^(-4)s^(-1)`
T = 60 min = 3600 s
`[A]_(0)=0.01 M`
`k=(2.303)/(t)"log"([A]_(0))/([A])`
`rArr 2.3xx10^(-4)s^(-1)=(2.303)/(3600s)"log"(0.01)/([A])`
`rArr 2.3xx10^(-4)s^(-1)=(2.303)/(3600s)(-log[A])`
`rArr - log[A]=(2.3xx10^(-4)xx3600)/(2.303)`
[A] = ANTILOG `[-(2.3xx10^(-4)xx3600)/(2.303)]`
`= 0.142 "MOL L"^(-1)` (APPROXIMATELY)


Discussion

No Comment Found

Related InterviewSolutions