1.

A first order reaction has rate constant 1.15xx10^(-3) s^(-1) How long will 5 g of this reactant take to reduce to 3g?

Answer»

Solution :SUPPOSE molecular MASS of reactant =M g `mol^(-1)`
Rate constant K=`1.15xx10^(-3)`
INITIAL mass =5 g
`therefore"Initial mole"=(5g)/(M g mol^(-1))`
`therefore [R]_(0)=(5)/(M)mol`
Final mass =3g
`therefore` Final mole =`(3g)/(MG mol^(-1))`
`therefore [R]_(t)=(3)/(M) mol`
Change in time `Deltat=(?)`
For first order reaction ,
Rate constant k`=(2.303)/(t)` log `([R]_(0))/([R]_(t))`
`therefore t=(2.303)/(k)` log `([R]_(0))/([R]_(t))`
`=(2.303)/(1.15xx10^(-3)s^(-1))` log `((5)/(M))/((3)/(M))`
`(2.303)/(1.15xx10^(-3)s^(-1))` log `(5)/(3)`
`(2.303)/(1.15xx10^(-3)s^(-1))xx0.2219=444.3s`


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