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Consider a certain reaction A to products with k=2.0xx10^(-2)s^(-1).Calculate the concentration of A remining after 100 s if the initial concentration of A is 1.0 mol L^(-1)

Answer»

Solution :Reaction A `to` product
K=`2.0xx10^(-2)s^(-1)`
t=100 s
Initial concentration =`[A]_(0)=1.0 mol L^(-1)`
After 100S, `[R]_(t)`=(?)
The unit of K is `SEC^(-1)`
`therefore` The reaction will be first order
`K=(2.303)/(t)`LOG `([R]_(0))/([R]_(t))`
`therefore 2.0xx10^(-2)=(2.303)/(100s)` log `((1.0))/(log[R]_(t))`
`therefore (2.0xx10^(-2)xx100)/(2.303)`=log 1=log `[R]_(t)`
`therefore0.8684=-log[R]_(t)`
`therefore log [R]_(t)=-0.8684`
`therefore[R]_(t)`=Antilog (1-0.8684)=0.1354 mol `L^(-1)`


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