The rate of a reaction `A to B` has the rate law, rate `= k [A]` with the rate constant, `= k = 0. 50 sec ^(-1) Q.`
What concentration of A would give a rate of `2. 4 xx 10 ^(-2) mol L ^(-1) sec ^(-1)?`

Explanation

Rate=k [A]
Where=k=0.50 `s^(-1)`
`therefore` Rate=`0.50s^(-1)[A]`
(a) Rate=? [A]=0.60 mol `L^(-1)`
Rate=0.50 `s^(-1)xx0.60` mol `L^(-1)`
=0.30 mol `L^(-1)s^(-1)=3xx10^(-2)" mol "L^(-1)s^(-1)`
(b) Rate=`2.4xx10^(-2)`,[A]=?
[A]=`("Rate")/(k)=(2.4xx10^(-2))/(0.50 s^(-1))` mol `L^(-1)s^(-1)`
`=4.8xx10^(-2)" mol L"^(-1)s^(-1)`