Consider a certain reaction `A to` products, with `k = 2.0 xx 10^(-2)" s"^(-1)`. Calculate the concentration of A remaining after 100 s if the initial concentration of `A` is `1.0` mol `"L"^(-1)`.

Explanation

`k= 2.0 xx 10^(-2)" s"^(-1)`
`[A]_(0) = 1.0" mol L"^(-1), t= 100" s", k = (2.303)/(t)" log" ([A]_0)/([A])`
Therefore,
`2.0 xx 10^(-2)" s"^(-1) = (2.303)/( 100" s") log "" (1.0" mol L"^(-1) )/( [A])`
`2.0 xx 10^(-2)" s"^(-1) = (2.303)/( 100" s" ) ( log1 - log [A] )`
`2.0 xx 10^(-2)" s"^(-1) = (2.303)/( 100" s")( 0- log[A])`,
`- log [A] = (2.0 xx 10^(-2)" s"^(-1)xx 100 "s")/(2.303)`
or `[A]="antilog" (-(2.0 xx 10^(-2)" s"^(-1) xx 100" s")/(2.303))`
or `[A] = 0.135" mol L"^(-1)`