Question

The rate constant is k =2.30×10^-3 s^-1. The concentration is 0.150 M, what will the concentration be after 15.0 minutes?

Answer 1

`"0.0189 M"`

Explanation:

Unit of Rate constant in the question is `"s"^-1`. With rate in `"M/s"`, these units suggest a first order reaction.

For a first order reaction

`ln[A]_t = -kt + ln[A]_@`

Upon rearrangement,

`t = 1/k ln (([A]_@)/([A]_t)) = (2.303)/klog(([A]_@)/([A]_t))`

Where

• `[A]_@ =` Initial concentration
• `[A]_t =` Concentration after time `t`
• `t =` Time taken
• `k =` Rate constant

`15.0 cancel"min" × "60 s"/(1 cancel"min") = (2.303)/(2.30 × 10^-3\ "s"^-1) log("0.150 M"/([A]_t))`

`900 cancel"s" = (2.303)/(2.30) cdot 10^3 cancel"s" × log("0.150 M"/([A]_t))`

`log("0.150 M"/([A]_t)) = 0.899`

`"0.150 M"/([A]_t) = 10^0.899`

`[A]_t = "0.150 M"/10^0.899 = "0.0189 M"`