# Question 59ea6

Sep 28, 2017

A: t_½ = "788 s"; B: k = 1.71 × 10^"-3" color(white)(l)"s"^"-1"; C: $t = \text{630. s}$.

#### Explanation:

A: Half-Life

The equation for the half-life t_½ of a first-order reaction is

color(blue)(bar(ul(|color(white)(a/a)t_½ = (ln2)/kcolor(white)(a/a)|)))" "

where $k$ is the rate constant for the reaction.

t_½ = ln2/(8.80 × 10^"-4"color(white)(l)"s"^"-1") = 0.693/(8.80 × 10^"-4"color(white)(l)"s"^"-1") = "788 s"

B) Half-life

t_text(½) = "405 s"

Also, t_text(½) = ln2/k, so

k = ln2/t_text(½) = ln2/"405 s" = 0.693/"405 s" = 1.71 × 10^"-3" color(white)(l)"s"^"-1"

C) Time required

If a reaction has a certain half-life,

• After one half-life, the concentration of the reactant will be $\frac{1}{2}$ of the original value.
• After two half-lives, the concentration of the reactant will be $\frac{1}{4}$ of the original value.
• After three half-lives, the concentration of the reactant will be $\frac{1}{8}$ of the original value, and so on.

The final concentration is $\frac{1}{8}$ of the initial concentration, so the reaction has taken three half-lives.

t_½ = ln2/k = ln2/(3.30 × 10^"-3"color(white)(l)"s"^"-1") = "210 s"

t = 3t_½ = "3 × 210 s" = "630. s"#