Question

Question #59ea6

Answer 1

A: `t_½ = "788 s"`; B: `k = 1.71 × 10^"-3" color(white)(l)"s"^"-1"`; C: `t = "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 `1/2` of the original value.
• After two half-lives, the concentration of the reactant will be `1/4` of the original value.
• After three half-lives, the concentration of the reactant will be `1/8` of the original value, and so on.

The final concentration is `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"`