Question #59ea6

1 Answer

Answer:

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"#