How many half-lives will it take to reach 6.25% of its original concentration?

If A(g) #-># B(g) has k=#2.3*10^-3# #s^-1#, how many half-lives will it take for the concentration of A to reach 6.25% of its original concentration?

2 Answers
Mar 28, 2018

Answer:

It will take 4 half lives.

Explanation:

1st gets you to 50%
2nd gets you to 25%
3rd gets you to 12.5%
4th gets you to 6.25%

#(1/2)^n, n = 4; (1/16)xx100 = 6.25%#

Mar 28, 2018

Answer:

#"4 half lives"#

Explanation:

#"A"_"(g)" -> "B"_"(g)" color(white)(...)"k" = 2.3 × 10^-3\ "s"^-1#

Here, unit of Rate constant #("k")# is #"s"^-1#. Therefore, it’s a first order reaction

Units of Rate constant in zero, first and second order reactions.
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For a first order reaction, we can write

#"N" = "N"_0/2^"n"#

Where

  • #"N"_0 =# Initial amount of substance
  • #"N ="# Amount of substance after #"n"# half lives
  • #"n ="# Number of half lives

#6.25% = (100%)/2^"n"#

#2^"n" = (100%)/(6.25%)#

#2^"n" = 16#

#2^"n" = 2^4#

#"n" = 4#

∴ It takes #4# half lives for concentration of #"A"# to reach #6.25%#