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

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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?

If A(g)

##### 2 Answers

Mar 28, 2018

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%

Mar 28, 2018

#### Explanation:

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

Here, unit of Rate constant

*Units of Rate constant in zero, first and second order reactions.*

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

∴ It takes