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

Question

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

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Answer 1 · 2018-03-28T02:41:04

It will take 4 half lives.

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

${(\frac{1}{2})}^{n}, n = 4; (\frac{1}{16}) \times 100 = 6.25 %$ $(1/2)^n, n = 4; (1/16)xx100 = 6.25%$

Answer 2 · 2018-03-28T02:41:09

$4 half lives$ $"4 half lives"$

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

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 $6.25%$