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

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

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%

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

Mar 28, 2018

$\text{4 half lives}$

#### Explanation:

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

Here, unit of Rate constant $\left(\text{k}\right)$ is ${\text{s}}^{-} 1$. Therefore, it’s a first order reaction

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

For a first order reaction, we can write

$\text{N" = "N"_0/2^"n}$

Where

• ${\text{N}}_{0} =$ Initial amount of substance
• $\text{N =}$ Amount of substance after $\text{n}$ half lives
• $\text{n =}$ Number of half lives

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

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

${2}^{\text{n}} = 16$

${2}^{\text{n}} = {2}^{4}$

$\text{n} = 4$

∴ It takes $4$ half lives for concentration of $\text{A}$ to reach 6.25%