Question

How many half-lives will it take for the concentration of the n2o to reach 30 % of its original concentration?

Answer 1

We have that after 1 half-life, the reactant reaches half its concentration $(\frac{1}{2})^1$. After 2 half-lives it reaches $(\frac{1}{2})^2$ of its concentration, namely one-fourth. After n half lives we have that it reaches $(\frac{1}{2})^n$ of its initial concentration. Thus, we have to solve the equation $(\frac{1}{2})^n=0,30$. Taking the log base 2 of each part of the equation we get from logarithm properties: $log_{2}[(1/2)^n]=log_{2}(2^{-n})=-n$. The other hand yields: $-n=log_{2}(0,3)$. Solving for n we get that n=1.74. Thus, after around 1.74 half lives the concentration becomes 30% of the initial one. Note how this is consistent with our previous analysis; if we let it a little bit more, 2 half-lives, the concentration will become 25% of the initial.

Answer 2

Answer:

Number of lives required to reach 30% of the its original concentration is 2.

Explanation:

Number of half lives are calculated by using formula :

$a=\frac{a_o}{2^n}$        ............(1)

where,

a = amount of reactant left after n-half lives

$a_o$ = Initial amount of the reactant

n = number of half lives

Here we have:

Initial concentration of nitrogen dioxide =$a_o=x$

Concentration of nitrogen dioxide after n number of half lives = a

a = 30% of x :

$\frac{30}{100}\times x= 0.3x$

$a=\frac{a_o}{2^n}$

$0.3x=\frac{x}{2^n}$

$2^n=\frac{1}{0.3}$

n = 1.736 ≈ 2

Number of lives required to reach 30% of the its original concentration is 2.