# For first degree equation , time for 75% completion of reaction is X times half time of that reaction , where X is ?

Jan 12, 2016

$X = 2$

#### Explanation:

As you know, the rate of a first-order reaction depends linearly on the concentration of a single reactant. The rate of a first-order reaction that takes the form

$\textcolor{b l u e}{A \to \text{products}}$

can thus be written as

color(blue)("rate" = - (d["A"])/dt = k * ["A"])" ", where

$k$ - the rate constant for the reaction

Now, in Integral form (I won't go into the derivation here), the rate law for a first-order reaction is equal to

color(blue)(ln( (["A"])/(["A"_0])) = - k * t)" ", where

$\left[\text{A}\right]$ - the concentration of the reactant after the passing of time $t$
$\left[{\text{A}}_{0}\right]$ - the initial concentration of the reactant

Now, the half-life of the reaction is the time needed for the concentration of the reactant to reach half of its initial value.

You can say that

t = t_"1/2" implies ["A"] = 1/2 * ["A"_0]

Plug this into the equation for the rate law to get

ln( (1/2 * color(red)(cancel(color(black)(["A"_0]))))/(color(red)(cancel(color(black)(["A"_0]))))) = -k * t_"1/2"

This means that you have

${t}_{\text{1/2}} = \ln \frac{\frac{1}{2}}{- k} = \frac{\ln \left(1\right) - \ln \left(2\right)}{- k} = \ln \frac{2}{k}$

In your case, you want to figure out how many half0lives must pass in order for 75% of the reactant to be consumed. In other words, you need 25% of the reactant to remain after a time ${t}_{x}$

t = t_"x" implies ["A"] = 25/100 * ["A"_0] = 1/4 * ["A"_0]

Once again, plug this into the rate law equation to get

ln( (1/4 * color(red)(cancel(color(black)(["A"_0]))))/(color(red)(cancel(color(black)(["A"_0]))))) = - k * t_"x"

This means that you have

${t}_{\text{x}} = \ln \frac{\frac{1}{4}}{- k} = \frac{\ln \left(1\right) - \ln \left(4\right)}{- k} = \ln \frac{4}{k}$

But since

$\ln \left(4\right) = \ln \left({2}^{2}\right) = 2 \cdot \ln \left(2\right)$

you can say that

t_"x" = 2 * overbrace(ln(2)/k)^(color(red)(t_text(1/2))) = color(green)(2 * t_"1/2")

Therefore,

$X = 2$

Simply put, the reaction will be 75% completed after the passing of two half-lives.