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

##### 1 Answer

#### 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

#color(blue)(A -> "products")#

can thus be written as

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

**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

*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_"1/2" = ln(1/2)/(-k) = (ln(1) - ln(2))/(-k) = ln(2)/k#

In your case, you want to figure out how many half0lives must pass in order for **consumed**. In other words, you need

#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_"x" = ln(1/4)/(-k) = (ln(1) - ln(4))/(-k) = ln(4)/k#

But since

#ln(4) = ln(2^2) = 2 * ln(2)#

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 **two half-lives**.