Question

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

Answer 1

`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

`color(blue)(A -> "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

`["A"]` - the concentration of the reactant after the passing of time `t`
`["A"_0]` - 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_"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 `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_"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 `75%` completed after the passing of two half-lives.