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

Md Ariful Islam

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Md Ariful Islam

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

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