A first order reaction is 50% completed after 30 minutes.This implies that the time required to complete 90% of the reaction is?

Md Ariful Islam

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Md Ariful Islam

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The integrated for a first-order reaction looks like this

$color(blue)(ln( A/A_0) = - k * t)" "$, where$

$A$ - the concentration at a given time $t$
$A_0$ - the initial concentration
$k$ - the rate constant, usually expressed in $"s"^(-1)$ for first-order reactions

Now, I'll assume that you're not familiar with the equation that describes the half-life of a first-order reaction.

As you know, the half-life of a chemical reaction tells you how much time is needed for the concentration of a reactant to reach half of its initial value.

Simply put, the time needed for $50%$ of a reaction to be completed will represent that reaction's half-life.

In your case, you know that your first-order reaction is $50%$ completed after $30$ minutes. This means that after $30$ minutes, you will have

$A = 1/2 * A_0 ->$ half of the initial concentration remains after one half-life

Plug this into the above equation and solve for $k$, the rate constant

$ln( (1/2 * color(red)(cancel(color(black)(A_0))))/(color(red)(cancel(color(black)(A_0)))) ) = - k * "30 min"$

This is equivalent to

$- k* "30 min" = ln(1/2)$

$-k * "30 min" = overbrace(ln(1))^(color(blue)(=0)) - ln(2)$

$k = ln(2)/"30 min" = 2.31 * 10^(-2)"min"^(-1)$

Now that you know the rate constant for this reaction, use the same equation again, only this time use

$A = 1/10 * A_0 ->$ equivalent to $90%$ completion

and solve for $t$, the time needed to get here

$ln( (1/10 * color(red)(cancel(color(black)(A_0))))/(color(red)(cancel(color(black)(A_0)))) ) = - 2.31 * 10^(-2)"min"^(-1) * t$

This is equivalent to

$t = (color(red)(cancel(color(black)(-))) ln(10))/(color(red)(cancel(color(black)(-))) 2.31 * 10^(-2)"min"^(-1)) = "99.68 min"$

Rounded to one , the answer will be

$t = color(green)("100 min")$

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