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

Dec 19, 2015

$\text{100 min}$

#### Explanation:

The integrated rate law for a first-order reaction looks like this

$\textcolor{b l u e}{\ln \left(\frac{A}{A} _ 0\right) = - k \cdot t} \text{ }$, where

$A$ - the concentration at a given time $t$
${A}_{0}$ - the initial concentration
$k$ - the rate constant, usually expressed in ${\text{s}}^{- 1}$ 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

$A = \frac{1}{2} \cdot {A}_{0} \to$ half of the initial concentration remains after one half-life

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

$\ln \left(\frac{\frac{1}{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}}}\right) = - k \cdot \text{30 min}$

This is equivalent to

$- k \cdot \text{30 min} = \ln \left(\frac{1}{2}\right)$

$- k \cdot \text{30 min} = {\overbrace{\ln \left(1\right)}}^{\textcolor{b l u e}{= 0}} - \ln \left(2\right)$

$k = \ln \frac{2}{\text{30 min" = 2.31 * 10^(-2)"min}} ^ \left(- 1\right)$

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

$A = \frac{1}{10} \cdot {A}_{0} \to$ equivalent to 90% completion

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

$\ln \left(\frac{\frac{1}{10} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}}}\right) = - 2.31 \cdot {10}^{- 2} {\text{min}}^{- 1} \cdot t$

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 sig fig, the answer will be

$t = \textcolor{g r e e n}{\text{100 min}}$