# Question 91357

Apr 13, 2015

You're dealing with a second order reaction that has the rate constant equal to ${\text{0.15 M"^(-1)"min}}^{- 1}$ and the initial concentration concentration of your reactant, ${\left[B\right]}_{0}$, equal to 0.5 M.

The generic form of your reaction is

$B \to P$

The rate of reaction for a second order reaction is

$\text{rate} = k {\left[B\right]}^{2}$

The integrated rate law for this reaction will be

$\frac{1}{\left[B\right]} - \frac{1}{{\left[B\right]}_{0}} = k \cdot t$ $\textcolor{b l u e}{\left(1\right)}$, where

$\left[B\right]$ - the concentration of $B$ after the passing of a certain amount of time, $t$;
${\left[B\right]}_{0}$ - the initial concentration of $B$;
$k$ - the rate constant - ${\text{0.15 M"^(-1)"min}}^{- 1}$;
$t$ - time.

So, you have to calculate the half-life of the reaction, which expresses the amount of time needed for the concentration of B to be halved. Mathematically, this can be written as

$\left[B\right] = \frac{{\left[B\right]}_{0}}{2}$

Plug this equation into equation $\textcolor{b l u e}{\left(1\right)}$ and solve for $t$

$\frac{1}{\frac{{\left[B\right]}_{0}}{2}} - \frac{1}{{\left[B\right]}_{0}} = k \cdot {t}_{\text{1/2}}$

$\frac{2}{{\left[B\right]}_{0}} - \frac{1}{{\left[B\right]}_{0}} = k \cdot {t}_{\text{1/2" => t_"1/2}} = \frac{1}{k \cdot {\left[B\right]}_{0}}$

t_"1/2" = 1/(0.15cancel("M"^(-1))"s"^(-1) * "0.5"cancel("M")) = color(green)("13 min")

Now you have to calculate how much time must pass for the concentration of $B$ to drop to 45% of the initial concentration.

Even before doing any calculations, you can predict that more than the half-life time must pass, since 45% of ${\left[B\right]}_{0}$ is a smaller amount than 50% of ${\left[B\right]}_{0}$.

The same approach can be used in this case as well

$\left[B\right] = \frac{45}{100} \cdot {\left[B\right]}_{0}$

Once again, plug this into equation $\textcolor{b l u e}{\left(1\right)}$ and solve for $t$

$\frac{1}{\frac{45}{100} \cdot {\left[B\right]}_{0}} - \frac{1}{{\left[B\right]}_{0}} = k \cdot {t}_{\text{45%}}$

t_"45%" = (1/(45/100 * 0.5) - 1/0.5)cancel("M"^(-1)) * 1/(0.15cancel(M^(-1))"s"^(-1))

t_"45%" = color(green)("16 min")#

To get the graph of a second-order integrated rate law, rearrange equation $\textcolor{b l u e}{\left(1\right)}$ to get

${\underbrace{\frac{1}{\left[B\right]}}}_{\textrm{y}} = {\underbrace{k}}_{\textrm{s l o p e}} \cdot t + {\underbrace{\frac{1}{{\left[B\right]}_{0}}}}_{\textrm{y - \int e r c e p t}}$

Your graph will look like this