# Question 8a56f

Mar 15, 2015

Let's start with point a). If you look at the graphs for zeroth, first, and second order reactions, you'll notice that only one of them has $\text{1/concentration}$ plotted against time with a resulting positive slope. This means that, because the $\frac{1}{{\left[A\right]}_{0}}$ graph you plotted has a positive slope, your reaction can only be second-order. Since there is only one reactant, $\text{A}$, the rate law for your reaction can be written as

$\text{rate} = k \cdot {\left[A\right]}^{2}$, where

$\text{k}$ - the rate constant.

The trick here is to realize that the slope of the graph is equal to the rate constant k for this reaction. This means that your rate constant will be

$k = {\text{slope" = 3.6 * 10^(-2)"L" * "mol"^(-1) * "s}}^{- 1}$

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

$\frac{1}{\left[A\right]} - \frac{1}{{\left[A\right]}_{0}} = k \cdot t$, where

$\left[A\right]$ - the concetration of $\text{A}$ after a time t has passed;
${\left[A\right]}_{0}$ - the initial concentration of $\text{A}$;

Moving on to point b). To determine the half-life of this reaction, replace $\text{t}$ with "t"_("1/2") and $\left[A\right]$ with ${\left[A\right]}_{0} / 2$ in the above equation. (Remember that, by definition, the reaction's half-life is the time needed for the reactant's initial concentration to be halved).

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

Solving this for ${t}_{\text{1/2}}$ will give you

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

Plug in the values you have for k and ${\left[A\right]}_{0}$ and you'll get

${t}_{{\text{1/2") = 1/(3.6 * 10^(-2)"L" * "mol"^(-1) * "s"^(-1) * 2.8 * 10^(-3)"mol" * "L}}^{- 1}}$

t_("1/2") = 9.92 * 10^(3)"s"

Finally, point c). Use the integrated rate law to solve for $\text{t}$, since you know that the value of $\left[A\right]$ must now be $7.00 \cdot {10}^{- 4} \text{M}$. So,

1/(7.00 * 10^(-4)"M") - 1/(2.8 * 10^(-3)"M") = 3.6 * 18^(-2)"M"^(-1)"s"^(-1) * t

$1071.43 {\text{M"^(-1) = 3.6 * 18^(-2)"M"^(-1)"s}}^{- 1} \cdot t$

t = 1071.43/(3.6 * 10^(-2)"s"^(-1)) = "29761.94 s"#, or

$t = 2.976 \cdot {10}^{4} \text{s}$

Rounded to two sig figs, the answer will be

$t = 3.0 \cdot {10}^{4} \text{s}$

SIDE NOTE I didn't derive the integrated rate law for this second-order reaction because the answer would have been even longer, so I'll leave that to you as practice.