# What is an example of a rate of reactions practice problem?

Jan 4, 2015

Here's an example of a rate of reaction problem:

What is the rate constant k for the first-order decomposition of ${N}_{2} {O}_{5 \left(g\right)}$ at ${25}^{\circ} C$ if the half-life of ${N}_{2} {O}_{5 \left(g\right)}$ at that temperature is $4.03 \cdot {10}^{4}$ seconds? Under these conditions, what percentage of ${N}_{2} {O}_{5}$ molecules will not have reacted after 1 day?

We know that half-life represents the time needed for a reactant's concentration to reach half of its initial value. The equation for the half-life of a reactant in a first-order reaction is given by

${t}_{\text{1/2}} = \frac{0.6931}{k}$, where $k$ represents the rate constant.

Since we know that ${N}_{2} {O}_{5}$ has a half-life of $4.03 \cdot {10}^{4} s$ under these conditions, the rate constant will be

$k = \frac{0.6931}{t} _ \left(\text{1/2}\right) = \frac{0.6931}{4.03 \cdot {10}^{4} s} = 1.72 \cdot {10}^{- 5} {s}^{- 1}$

Since this is a first-order reaction, the equation we'll use to determine what percentage is left unreacted is

$\ln \left(\frac{\left[{N}_{2} {O}_{5}\right]}{{\left[{N}_{2} {O}_{5}\right]}_{0}}\right) = - k \cdot t$, where

$\left[{N}_{2} {O}_{5}\right]$ - the amount that does not react;
${\left[{N}_{2} {O}_{5}\right]}_{0}$ - the initial amount;
$t$ - the time of the reaction - in this case, 1 day;

Since $\left[{N}_{2} {O}_{5}\right]$ represents the amount that does not react, we can say that ([N_2O_5])/([N_2O_5]_0) * 100% will be the percentage of ${N}_{2} {O}_{5}$ that does not react.

Converting 1 day to seconds will give

$1$ "day" * ("24 hours")/("1 day") * ("60 minutes")/("1 hour") * ("60 seconds")/("1 minute") = 86400s

$\ln \left(\frac{\left[{N}_{2} {O}_{5}\right]}{{\left[{N}_{2} {O}_{5}\right]}_{0}}\right) = - 1.72 \cdot {10}^{- 5} {s}^{- 1} \cdot 86400 s = - 1.49$

This means that

$\frac{\left[{N}_{2} {O}_{5}\right]}{{\left[{N}_{2} {O}_{5}\right]}_{0}} = {e}^{- 1.49} = 0.225$

Therefore, the percentage left unreacted is

0.225 * 100% = 22.5%