3 Engineering Example 1

3.1 Arrhenius’ law

Introduction

Chemical reactions are very sensitive to temperature; normally, the rate of reaction increases as temperature increases. For example, the corrosion of iron and the spoiling of food are more rapid at higher temperatures. Chemically, the probability of collision between two molecules increases with temperature, and an increased collision rate results in higher kinetic energy, thus increasing the proportion of molecules that have the activation energy for the reaction, i.e. the minimum energy required for a reaction to occur. Based upon his observations, the Swedish chemist, Svante Arrhenius, proposed that the rate of a chemical reaction increases exponentially with temperature. This relationship, now known as Arrhenius’ law, is written as

$k=k_{0}exp\left(\frac{-E_a}{RT}\right)$ (1)

where $k$ is the reaction rate constant, $k_0$ is the frequency factor, $E_a$ is the activation energy, $R$ is the universal gas constant and $T$ is the absolute temperature. Thus, the reaction rate constant, $k$ , depends on the quantities $k_0$ and $E_a$ , which characterise a given reaction, and are generally assumed to be temperature independent.

Problem in words

In a laboratory, ethyl acetate is reacted with sodium hydroxide to investigate the reaction kinetics. Calculate the frequency factor and activation energy of the reaction from Arrhenius’ Law, using the experimental measurements of temperature and reaction rate constant in the table:

Mathematical statement of problem

Given that $k=7.757192{\text{s}}^{-1}$ at $T$ = 310 K and $k=110.9601{\text{s}}^{-1}$ at $T$ = 350 K, use Equation (1) to produce two linear equations in $E_a$ and $k_0$ . Solve these to find $E_a$ and $k_0$ . (Assume that the gas constant $R=8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ .)

Mathematical analysis

Taking the natural logarithm of both sides of (1)

$lnk=ln\left\{k_{0}exp\left(\frac{-E_a}{RT}\right)\right\}=ln{k}_0-\frac{E_a}{RT}$

Now inserting the experimental data gives the two linear equations in $E_a$ and $k_0$

$ln{k}_1=ln{k}_0-\frac{E_a}{R{T}_1}$ (2)

$ln{k}_2=ln{k}_0-\frac{E_a}{R{T}_2}$ (3)

where $k_1$ = 7.757192, $T_1$ = 310 and $k_2$ = 110.9601, $T_2$ = 350.

Firstly, to find $E_a$ , subtract Equation (2) from Equation (3)

$ln{k}_2-ln{k}_1=\frac{E_a}{R{T}_1}-\frac{E_a}{R{T}_2}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

so that

$E_a=\frac{R\left(ln{k}_2-ln{k}_1\right)}{\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

and substituting the values gives

$E_a=60000{\text{ J mol}}^{-1}=60{\text{ kJ mol}}^{-1}$

Secondly, to find $k_0$ , from (2)

$ln{k}_0=ln{k}_1+\frac{E_a}{R{T}_1}\Rightarrow k_0=exp\left(ln{k}_1+\frac{E_a}{R{T}_1}\right)=k_{1}exp\left(\frac{E_a}{R{T}_1}\right)$

and substituting the values gives

$k_0=1.0\times 10^{11}{\text{ s}}^{-1}$

Task!

The reaction

has a reaction rate constant of $1.0\times 10^{-10}{\text{s}}^{-1}$ at 300 K and activation energy of $111\text{kJ}{\text{mol}}^{-1}=111000\text{J}{\text{mol}}^{-1}$ . Use Arrhenius’ law to find the reaction rate constant at a temperature of 273 K.

Answer

Task!

For a chemical reaction with frequency factor $k_0=0.5{\text{s}}^{-1}$ and ratio $E_a∕R$ = 800 K, use Arrhenius’ law to find the temperature at which the reaction rate constant would be equal to $0.1{\text{s}}^{-1}$ .

Answer

As a final example we consider equations involving the hyperbolic functions.

Example 8

Solve the equations

1. $cosh3x=1$
2. $cosh3x=2$
3. $2{cosh}^{2}x=3cosh2x-3$

Solution

1. From its graph we know that $coshx=0$ only when $x=0$ , so we need $3x=0$ which implies

   $x=0$ .

2. $cosh3x=2\text{implies}\frac{{\text{e}}^{3x}+{\text{e}}^{-3x}}{2}=2\text{or}{\text{e}}^{3x}+{\text{e}}^{-3x}-4=0$

   Now multiply through by ${\text{e}}^{3x}$ (to eliminate the term ${\text{e}}^{-3x}$ ) to give

   ${\text{e}}^{3x}{\text{e}}^{3x}+{\text{e}}^{3x}{\text{e}}^{-3x}-4{\text{e}}^{3x}=0\text{or}{\left({\text{e}}^{3x}\right)}^2-4{\text{e}}^{3x}+1=0$

   This is a quadratic equation in the variable ${\text{e}}^{3x}$ so substituting $y={\text{e}}^{3x}$ gives

   $y^2-4y+1=0\text{implying}y=2\pm \sqrt{3}\text{so}y=3.7321\text{or}0.26795$

   ${\text{e}}^{3x}=3.7321\text{implies}x=\frac{1}{3}ln3.7321=0.439\text{to 3 d.p.}$

   ${\text{e}}^{3x}=0.26795\text{implies}x=\frac{1}{3}ln0.26795=-0.439\text{to 3 d.p.}$

3. We first simplify this expression by using the identity: $cosh2x=2{cosh}^2-1$ . Thus the original equation $2{cosh}^{2}x=3cosh2x-3$ becomes $cosh2x+1=3cosh2x-3$ or, when written in terms of exponentials:

   $\frac{{\text{e}}^{2x}+{\text{e}}^{-2x}}{2}=3\left(\frac{{\text{e}}^{2x}+{\text{e}}^{-2x}}{2}\right)-4$

   Multiplying through by $2{\text{e}}^{2x}$ gives ${\text{e}}^{4x}+1=3\left({\text{e}}^{4x}+1\right)-8{\text{e}}^{2x}$ or, after simplifying:

   ${\text{e}}^{4x}-4{\text{e}}^{2x}+1=0$

   Writing $y={\text{e}}^{2x}$ we easily obtain $y^2-4y+1=0$ with solution (using the quadratic formula):

   $y=\frac{4\pm \sqrt{16-4}}{2}=2\pm \sqrt{3}$

   If $y=2+\sqrt{3}$ then $2+\sqrt{3}={\text{e}}^{2x}$ implying $x=0.65848$ to 5 d.p.

   If $y=2-\sqrt{3}$ then $2-\sqrt{3}={\text{e}}^{2x}$ implying $x=-0.65848$ to 5 d.p.

Task!

Find the solution for $x$ if $tanhx=0.5$ .

First re-write $tanhx$ in terms of exponentials:

Answer

Now substitute into $tanhx=0.5$ :

Answer

Now complete your solution by finding $x$ :

Answer

Example 9

Solve for $x$ if $3lnx+4logx=1$ .

Solution

This has logs to two different bases. So we must first express each logarithm in terms of logs to the same base, $\text{e}$ say. From Key Point 8

$logx=\frac{lnx}{ln10}$

So $3lnx+4logx=1$ becomes

$3lnx+4\frac{lnx}{ln10}=1\text{or}\left(3+\frac{4}{ln10}\right)lnx=1$

leading to $lnx=\frac{ln10}{3ln10+4}=\frac{2.302585}{10.907755}=0.211096$ and so

$x={\text{e}}^{0.211096}=1.2350311$

Exercises

1. Solve for the variable $x$ :
   1. $\pi =10^x$
   2. $10^{-x∕2}=3$
   3. $\frac{1}{17-{\pi }^x}=4$
2.  Solve the equations
   1. ${\text{e}}^{2x}=17{\text{e}}^x$ ,
   2. ${\text{e}}^{2x}-2{\text{e}}^x-6=0$ ,
   3. $coshx=3$ .

Answer