Question

Explain the derivative and integral forms of second-order reactions

Answer 1

The order of chemical reactions:

• At a fixed temperature the rate of a given reaction depends on the concentration of reactants.
• An Equation that expresses the experimentally observed rate of a reaction in terms of the molar concentrations of the reactants that determine the rate of the reaction is called the rate law or rate equation.
• Thus, for a general reaction $\mathrm{aA}+\mathrm{bB}\to \mathrm{cC}+\mathrm{dD}$ , then $\mathrm{rate}=\mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{x}}{\left[\mathrm{B}\right]}^{\mathrm{y}}$ where $\mathrm{k}$ is the rate constant, $\left[\mathrm{A}\right]$and $\left[\mathrm{B}\right]$ is the concentrations of the reactants $\mathrm{A}$ and $\mathrm{B}$, $\mathrm{x}$ and $\mathrm{y}$ denote the partial reaction orders for reactants $\mathrm{A}$ and $\mathrm{B}$, (which may or may not be equal to their stoichiometric coefficients $\mathrm{a}$ & $\mathrm{b}$).
• Therefore, the order of reaction will be $\mathrm{x}+\mathrm{y}$ and $\mathrm{x}$ is the order of reaction with respect to $\mathrm{A}$ and $\mathrm{y}$ is the order of reaction with respect to $\mathrm{B}$.

Second-order reaction:

• A reaction can be called a second-order reaction when the overall order is two.
• If suppose $\mathrm{x}=1$ and $y=1$ then the reaction will be a second-order reaction. Reactions in which reactants are identical and form a product can also be second-order reactions.

The derivative and integral forms of second-order reactions:

• Case 1: Identical Reactants

Two of the same reactant ( A ) combine to form products,

$$\begin{gathered} \mathrm{A}+\mathrm{A}\to \mathrm{Product} \\ 2\mathrm{A}\to \mathrm{Product} \end{gathered}$$

If the initial concentration of the reactant $\left[\mathrm{A}\right]$or each of the two reactants $\left[\mathrm{A}\right]$and $\left[\mathrm{B}\right]$ be '$\mathrm{a}\mathrm{mol}{\mathrm{L}}^{-1}$'.

If after time $\mathrm{t}$, $x$ mole of $\mathrm{A}$ have reacted, the concentration of $\mathrm{A}$ is $\mathrm{a}-\mathrm{x}$

In a second-order reaction, the rate of reaction is proportional to the square of the concentration of the reactants,

$$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}(\mathrm{a}-\mathrm{x})(\mathrm{a}-\mathrm{x}) \\ \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}{(\mathrm{a}-\mathrm{x})}^2.........\left(1\right) \end{gathered}$$

By rearranging equation (1) we get;

$\frac{\mathrm{dx}}{{(\mathrm{a}-\mathrm{x})}^2}=\mathrm{kdt}.........\left(2\right)$

On integrating this equation (2) we get;

$$\begin{gathered} {\int }_0^{\mathrm{x}}\frac{dx}{{\left(\mathrm{a}-\mathrm{x}\right)}^2}=\mathrm{k}{\int }_0^{\mathrm{t}}\mathrm{dt}.....\left(3\right) \\ \left[\frac{1}{\mathrm{a}-\mathrm{x}}-\frac{1}{\mathrm{a}-0}\right]=k\left[t-0\right]+I.....\left(4\right) \end{gathered}$$

Where $\mathrm{I}$ is the constant of integration, to determine $\mathrm{I}$, by putting $\mathrm{t}=0,\mathrm{x}=0$

Thus $\mathrm{I}=\frac{1}{\mathrm{a}}$

Substituting the value of $\mathrm{I}$ in equation (4), we get,

$$\begin{gathered} \frac{1}{\mathrm{a}-\mathrm{x}}=kt+\frac{1}{a} \\ \mathrm{kt}=\frac{1}{\mathrm{a}-\mathrm{x}}-\frac{1}{a} \\ \mathrm{kt}=\frac{\mathrm{a}-\mathrm{a}+\mathrm{x}}{\mathrm{a}(\mathrm{a}-\mathrm{x})} \\ \mathrm{kt}=\frac{\mathrm{x}}{\mathrm{a}(\mathrm{a}-\mathrm{x})} \\ \mathrm{k}=\frac{1}{\mathrm{t}}\frac{\mathrm{x}}{\mathrm{a}(\mathrm{a}-\mathrm{x})} \end{gathered}$$

Therefore, the integrated rate equation for a second-order reaction is $k=\frac{1}{\mathrm{t}}\frac{\mathrm{x}}{\mathrm{a}(\mathrm{a}-\mathrm{x})}$

• Case 2: Second-Order Reaction with Multiple Reactants:

Two different reactants ($\mathrm{A}$ and $\mathrm{B}$) combine in a single elementary step,

$\mathrm{A}+\mathrm{B}\to \mathrm{Product}$

If the initial concentration of the reactant $\mathrm{A}$ is $\mathrm{a}\mathrm{mol}{\mathrm{L}}^{-1}$and that of $\mathrm{B}$ is $\mathrm{b}\mathrm{mol}{\mathrm{L}}^{-1}$

After time $\mathrm{t}$, $\mathrm{x}\mathrm{mol}{\mathrm{L}}^{-1}$of $\mathrm{A}$ and $\mathrm{x}\mathrm{mol}{\mathrm{L}}^{-1}$ of $\mathrm{B}$ react to form $\mathrm{x}\mathrm{mol}{\mathrm{L}}^{-1}$of the product.

Thus the reactant concentration at time t are $(\mathrm{a}-\mathrm{x})$ and $(b-x)$ respectively

The differential rate expression for the second-order reaction is:

$$\begin{gathered} r=-\frac{d\left[A\right]}{dt}=-\frac{d\left[B\right]}{dt}=\frac{dP}{dt} \\ r=k_2\left[A\right]\left[B\right].....\left(4\right) \\ \mathrm{r}=\frac{dx}{dt}={\mathrm{k}}_2(\mathrm{a}-\mathrm{x})(\mathrm{b}-\mathrm{x})....\left(5\right) \end{gathered}$$

Where ${\mathrm{k}}_2$ is the second-order constant separating the variables, we have,

$\frac{\mathrm{dx}}{(\mathrm{a}-\mathrm{x})(\mathrm{b}-\mathrm{x})}={\mathrm{k}}_2\mathrm{dt}$

Resolving into partial functions$(\mathrm{a}>\mathrm{b})$, we have

$\frac{1}{(a-x)(b-x)}=\frac{1}{(a-b)}\left[\frac{1}{b-x}-\frac{1}{a-x}\right].....\left(6\right)$

Integrating this equation (6)

$$\begin{gathered} \int \frac{dx}{(a-x)(b-x)}=\frac{1}{(a-b)}\int \left(\frac{1}{b-x}-\frac{1}{a-x}\right)dx=k_2\int dt \\ \frac{1}{a-b}\left[-\ln \left(b-x\right)-(-\ln \left(a-x)\right)\right]=k_{2}t+I \\ \frac{1}{a-b}\ln \left(\frac{a-x}{b-x}\right)=k_{2}t+I......\left(7\right) \end{gathered}$$

Where $\mathrm{I}$ is the constant of integration, to determine $\mathrm{I}$, by putting $\mathrm{t}=0,\mathrm{x}=0$

From equation (7) ,$\mathrm{I}=\frac{1}{\mathrm{a}-\mathrm{b}}\ln \left(\frac{\mathrm{a}-\mathrm{x}}{\mathrm{b}-\mathrm{x}}\right)$

Substituting the value of $\mathrm{I}$ in equation (7), we get

$\frac{1}{a-b}\ln \left(\frac{a-x}{b-x}\right)=k_{2}t+\frac{1}{\mathrm{a}-\mathrm{b}}\ln \left(\frac{\mathrm{a}-\mathrm{x}}{\mathrm{b}-\mathrm{x}}\right).....\left(8\right)$

By rearranging equation (8), we get;

$\frac{1}{a-b}\ln \left[\frac{b\left(a-x\right)}{b\left(b-x\right)}\right]=k_{2}t$

Solving for ${\mathrm{k}}_2$, we get

$$\begin{gathered} k_2=\frac{1}{(a-b)t}\ln \left(\frac{b(a-x)}{a(b-x)}\right) \\ {k}_2=\frac{1}{t}\frac{2.303}{(a-b)}\log \left(\frac{b(a-x)}{a(b-x)}\right) \end{gathered}$$

This is the integrated equation for the rate constant of a second order reaction.

Characteristics of second order reactions:

• The unit of rate constant for a second order reaction is ${\mathrm{mol}}^{-1}\mathrm{L}{\mathrm{s}}^{-1}$
• For all second order reactions involving a single reactant or two reactant of equal initial concentration, $k=\frac{1}{\mathrm{t}}\frac{\mathrm{x}}{\mathrm{a}(\mathrm{a}-\mathrm{x})}$.

On rearranging we get,$\frac{1}{a-x}=kt+\frac{1}{a}$ this is the equation for straight line$(y=mx+b)$

• For a second order reaction involving two reactants having different initial concentration, $k_2=\frac{1}{t}\frac{2.303}{(a-b)}\log \left(\frac{b(a-x)}{a(b-x)}\right)$.

On rearranging we get, $k_{2}t=\frac{2.303}{(a-b)}\log \left(\frac{b(a-x)}{a(b-x)}\right)$,this is also the equation for straight line