6.3 Second Order Reaction

  Chemical Kinetics

There are two different types of reaction exists for second order:

Case I: Two same reactants react with same initial concentration to form products.

Let us assume the concentration of reactant $A$ at time $t=0$ be $\left[ {{C_0}} \right]$ and at time $t=t\ sec$ be $\left[ {{C_t}} \right]$ which can be written as $$\begin{equation} \begin{aligned} \quad {\text{ }}A + A \to Products \\ At\;t = 0\sec \;\quad {C_0} = a\quad a\quad {\text{ }}{C_0}^P = 0 \\ At\;t = t\sec \;\quad {C_t} = a - x{\text{ }}a - x\quad {\text{ }}{C_t}^P = x \\\end{aligned} \end{equation} $$

Therefore, the rate of reaction can be written with respect to second order as $$\frac{{dx}}{{dt}} = k{\left( {a - x} \right)^2}$$ Integrating both sides, we get

$$\begin{equation} \begin{aligned} \int\limits_0^x {\frac{{dx}}{{{{\left( {a - x} \right)}^2}}}} = \int\limits_0^t {kdt} \\ \left[ { - \frac{{{{\left( {a - x} \right)}^{ - 2 + 1}}}}{{ - 2 + 1}}} \right]_0^x = k\left[ {t - 0} \right]_0^t \\ \left[ {\frac{1}{{a - x}}} \right]_0^x = kt \\ \frac{1}{{a - x}} - \frac{1}{a} = kt \\ \frac{1}{{a - x}} = kt + \frac{1}{a} \\ \frac{1}{{{C_t}}} = kt + \frac{1}{{{C_0}}} \\\end{aligned} \end{equation} $$

Interpretation of second order reaction

• To plot the graph between concentration $\frac{1}{{{C_t}}}$ and time $(t)$, compare the derived relationship $$\frac{1}{{{C_t}}} = kt + \frac{1}{{{C_0}}}$$
  with the equation of straight line $y=mx+c$, we get the straight line with negative slope $ k$ which is shown in figure.
  
  

• We can find the unit of rate constant for first order reaction $$k = \left( {\frac{1}{{{C_t}}} - \frac{1}{{{C_0}}}} \right)\frac{1}{t} = \frac{1}{{mollitr{e^{ - 1}}time}} = mo{l^{ - 1}}lit.tim{e^{ - 1}}$$

• Half life period: It means the time required to consume half of the reactant or we can say that half the reactant is converted to product.
  
  $\therefore$ At half life time, ${C_t} = \frac{{{C_0}}}{2}$. We get $$\begin{equation} \begin{aligned} \frac{2}{{{C_0}}} = k{t_{1/2}} + \frac{1}{{{C_0}}} \\ {t_{1/2}} = \left( {\frac{2}{{{C_0}}} - \frac{1}{{{C_0}}}} \right)\frac{1}{k} = \frac{1}{{{C_0}k}} \\\end{aligned} \end{equation} $$
  
  The graph between ${t_{1/2}}$ and ${{C_0}}$ is a rectangular hyperbola as shown in figure.
  
  
  

Case II: When two different reactants with different initial concentration react to form products

Let us assume the concentration of reactant $A$ at time $t=0$ be $\left[ {{C_0}} \right]$ and at time $t=t\ sec$ be $\left[ {{C_t}} \right]$ which can be written as $$\begin{equation} \begin{aligned} \quad {\text{ }}A + B \to Products \\ At\;t = 0\sec \;\quad {C_0} = a\quad b\quad {\text{ }}{C_0}^P = 0 \\ At\;t = t\sec \;\quad {C_t} = a - x{\text{ }}b - x\quad {\text{ }}{C_t}^P = x \\\end{aligned} \end{equation} $$

Therefore, the rate of reaction can be written with respect to second order as

$$\begin{equation} \begin{aligned} \frac{{dx}}{{dt}} = k\left( {a - x} \right)(b - x) \\ \int\limits_0^x {\frac{{dx}}{{\left( {a - x} \right)(b - x)}}} = \int\limits_0^t {kdt} \\ \int\limits_0^x {\frac{1}{{(a - b)}}\frac{{\left( {a - b} \right)}}{{\left( {a - x} \right)(b - x)}}dx} = \int\limits_0^t {kdt} \\ \frac{1}{{(a - b)}}\int\limits_0^x {\frac{{\left( {a - x) - (b - x} \right)}}{{\left( {a - x} \right)(b - x)}}dx} = \int\limits_0^t {kdt} \\ \frac{1}{{(a - b)}}\int\limits_0^x {\frac{{\left( {a - x} \right)}}{{\left( {a - x} \right)(b - x)}} - \frac{{\left( {b - x} \right)}}{{\left( {a - x} \right)(b - x)}}dx} = \int\limits_0^t {kdt} \\ \frac{1}{{(a - b)}}\int\limits_0^x {\frac{1}{{(b - x)}} - \frac{1}{{\left( {a - x} \right)}}dx} = \int\limits_0^t {kdt} \\ \frac{1}{{(a - b)}}\left[ {\int\limits_0^x {\frac{{dx}}{{(b - x)}} - \int\limits_0^x {\frac{{dx}}{{\left( {a - x} \right)}}} } } \right] = \int\limits_0^t {kdt} \\ \frac{1}{{(a - b)}}\left[ { - \ln (b - x) + \ln (a - x)} \right]_0^x = k\left[ {t - 0} \right]_0^t \\ \frac{1}{{(a - b)}}\left[ {\ln \frac{{(a - x)}}{{(b - x)}}} \right]_0^x = kt \\ \frac{1}{{(a - b)}}\left[ {\ln \frac{{(a - x)}}{{(b - x)}} - \ln \frac{a}{b}} \right] = kt \\ \frac{1}{{(a - b)}}\left[ {\ln \frac{{b(a - x)}}{{a(b - x)}}} \right] = kt \\ k = \frac{{2.303}}{{t(a - b)}}\left[ {\ln \frac{{b(a - x)}}{{a(b - x)}}} \right] \\\end{aligned} \end{equation} $$