6.2 First Order Reaction

  Chemical Kinetics

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 $$\begin{equation} \begin{aligned} \quad \quad \quad \quad \quad \quad \quad A \to Products \\ At\;t = 0\sec \;\quad {C_0} = a\quad \quad {\text{ }}{C_0}^P = 0 \\ At\;t = t\sec \;\quad {C_t} = a - x\quad {C_t}^P = x \\\end{aligned} \end{equation} $$

Therefore, the rate of reaction can be written with respect to first order as $$\frac{{dx}}{{dt}} = k\left( {a - x} \right)$$ Integrating both sides, we get $$\begin{equation} \begin{aligned} \int\limits_0^x {\frac{{dx}}{{\left( {a - x} \right)}}} = \int\limits_0^t {kdt} \\ \left[ { - \ln \left( {a - x} \right)} \right]_0^x = k\left[ t \right]_0^t \\ \ln \left( {a - x} \right) - \ln \left( a \right) = - k\left( {t - 0} \right) \\ \ln \left( {\frac{{a - x}}{a}} \right) = - kt \\ \ln \left( {\frac{a}{{a - x}}} \right) = kt \\ k = \frac{{2.303}}{t}\log \left( {\frac{a}{{a - x}}} \right)\ \ \ \ \ ...(1) \\\end{aligned} \end{equation} $$

Now, from the rate law equation, we can write $x$ as $${C_t} = a - x{\text{ and }}{C_0} = a$$

Put the values in $(1)$, we get $$k = \frac{{2.303}}{t}\log \left( {\frac{{{C_0}}}{{{C_t}}}} \right)$$

Interpretation of first order reaction

• To plot the graph between concentration $(\log {C_t})$ and time $(t)$, compare the derived relationship $$\begin{equation} \begin{aligned} k = \frac{{2.303}}{t}\log \left( {\frac{{{C_0}}}{{{C_t}}}} \right) \\ t = \frac{{2.303}}{k}\log {C_0} - \frac{{2.303}}{k}\log {C_t} \\ t = - \frac{{2.303}}{k}\log {C_t} + \frac{{2.303}}{k}\log {C_0} \\\end{aligned} \end{equation} $$
  with the equation of straight line $y=mx+c$, we get the straight line with negative slope $ - \frac{{2.303}}{k}$ which is shown in figure.
  
  
  
  
  Also the graph between time $(t)$ and $\log \left( {\frac{{{C_0}}}{{{C_t}}}} \right)$ is a straight line passing through origin with slope $\frac{{2.303}}{k}$ as shown in figure.
  
  
  

• We can find the unit of rate constant for first order reaction $$\begin{equation} \begin{aligned} \frac{{dx}}{{dt}} = k\left( {a - x} \right) \\ k = \frac{{dx}}{{dt}}.\frac{1}{{\left( {a - x} \right)}} = \frac{{conc.}}{{time \times conc.}} = {\sec ^{ - 1}} \\\end{aligned} \end{equation} $$

• 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 $$\begin{equation} \begin{aligned} k = \frac{{2.303}}{{{t_{\frac{1}{2}}}}}\log \left( {\frac{{{C_0}}}{{{C_0}/2}}} \right) \\ {t_{\frac{1}{2}}} = \frac{{2.303}}{k}\log \left( {\frac{{{C_0}}}{{{C_0}/2}}} \right) = \frac{{2.303\log (2)}}{k} = \frac{{0.693}}{k} \\\end{aligned} \end{equation} $$

Example 1. Calculate $\frac{{{t_{0.75}}}}{{{t_{0.50}}}}$ for a first order reaction.

Solution: As given in the question, we have to find the time in terms of rate constant $k$ when $0.75$ or $\frac{3}{4}th$ of the reaction completes and $0.50$ or $\frac{1}{2}th$ of the reaction completes. For a first order reaction, $$k = \frac{{2.303}}{t}\log \left( {\frac{{{C_0}}}{{{C_t}}}} \right)\ \ \ ...(1)$$ Therefore, for $\frac{3}{4}th$ of the reaction to be completed means $${{C_{0.75}} = \frac{{{C_0}}}{4}}$$ Put the value in $(1)$, we get $${t_{0.75}} = \frac{{2.303}}{k}\log \left( {\frac{{{C_0}}}{{{C_0}/4}}} \right) = \frac{{2.303}}{k}\log 4$$ Similarly, $${t_{0.50}} = \frac{{2.303}}{k}\log \left( {\frac{{{C_0}}}{{{C_0}/2}}} \right) = \frac{{2.303}}{k}\log 2$$ Divide both the equations, we get $$\frac{{{t_{0.75}}}}{{{t_{0.50}}}} = \frac{{\frac{{2.303}}{k}\log 4}}{{\frac{{2.303}}{k}\log 2}} = \frac{{\log 4}}{{\log 2}} = \frac{{2\log 2}}{{\log 2}} = 2$$

Example 2. Find the minimum half-lives to be elapsed for a first order reaction so that the reaction is at least $95\%$ completed. Consider the value of $log2=0.3$.

Solution: We can easily solve this question by using half life concept. As we know that in every half life, half the reaction is completed. Consider a value say $100$ and apply half life on it and increase the percentage of reaction completion every time as shown. There will be one time in it when he reaction is completed more than $95\%$.

$$\begin{equation} \begin{aligned} 100\mathop \to \limits^{{t_{1/2}}} 50\mathop \to \limits^{{t_{1/2}}} 25\mathop \to \limits^{{t_{1/2}}} 12.5\mathop \to \limits^{{t_{1/2}}} 6.25\mathop \to \limits^{{t_{1/2}}} 3.125 \\ 0\% \quad 50\% {\text{ }}75\% {\text{ }}87.5\% {\text{ }}93.75\% {\text{ 96}}{\text{.87% }} \\\end{aligned} \end{equation} $$ Therefore, $5$ half-lives should elapse to complete at least $95\%$ completion of any first order reaction.