6.1 Zero 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 zero order as $$\begin{equation} \begin{aligned} \frac{{dx}}{{dt}} = k{\left( {a - x} \right)^0} \\ dx = kdt \\\end{aligned} \end{equation} $$ Integrating both sides, we get

$$\begin{equation} \begin{aligned} \int\limits_0^x {dx} = \int\limits_0^t {kdt} \\ \left[ x \right]_0^x = k\left[ t \right]_0^t \\ x - 0 = k\left( {t - 0} \right) \\ x = kt\ \ \ \ \ ...(1) \\\end{aligned} \end{equation} $$ Now, from the rate law equation, we can write $x$ as

$$\begin{equation} \begin{aligned} {C_t} = a - x = {C_0} - x \\ \Rightarrow x = {C_0} - {C_t} \\\end{aligned} \end{equation} $$

Put the value of $x$ in $(1)$, we get $$\begin{equation} \begin{aligned} \therefore {C_0} - {C_t} = kt \\ {C_t} = {C_0} - kt \\\end{aligned} \end{equation} $$

Interpretation of zero order reaction

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

• From the above derived rate equation, we can calculate the time required to complete the reaction i.e., all the reactants converted to product. When it happens means concentration at time $t$, ${C_t} = 0$. We get $$\begin{equation} \begin{aligned} {C_t} = {C_0} - kt \\ 0 = {C_0} - kt \\ t = \frac{{{C_0}}}{k} \\\end{aligned} \end{equation} $$

• We can find the unit of rate constant for zero order reaction $$\begin{equation} \begin{aligned} {C_t} = {C_0} - kt \\ k = \frac{{{C_0} - {C_t}}}{t} = \frac{{conc.}}{{time}} = \frac{{mol/litre}}{{\sec }} = molli{t^{ - 1}}{\sec ^{ - 1}} \\\end{aligned} \end{equation} $$

• 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 $$\begin{equation} \begin{aligned} {C_t} = {C_0} - kt \\ \frac{{{C_0}}}{2} = {C_0} - k{t_{\frac{1}{2}}} \\ k{t_{\frac{1}{2}}} = {C_0} - \frac{{{C_0}}}{2} = \frac{{{C_0}}}{2} \\ {t_{\frac{1}{2}}} = \frac{{{C_0}}}{{2k}} \\\end{aligned} \end{equation} $$ Half life period for zero order reaction is constant and directly proportional to the initial concentration of the reactant.

• Generally, decomposition of gases on metal surface at high concentration follows zero order reaction. For example $${H_2} + C{l_2} \to 2HCl$$