Second Order Reaction - Practice Questions & MCQ

Consider the reaction

$\mathrm{A} \rightarrow \mathrm{P}$

For second-order reaction, the rate law is given as follows:

Rate $=\mathrm{k}[\mathrm{A}]^2=-\frac{\mathrm{dA}}{\mathrm{dt}}$

• Integrated Rate Law: Since it is a second-order reaction, thus:
  $\frac{-\mathrm{dA}}{\mathrm{dt}}=\mathrm{k}[\mathrm{A}]^2$
  Integrating both sides, we get:
  $\int_{\mathrm{A}_{\mathrm{o}}}^{\mathrm{A}} \frac{\mathrm{dA}}{\mathrm{A}^2}=-\int_0^{\mathrm{t}} \mathrm{kdt}$
  
  where A_o is the initial concentration of A at time t = 0
  A is the remaining concentration of A after time 't'
  
  $\Rightarrow\left[\frac{-1}{\mathrm{~A}}\right]_{\mathrm{A}_{\mathrm{o}}}^{\mathrm{A}}=-\mathrm{k}(\mathrm{t})_0^{\mathrm{t}}$
  
  $\Rightarrow\left[\frac{1}{\mathrm{~A}}\right]_{\mathrm{A}_{\mathrm{o}}}^{\mathrm{A}}=\mathrm{kt}$
  
  $
  \Rightarrow \frac{1}{\mathrm{~A}}-\frac{1}{\mathrm{~A}_{\mathrm{o}}}=\mathrm{kt}
  $
  
  
  Thus, $\frac{1}{[\mathbf{A}]}=\mathbf{k t}+\frac{\mathbf{1}}{\left[\mathbf{A}_{\mathbf{o}}\right]}$
• Half-life(t_1/2): We know that the half-life for a reaction is the time when the concentration of the reactant(A) is half of its initial value.
  Thus, at time t = t_1/2, A = A_o/2
  From equation (i) we have:
  
  $\mathrm{t}=\frac{1}{\mathrm{k}}\left[\frac{1}{\mathrm{~A}}-\frac{1}{\mathrm{~A}_{\mathrm{o}}}\right]$
  
  $\Rightarrow \mathrm{t}_{1 / 2}=\frac{1}{\mathrm{k}}\left[\frac{1}{\mathrm{~A}_{\mathrm{o}} / 2}-\frac{1}{\mathrm{~A}_{\mathrm{o}}}\right]$
  
  Thus, $\mathbf{t}_{1 / 2}=\frac{1}{\mathrm{kA}_{\mathrm{o}}}$
  This is the half-life for the second-order reaction.