Question

Consider the following initial rate data for the decomposition

of compound AB to give A and B:

${\left[\mathrm{AB}\right]}_{\mathbf{0}}\left(\frac{\mathrm{mol}}{l}\right)$	$\mathbf{Initialrate}\left({\mathrm{molL}}^{-1}{s}^1\right)$
0.200	$\mathbf{3}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$
0.400	$\mathbf{1}\mathbf{.}\mathbf{28}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}$
0.600	$\mathbf{2}\mathbf{.}\mathbf{88}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}$

Determine the half-life for the decomposition reaction initially having 1.00 M AB present.

Short answer

The half-life (t ½) for second order reaction is 12.195 s.

Step-by-step solution

Calculating the order of the reaction

From the given data,

Half-life to calculate from 1.00 M AB.

The rate law can be given as

$\mathit{R}\mathit{a}\mathit{t}\mathit{e}\mathbf{=}\mathit{k}{\left[AB\right]}^{\mathbf{2}}$

So, by substituting the values,

$3.2\times {10}^{-3}=k{\left[0.200\right]}^n$ …… (1)

$1.28\times {10}^{-3}=k{\left[0.400\right]}^n$ …… (2)

Dividing the (2) by (1),

$\frac{1.28\times {10}^{-3}}{3.2\times {10}^{-3}}=k{\left[\frac{0.400}{k200}\right]}^n$

$\begin{array}{l}4=2^n\\ n=2\end{array}$

Therefore, the order of the reaction is second.

Calculating the rate constant for the reaction

The rate constant for second order reaction

$$\begin{gathered} 3.28\times {10}^{-3}=k0.{200}^2 \\ k=0.082{\mathrm{M}}^{-1}{\mathrm{s}}^{-1} \end{gathered}$$

The rate constant for the reaction is $0.082{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$.

Calculating the half-life (t ½) for the reaction

For half-life of second order reaction,

$\begin{array}{c}{t}_{\frac{1}{2}}=\frac{1}{\left[A\right]0k}\\ =\frac{1}{1\left(0.082\right)}\\ =12.195\mathrm{s}\end{array}$

The half-life (t ½) for second order reaction is 12.195 s.