Question

The dimerization of butadiene was studied at 500K.

$\mathbf{2}{\mathbf{C}}_{\mathbf{4}}{\mathbf{H}}_{\mathbf{6}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }{\mathbf{C}}_{\mathbf{8}}{\mathbf{H}}_{\mathbf{12}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

The following data were obtained where $\mathbf{Rate}\mathbf{=}\frac{\mathbf{-}\mathbf{d}\mathbf{\left[}{\mathbf{C}}_{\mathbf{4}}{\mathbf{H}}_{\mathbf{6}}\mathbf{\right]}}{\mathbf{dt}}$

Determine the forms of integrated rate law, the differential rate law, and the value of the rate constant for their reaction.

Short answer

The integrated rate law will be $\frac{1}{\left[C_4{H}_6\right]}=Kt+\frac{1}{{\left[C_4{H}_6\right]}_0}$

The differential rate law will be $\frac{d\left[C_4{H}_6\right]}{dt}=-K{\left[C_4{H}_6\right]}^2$

The rate constant will be $0.014\mathrm{L}/\mathrm{mol}$

Step-by-step solution

Rate law

Dimerization reaction: -

A dimerization is an addition reaction in which two molecules of the same compound react with each other.

$2C_4{H}_6\left(g\right)\to C_8{H}_{12}\left(g\right)$

To decide whether the rate law for this reaction is first order or second order. We must see whether the plot of $\ln \left[C_4{H}_6\right]$versus time is a straight line [for first order] or the plot of $\frac{1}{\left[C_4{H}_6\right]}$ versus time is a straight line [for second order]

Since the second order plot of $\frac{1}{\left[C_4{H}_6\right]}$ versus time is a straight line, we can conclude the reaction is second order.

$\mathrm{Rate}=\mathrm{K}{\left[{\mathrm{C}}_4{\mathrm{H}}_6\right]}^2$

The differential rate law will be

$\frac{d\left[C_4{H}_6\right]}{dt}=-K{\left[C_4{H}_6\right]}^2$.

Integrated rate law.

$\begin{array}{rcl}\mathrm{Rate}& =& \mathrm{K}{\left[{\mathrm{C}}_4{\mathrm{H}}_6\right]}^2\\ \frac{\mathrm{d}\left[{\mathrm{C}}_4{\mathrm{H}}_6\right]}{\mathrm{dt}}& =& -\mathrm{K}{\left[{\mathrm{C}}_4{\mathrm{H}}_6\right]}^2\\ \underset{{\left[{\mathrm{C}}_4{\mathrm{H}}_6\right]}_0}{\overset{\left[{\mathrm{C}}_4{\mathrm{H}}_6\right]}{\int }}\frac{\mathrm{d}\left[{\mathrm{C}}_4{\mathrm{H}}_6\right]}{{\left[{\mathrm{C}}_4{\mathrm{H}}_6\right]}^2}& =& -\mathrm{K}\underset{0}{\overset{\mathrm{t}}{\int }}\mathrm{dt}\end{array}$

After integrating, we have

$$\begin{gathered} \frac{1}{{\left[C_4{H}_6\right]}_0}-\frac{1}{\left[C_4{H}_6\right]}=-Kt \\ \frac{1}{\left[C_4{H}_6\right]}=Kt+\frac{1}{{\left[C_4{H}_6\right]}_0} \end{gathered}$$

Rate constant calculation.

As we know, the slope of the straight line equals the value of the rate constant (from the graph).

The slope

$$\begin{gathered} \frac{\mathrm{\Delta y}}{\mathrm{\Delta x}}=\frac{76.92-66.66}{1246-604} \\ =0.014\mathrm{L}/\mathrm{mol} \end{gathered}$$

Here for second-order reactions slope is equal to the rate constant. Therefore, the rate constant will be $0.014\mathrm{L}/\mathrm{mol}$.