Question

The decomposition of ethanol $\left(C_2{H}_5\mathrm{OH}\right)$on an alumina surface

${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\left(g\right)\mathbf{\to }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\left(g\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)$

was studied at 600K. Concentration versus time data were collected for this reaction and a plot of$\left[A\right]$versus time resulted in a straight line with a slope value of$\mathbf{-}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$.

a. Determine the rate law, integrated rate law and the value of rate constant for this reaction?

b. If the initial concentration of${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}$ was$\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{M}$, calculate the half-life for this reaction.

c. How much time is required for all of the$\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{M}$ to decompose?

Short answer

Zero order reaction:-

Those reactions for which the rate of reaction is directly proportional to zero power of reactant concentration are known as zero order reactions.

Therefore decomposition of ethanol $\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right)$on an alumina surface is a zero order reaction because most heterogeneous catalysis reactions are zero order.

Step-by-step solution

a. Rate law for the reaction

${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{g}\right)\to {\mathrm{C}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

At $$\begin{gathered} \mathrm{t}=0\mathrm{a}0 \\ \mathrm{a}-\mathrm{x}\mathrm{x} \end{gathered}$$

$\mathrm{Rate}=\frac{\mathrm{dx}}{\mathrm{dt}}\mathrm{\alpha }{\left(\mathrm{a}-\mathrm{x}\right)}^0$

$\frac{\mathrm{dx}}{\mathrm{dt}}={\mathrm{K}}_{\circ }{\left(\mathrm{a}-\mathrm{x}\right)}^0$

$\frac{\mathrm{dx}}{\mathrm{dt}}={\mathrm{K}}_{\circ }$

$\mathrm{dx}={\mathrm{K}}_{\circ }\mathrm{dt}$

Integrated rate law

On integration

$\int \mathrm{dx}={\mathrm{K}}_{\circ }\int \mathrm{dt}$

$\mathrm{x}={\mathrm{K}}_{\circ }\mathrm{t}+\mathrm{C}$

At $t=0,x=0$, therefore

$$\begin{gathered} 0=K_{\circ }t+C \\ C=0 \end{gathered}$$

So,

$$\begin{gathered} x=K_{\circ }t \\ {K}_{\circ }=\frac{x}{t} \end{gathered}$$

Or

Where,$K_{\circ }=$ Rate constant for zero order

$C=$Intercept

$x=$Reactant, which is converted into a product

We know

$$\begin{gathered} x=\left[A_{\circ }\right]-\left[A_t\right] \\ \left[A_{\circ }\right]-\left[A_t\right]=K_{\circ }t \\ \left[A_t\right]=-K_{\circ }t+\left[A_{\circ }\right] \end{gathered}$$

Plot the graph

The plot of $\left[A_t\right]$versus time

Therefore the value of the rate constant for this reaction is ${\mathrm{K}}_{\mathrm{o}}=-4\times {10}^{-5}{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}$because.$\mathrm{slope}=-{\mathrm{K}}_{\circ }$.

b Given condition

Given, the initial concentration of${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}=1.25\times {10}^{-2}\mathrm{M}$

${\mathrm{K}}_{\circ }=-4\times {10}^{-5}{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}$

Half-life and zero-order reaction

The time required for a reactant to reach half of its original concentration is known as the half-life.

Or zero order reaction

$t\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\frac{9}{2K_{\circ }}${a= Initial concentration, $K_{\circ }=$Rate constant for Zero order}

Put the given values

$\begin{array}{rcl}t\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& =& \frac{1.25\times {10}^{-2}}{2\times \left(4\times {10}^{-5}\right)}\\ & =& \frac{1.25\times {10}^{-2}}{8\times {10}^{-5}}\\ & =& \frac{1250}{8}\\ & =& 156.25\sec \end{array}$

c. Find the required time

Time required to decompose all of ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(1.25\times {10}^{-2}\right)$.

For 100% decomposition

$$\begin{gathered} t=? \\ x=9 \end{gathered}$$

We know $K_{\circ }=\frac{x}{t}$ (Also, know $K=\frac{9}{2t_2^1}$)

role="math" localid="1663848376742" $$\begin{gathered} t=\frac{x}{K_0}=\frac{9}{K_0}=\frac{9}{9}\times 2t\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ t=2t\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ t=156.25\times 2 \\ t=312.50\sec \end{gathered}$$