Q17.93CP - Chemistry: Molecular Nature Of Matter And Change

Question

Hydrogenation of carbon-carbon $\mathbf{\pi }$ bonds is important in the petroleum and food industries. The conversion of acetylene to ethylene is a simple example of the process:

 ${\mathbf{\text{C}}}_{\mathbf{2}}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{\text{C}}}_{\mathbf{2}}{\mathbf{\text{H}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

The calculated ${\mathbf{K}}_{\mathbf{c}}$  at 2000K is $\mathbf{2}\mathbf{.9}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}$ . But the process is run at lower temperatures with the aid of a catalyst to prevent decomposition. Use  ${\mathbf{\Delta H}}_{\mathbf{\text{f}}}^{\mathbf{°}}$  values to calculate the ${\mathbf{K}}_{\mathbf{C}}$   at 300K.

Step-by-Step Solution

Answer

The required value of the given problem is:  $K_{\text{c}}=2.0\cdot {10}^{34}$.

Step 1: Definition of Pi bond

Pi Bonds are covalent chemical bonds between two different atoms that include the lateral overlapping of two lobes of the same orbital.

Step 2: Calculate the value of ∆ H

Hydrogenation of carbon-carbon p bonds:

 ${\text{C}}_2{\text{H}}_2\left(\mathrm{g}\right)+{\text{H}}_2\left(\mathrm{g}\right)\rightleftharpoons {\text{C}}_2{\text{H}}_4\left(\mathrm{g}\right)$

To calculate  data-custom-editor="chemistry" ${\mathrm{K}}_{\mathrm{C}}$ we will use Van't Hoff equation:

 data-custom-editor="chemistry" $\ln \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=-\frac{{\mathrm{\Delta H}}_{\text{x}\times \text{n}}^{°}}{\mathrm{R}}\left[\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right]$

Therefore, firstly we have to calculate the value of  data-custom-editor="chemistry" $\Delta H_{\text{rxn}}^{°}$ as:

 data-custom-editor="chemistry" $\begin{array}{c}{\mathrm{\Delta H}}_{\text{rxn}}^{°}={\mathrm{\Delta H}}^{°}\text{ (products) }-{\mathrm{\Delta H}}^{°}\text{ (reactants)}\\ {\mathrm{\Delta H}}_{\text{rxn}}^{°}={\mathrm{\Delta H}}^{°}\left({\text{C}}_2{\text{H}}_4\right)-({\mathrm{\Delta H}}^{°}\left({\text{C}}_2{\text{H}}_2\right)+{\mathrm{\Delta H}}^{°}\left({\text{H}}_2\right))\\ {\mathrm{\Delta H}}_{\text{rxn}}^{°}=52.47 \text{kJ}/\text{mol}-227 \text{kJ}/\text{mol}-0 \text{kJ}/\text{mol}\\ {\mathrm{\Delta H}}_{\text{rxn}}^{°}=-174.53 \text{kJ}/\text{mol}\end{array}$

Step 3: Use Van’t Hoff equation

Let us solve this further,

 ${\mathrm{K}}_{\text{c}}=?$

At  ${\text{T}}_2=300 \text{K}$

 ${\mathrm{K}}_1=2.9\cdot {10}^8$

At  ${\text{T}}_1=2000 \text{K}$

 $\text{R}=8.314 \text{J}/\text{mol}\cdot \text{L}$

So, use the Van't Hoff equation to express and calculate  data-custom-editor="chemistry" $\Delta H_{\text{rxn}}^{°}$ :

ln$\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=-\frac{{\mathrm{\Delta H}}_{\text{rxn}}^{\mathrm{o}}}{\mathrm{R}}\left[\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right]$

In  $\frac{K_c}{2.9\cdot {10}^8}=-\frac{-174.53 \text{kJ}/\text{mol}}{8.314 \text{J}/\text{mol}\cdot \text{L}}\left[\frac{1}{300 \text{K}}-\frac{1}{2000 \text{K}}\right]$

In  data-custom-editor="chemistry" $\frac{{\mathrm{K}}_{\mathrm{c}}}{2.9\cdot {10}^8}=\frac{174.53 \text{K}}{8.314}\left(\frac{1700 \text{K}}{600\cdot {10}^3 {\text{K}}^2}\right)\cdot {10}^3$

In

 $\begin{array}{c}\frac{{\mathrm{K}}_{\text{c}}}{2.9\cdot {10}^8}\\ =59.48\frac{{\mathrm{K}}_{\text{c}}}{2.9\cdot {10}^8}\\ ={\text{e}}^{59.48}\\ {\mathrm{K}}_{\text{c}}=2.0\cdot {10}^{34}\end{array}$

 

Hence, the required value of the given problem is: ${\mathrm{K}}_{\text{c}}=2.0\cdot {10}^{34}$  .