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.

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

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

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:${\mathrm{K}}_{\text{c}}=2.0\cdot {10}^{34}$ .