91影视

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:

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

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

data-custom-editor="chemistry" $\ln \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=鈭�\frac{{\mathrm{螖贬}}_{\text{x}脳\text{n}}^{掳}}{\mathrm{R}}\left[\frac{1}{{\mathrm{T}}_2}鈭�\frac{1}{{\mathrm{T}}_1}\right]$

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

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

Let us solve this further,

${\mathrm{K}}_{\text{c}}=?$

At ${\text{T}}_2=300\text{K}$

${\mathrm{K}}_1=2.9鈰�{10}^8$

At ${\text{T}}_1=2000\text{K}$

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

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

ln$\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=鈭�\frac{{\mathrm{螖贬}}_{\text{rxn}}^{\mathrm{o}}}{\mathrm{R}}\left[\frac{1}{{\mathrm{T}}_2}鈭�\frac{1}{{\mathrm{T}}_1}\right]$

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

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

In

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

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