Q95CP - Chemistry: Molecular Nature Of Matter And Change

Question

Hydrogenation is the addition of${\mathit{H}}_{\mathbf{2}}$   to double (or triple) carbon-carbon bonds. Peanut butter and most commercial baked goods include hydrogenated oils. Find   $\mathbf{∆}{\mathit{H}}^{\mathbf{°}}\mathbf{,}\mathbf{ }\mathbf{∆}{\mathit{S}}^{\mathbf{°}}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathbf{∆}{\mathit{G}}^{\mathbf{°}}$for the hydrogenation of ethene    to ethane$\left(C_2{H}_6\right)$   at${\mathbf{25}}^{\mathbf{°}}\mathit{C}$  .

Step-by-Step Solution

Answer

The enthalpy change of the reaction at$298K at -137.137\frac{k·J}{mol}$  .

The entropy change of the reaction at  $298K at -120.32\frac{J}{mol\times K}$ .

The Gibb's free energy change of the reaction  $298K is -101.282\frac{k·J}{mol}$

Step 1: Concept Introduction

The total heat present in a thermodynamic system with constant pressure is measured by enthalpy.

The measure of disorder in a thermodynamic system is entropy.

Gibbs free energy is a thermodynamic quantity equal to the enthalpy (of a system or process) minus the entropy-absolute-temperature product.

Step 2: Calculating the enthalpy change of the reaction at 298 K

Let us write the chemical reaction for hydrogenation of ethene:

 ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(g) + H}}_{\text{2}}\text{(g) }\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}\text{(g)}$

The enthalpy of the reaction occurring at ${25}^{°}C\left(298K\right)$  can be calculated using Appendix B:

 $$\begin{gathered} ∆H_{rxn}^{°}=∆H_{product}^{°}-∆H_{reac\tan t}^{°} \\ ∆H_{Txn}^{°}=\left[1∆H^{°}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{(g)}\right)\right]-\left[1∆H^{°}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{(g)}\right)+1∆H^{°}\left({\text{H}}_{\text{2}}\text{(g)}\right)\right] \\ ∆H_{Txn}^{°}=\left[1mol\times \left(-84.667\frac{kJ}{mol}\right)\right]-\left[1mol\times \left(52.47\frac{kJ}{mol}\right)\_1mol\times \left(0.00\frac{kJ}{mol}\right)\right] \\ ∆H_{rxn}^{°}=-137.137\frac{k.J}{mol} \end{gathered}$$

The enthalpy change of the reaction at$298K is -137.137\frac{k.J}{mol}$  .

Step 3: Calculating the entropy change of the reaction at 298 K

The enthalpy change of the reaction at   $298K is -137.137\frac{K.J}{mol}$

Similarly, the entropy of the reaction at ${25}^{°}C\left(298K\right)$ can be calculated using Appendix B:

 $$\begin{gathered} ∆S_{ren}^{°}=∆S_{product}^{°}-∆S_{reac\tan t}^{°}∆S_{Txn}^{°} \\ =\left[1∆S^{°}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{(g)}\right)\right]-\left[1∆S^{°}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{(g)}\right)+1∆S^{°}\left({\text{H}}_{\text{2}}\text{(g)}\right)\right] \\ ∆S_{Txn}^{°}=\left[1mol\times \left(229.50\frac{J}{mol\times K}\right)\right]-\left[1mol\times \left(219.22\frac{J}{mol\times K}\right)+1mol\times \left(130.60\frac{J}{mol\times K}\right)\right] \\ ∆S_{Txn}^{°}=-120.32\frac{J}{mol\times K} \end{gathered}$$ 

The entropy change of the reaction at $298K is -120.32\frac{J}{mol\times K}$ .

Step 4: Calculating the Gibb's free energy change of the reaction at 298 K

Let us calculate Gibb’s free energy,

 $∆G_{rxn}^{°}$ can be calculated as,

$$\begin{gathered} ∆G_{rxn}^{°}=∆H_{rxn}^{°}-T\times ∆S_{rxn}^{°} \\ ∆G_{rxn}^{°}=-137.137\frac{kJ}{mol}-298K\times \left(-120.32\times {10}^{-3}\frac{kJ}{mol\times K}\right) \\ ∆G_{rxn}^{°}=-101.282\frac{k.J}{mol} \end{gathered}$$

 

Gibb's free energy change of the reaction at $298K is -101.282\frac{k.J}{mol}$