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${\mathbf{25}}^{\mathbf{°}}\mathit{C}$ .

Short answer

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

The entropy change of the reaction atrole="math" localid="1663902965633" $298Kat-120.32\frac{J}{mol\times K}$.

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

Step-by-step solution

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 at298K

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$298Kis-137.137\frac{k.J}{mol}$ .

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

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

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 $298Kis-120.32\frac{J}{mol\times K}$.

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

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 $298Kis-101.282\frac{k.J}{mol}$