Problem 34 - Chemistry and Chemical Reactivity

Question

Hydrogenation, the addition of hydrogen to an organic compound, is an industrially important reaction. Calculate \(\Delta_{\mathrm{r}} H^{\circ}, \Delta_{\mathrm{r}} S^{\circ},\) and \(\Delta_{\mathrm{r}} G^{\circ}\) for the hydrogenation of octene, \(\mathrm{C}_{8} \mathrm{H}_{16},\) to give octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) at \(25^{\circ} \mathrm{C} .\) Is the reaction product- or reactant-favored at equilibrium? $$\mathrm{C}_{8} \mathrm{H}_{16}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{g})$$ Along with data in Appendix \(L\), the following information is needed for this calculation. $$\begin{array}{lll} \text { Compound } & \Delta_{f} H^{\circ}(\mathrm{k} \mathrm{J} / \mathrm{mol}) & S^{\circ}(\mathrm{J} / \mathrm{K} \cdot \mathrm{mol}) \\ \hline \text { Octene } & -82.93 & 462.8 \\ \text { Octane } & -208.45 & 463.639 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

The reaction is product-favored with \(\Delta_{\mathrm{r}} G^{\circ} = -86.841 \ \mathrm{kJ/mol}\).

Step 1: Define the Reaction

The given reaction is the hydrogenation of octene (\(\mathrm{C}_8\mathrm{H}_{16}\)) to form octane (\(\mathrm{C}_8\mathrm{H}_{18}\)). The balanced chemical equation for this reaction is: \[ \mathrm{C}_8\mathrm{H}_{16}(\mathrm{g}) + \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{C}_8\mathrm{H}_{18}(\mathrm{g}) \] The task is to calculate the standard enthalpy change (\(\Delta_{\mathrm{r}} H^{\circ}\)), the standard entropy change (\(\Delta_{\mathrm{r}} S^{\circ}\)), and the standard Gibbs free energy change (\(\Delta_{\mathrm{r}} G^{\circ}\)).

Step 2: Calculate Standard Enthalpy Change (\(\Delta_{\mathrm{r}} H^{\circ}\))

Use the formula for the standard enthalpy change of a reaction: \[ \Delta_{\mathrm{r}} H^{\circ} = \sum \Delta_{f} H^{\circ} \text{(products)} - \sum \Delta_{f} H^{\circ} \text{(reactants)} \] Substitute the given values: \(\Delta_{f} H^{\circ}_{\mathrm{octane}} = -208.45 \ \mathrm{kJ/mol}\) and \(\Delta_{f} H^{\circ}_{\mathrm{octene}} = -82.93 \ \mathrm{kJ/mol}\). For hydrogen, \(\Delta_{f} H^{\circ}_{\mathrm{H}_2} = 0 \ \mathrm{kJ/mol}\) as it is an element in its standard state. Thus, \[ \Delta_{\mathrm{r}} H^{\circ} = \left( -208.45 \right) - \left( -82.93 + 0 \right) = -125.52 \ \mathrm{kJ/mol} \]

Step 3: Calculate Standard Entropy Change (\(\Delta_{\mathrm{r}} S^{\circ}\))

Use the formula for the standard entropy change of a reaction: \[ \Delta_{\mathrm{r}} S^{\circ} = \sum S^{\circ} \text{(products)} - \sum S^{\circ} \text{(reactants)} \] Given \(S^{\circ}_{\mathrm{octane}} = 463.639 \ \mathrm{J/K \, mol}\), \(S^{\circ}_{\mathrm{octene}} = 462.8 \ \mathrm{J/K \, mol}\), and \(S^{\circ}_{\mathrm{H}_2} = 130.68 \ \mathrm{J/K \, mol}\), calculate \(\Delta_{\mathrm{r}} S^{\circ}\): \[ \Delta_{\mathrm{r}} S^{\circ} = 463.639 - (462.8 + 130.68) = -129.841 \ \mathrm{J/K \, mol} \]

Step 4: Calculate Standard Gibbs Free Energy Change (\(\Delta_{\mathrm{r}} G^{\circ}\))

Use the relationship: \[ \Delta_{\mathrm{r}} G^{\circ} = \Delta_{\mathrm{r}} H^{\circ} - T \Delta_{\mathrm{r}} S^{\circ} \] where \(T = 298.15 \ \mathrm{K}\). First convert \(\Delta_{\mathrm{r}} S^{\circ}\) to \(\mathrm{kJ/K \, mol}\): \( -129.841 \ \mathrm{J/K \, mol} = -0.129841 \ \mathrm{kJ/K \, mol}\). Then calculate \(\Delta_{\mathrm{r}} G^{\circ}\): \[ \Delta_{\mathrm{r}} G^{\circ} = -125.52 - (298.15 \times -0.129841) \] \[ \Delta_{\mathrm{r}} G^{\circ} = -125.52 + 38.679 = -86.841 \ \mathrm{kJ/mol} \]

Step 5: Determine Reaction Favorability at Equilibrium

Since \(\Delta_{\mathrm{r}} G^{\circ} < 0\), the reaction is spontaneous under standard conditions, indicating that the formation of octane from octene and hydrogen is product-favored at equilibrium.

Key Concepts

Hydrogenation reaction

In the world of chemistry, hydrogenation is a fundamental reaction where hydrogen molecules ( H_2 ) are added to another chemical compound. This process is widely utilized in industries, especially those dealing with organic compounds, to transform unsaturated compounds into saturated ones.
For example, the conversion of octene ( C_8H_{16} ) to octane ( C_8H_{18} ) is a classic hydrogenation reaction. When hydrogen is added to octene, the double bond is broken, allowing two hydrogen atoms to attach and transform it into octane.
Hydrogenation reactions typically require a catalyst, such as nickel, palladium, or platinum, to initiate the reaction effectively.

• It is an exothermic reaction, meaning it releases energy in the form of heat.
• Hydrogenation is essential for producing saturated fats in the food industry.
• It is also vital in converting unsaturated hydrocarbons into saturated ones in petrochemical industries.

Understanding hydrogenation is crucial as it influences many industrial processes, contributing to the manufacture of various everyday products.

Standard enthalpy change

The standard enthalpy change, \( \Delta_{\mathrm{r}} H^{\circ} \), represents the heat change associated with a chemical reaction at constant pressure. It is calculated under standard conditions, typically at a temperature of 25°C and 1 atm pressure.
For the hydrogenation reaction from octene to octane, the equation to determine \( \Delta_{\mathrm{r}} H^{\circ} \) involves subtracting the sum of the standard enthalpies of the reactants from the sum of the standard enthalpies of the products. As shown in the solution:
\[ \Delta_{\mathrm{r}} H^{\circ} = \sum \Delta_{f} H^{\circ} \text{(products)} - \sum \Delta_{f} H^{\circ} \text{(reactants)} \]
Substituting the given values provides us with \( \Delta_{\mathrm{r}} H^{\circ} = -125.52 \ \mathrm{kJ/mol} \). This negative value indicates that the reaction is exothermic, meaning it releases heat.

• Standard enthalpy values can be found in tables, often organized by compound and state.
• The enthalpy of formation for hydrogen in its natural state is zero.
• The negative enthalpy change suggests the products have lower energy than the reactants.

Comprehending standard enthalpy change is key to predicting whether the reaction will release or absorb energy.

Standard entropy change

Entropy, denoted as \( \Delta_{\mathrm{r}} S^{\circ} \), measures the disorder or randomness in a system. Like enthalpy, the change in entropy for a reaction is calculated under standard conditions. For reactions such as the hydrogenation of octene to octane, the standard entropy change can significantly influence the overall reaction spontaneity.
Entropy change is determined using the formula:
\[ \Delta_{\mathrm{r}} S^{\circ} = \sum S^{\circ} \text{(products)} - \sum S^{\circ} \text{(reactants)} \]
From the provided data, the \( \Delta_{\mathrm{r}} S^{\circ} \) for the hydrogenation process is \(-129.841 \ \mathrm{J/K \, mol}\). This negative value suggests a decrease in randomness or disorder in the system as the reaction progresses.

• Systems tend to move towards higher entropy states.
• A negative entropy change indicates that the system becomes more ordered.
• This often occurs when gases form liquids or solids, reducing movement freedom.

Understanding entropy change assists in predicting whether a reaction will proceed spontaneously based solely on entropy considerations.

Gibbs free energy change

Gibbs free energy, represented as \( \Delta_{\mathrm{r}} G^{\circ} \), is a pivotal thermodynamic quantity that combines enthalpy and entropy to determine the reaction spontaneity. In simpler terms, it tells us whether a chemical reaction will occur on its own.
The formula to calculate Gibbs free energy is:
\[ \Delta_{\mathrm{r}} G^{\circ} = \Delta_{\mathrm{r}} H^{\circ} - T \Delta_{\mathrm{r}} S^{\circ} \]
For the hydrogenation of octene, substituting the known values gives \( \Delta_{\mathrm{r}} G^{\circ} = -86.841 \ \mathrm{kJ/mol} \), indicating the reaction is spontaneous under standard conditions.
This negative value implies the products, octane, are thermodynamically favored compared to the reactants, octene and hydrogen.

• \( \Delta_{\mathrm{r}} G^{\circ} < 0 \) means the reaction proceeds without external energy.
• \( \Delta_{\mathrm{r}} G^{\circ} > 0 \) would mean the reaction is non-spontaneous.
• This helps chemists understand and predict reaction pathways and feasibilities.

Grasping Gibbs free energy is vital as it combines all thermodynamic aspects to elucidate the driving forces behind chemical reactions.