Question

Values of the bond enthalpy terms for \(\mathrm{C}=\mathrm{C}, \mathrm{C}-\mathrm{C}\) \(\mathrm{H}-\mathrm{H}\) and \(\mathrm{C}-\mathrm{H}\) bonds are 598,346,436 and \(416 \mathrm{kJ} \mathrm{mol}^{-1} .\) Determine the enthalpy change associated with the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) Compare your answer with that determined using values of \(\Delta_{\mathrm{f}} H^{\circ}\) for \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})\) and \(\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})\) of +52.5 and \(-83.8 \mathrm{kJ} \mathrm{mol}^{-1}\)

Short answer

The enthalpy change is -490 kJ/mol using bond enthalpies and -136.3 kJ/mol using formation values.

Step-by-step solution

Write the balanced equation for hydrogenation

The hydrogenation reaction of ethene (C_2H_4) converts it into ethane (C_2H_6). Write the balanced chemical equation for this reaction:\[ \mathrm{C}_2\mathrm{H}_4 (g) + \mathrm{H}_2 (g) \rightarrow \mathrm{C}_2\mathrm{H}_6 (g) \]

Identify bonds broken and formed

In the C_2H_4 to C_2H_6 reaction, the following bonds are broken and formed:- Break 1 \(\mathrm{C} = \mathrm{C}\) bond and 1 \(\mathrm{H} - \mathrm{H}\) bond.- Form 2 \(\mathrm{C} - \mathrm{C}\) bonds and 2 \(\mathrm{C} - \mathrm{H}\) bonds.

Calculate the total energy for bonds broken

Add the bond enthalpies for bonds that need to be broken:- \(\mathrm{C} = \mathrm{C}\): 598 kJ/mol- \(\mathrm{H} - \mathrm{H}\): 436 kJ/molThe total energy required to break these bonds is:\[598 + 436 = 1034\ \mathrm{kJ/mol} \]

Calculate the total energy for bonds formed

Add the bond enthalpies for bonds that are formed:- 2 \(\mathrm{C} - \mathrm{C}\) bonds: 2(346 kJ/mol)- 2 \(\mathrm{C} - \mathrm{H}\) bonds: 2(416 kJ/mol)The total energy released in forming these bonds is:\[2(346) + 2(416) = 692 + 832 = 1524\ \mathrm{kJ/mol} \]

Determine the enthalpy change

Subtract the total energy for bonds formed from the total energy for bonds broken:\[\Delta H = 1034\ \mathrm{kJ/mol} - 1524\ \mathrm{kJ/mol} = -490\ \mathrm{kJ/mol}\]

Compare with standard enthalpy of formation values

Calculate the enthalpy of the reaction using standard formation values: - Enthalpy change using formation values:\[ \Delta H^\circ = \Delta_f H^\circ (\mathrm{C}_2 \mathrm{H}_6) - \Delta_f H^\circ (\mathrm{C}_2 \mathrm{H}_4) \]\[ \Delta H^\circ = -83.8 - 52.5 = -136.3\ \mathrm{kJ/mol} \]

Final Comparison

The bond enthalpy method gives an enthalpy change of -490 kJ/mol, while using standard enthalpy of formation values gives -136.3 kJ/mol. These differences typically arise due to the average values used in bond enthalpies.

Key concepts

Bond Enthalpy

Understanding bond enthalpy is essential for estimating the energy changes involved in chemical reactions. Bond enthalpy refers to the energy required to break one mole of a specific type of bond in a molecule, measured in kilojoules per mole (kJ/mol).

• It is an average value, often used for similar bonds in various molecules.
• Considered as positive because energy input is necessary to break bonds.

In our example involving ethene (0_2H4), we break a carbon-carbon double bond (2C) and a hydrogen-hydrogen bond (H-H). The energy required for these is calculated by summing up their bond enthalpies. For ethene, this sum is 1,034 kJ/mol. Breaking bonds always requires input energy, making this process endothermic. The released energy upon forming new bonds is calculated similarly, but results in a release, being an exothermic process.

Hydrogenation

Hydrogenation is a chemical reaction where hydrogen is added to a compound, typically involving unsaturated organic compounds. In this reaction, the pi bonds of unsaturated hydrocarbons, like alkenes, are replaced by sigma bonds with hydrogen.

• In the process, an unsaturated molecule becomes saturated.
• Catalysts are often used to facilitate the reaction under milder conditions.

For ethene hydrogenation, the balanced reaction is written as: 0_2H4 + H2 0 0_2H6. Here, ethene gains hydrogen, resulting in the formation of ethane. In this specific case, the more stable saturated molecule 0_2H6 is formed, and we have an associated enthalpy change due to bond breaking and formation.

Enthalpy of Formation

The enthalpy of formation (fH) of a substance indicates the change in enthalpy when one mole of compound is formed from its elements in their standard states.

• It is a critical factor in calculating the overall energy change of a reaction.
• Values can be positive or negative depending on whether energy is absorbed or released.

In hydrogenation, the enthalpy change can also be calculated by using the enthalpy of formation values. For instance, the standard enthalpy of formation of ethane (0_2H6) is given as -83.8 kJ/mol and of ethene (0_2H4) is +52.5 kJ/mol. Thus, the reaction's enthalpy change, when calculated with these standard enthalpy values, is found to be -136.3 kJ/mol.

Energy Calculation

Energy calculation in reactions involves determining the enthalpy change (H) by comparing energies required for breaking bonds and energies released in forming bonds. This calculation helps in understanding whether a reaction is exothermic or endothermic.

• An exothermic reaction releases energy, resulting in negative H.
• Conversely, an endothermic reaction absorbs energy, reflected by a positive H.

For the hydrogenation of ethene, the total energy required to break the 2C and H-H bonds was 1,034 kJ/mol. In contrast, 1,524 kJ/mol energy was released upon forming the new C-C and C-H bonds, making H = -490 kJ/mol. Comparing this energy calculation with the standard enthalpy of formation highlights differences mainly due to averaging in bond enthalpy techniques. These insights are vital for predicting reaction behavior and energy outcomes.