Question

The enthalpy of following reaction: H₂C=CH₂(g)+H₂(g)→CH₃ - CH₃(g) The bond energies of C - H,C - C,C = C, and H-H are 99,83,104 and 108 kcal respectively.

Answer 1

The enthalpy change for the given reaction is -4 kcal/mol.

Step-by-step explanation:

The enthalpy change of a reaction can be calculated using bond energies. Bond energy is the energy required to break one mole of a particular bond in a gaseous molecule. The given reaction involves the breaking and forming of several bonds.

Firstly, we identify the bonds broken and formed in the reaction. In the reactants, there is one C=C bond (104 kcal/mol) and one H-H bond (108 kcal/mol) broken. In the product, there are four C-H bonds (4 × 99 kcal/mol) formed.

The enthalpy change (ΔH) for the reaction is calculated using the formula:

`\[ΔH = \text{(sum of bond energies of bonds broken)} - \text{(sum of bond energies of bonds formed)}.\]`

Substituting the values:

`\[ΔH = (104 + 108) - (4 × 99) = -4 \, \text{kcal/mol}.\]`

The negative sign indicates that the reaction is exothermic, releasing 4 kcal/mol of energy. This means that the formation of CH₃-CH₃ is associated with the release of heat, and the reactants have higher energy compared to the products. The calculation demonstrates the application of bond energies in determining the enthalpy change of a chemical reaction, providing insights into the thermodynamics of the process.