Question

Hess' Law and bond energy data can be used to calculate the enthalpy change of a reaction. Bromoethane, CH2CH2Br, can be made by reacting ethene with hydrogen bromide. CH2=CH2+HBr→CH3CH2Br What is the enthalpy change for this reaction? A−674 kJ mol−1 B−64 kJ mol−1 C+186 kJ mol−1 D+346 kJ mol−1

Answer 1

To calculate the enthalpy change for the reaction between ethene and hydrogen bromide, we can use bond energy data and Hess' Law. The C=C bond and the H-Br bond are broken in the reactants, and a C-Br bond is formed in the product. By looking up the bond dissociation energies and applying the formula for enthalpy change, we can calculate the approximate enthalpy change for the reaction to be approximately +697 kJ/mol.

Step-by-step explanation:

To calculate the enthalpy change for the reaction between ethene and hydrogen bromide, we can use bond energy data and Hess' Law. First, we need to identify the bonds that are broken and formed in the reaction.

In the reaction, the ethene molecule has a C=C double bond which is broken, and the hydrogen bromide molecule has a H-Br bond which is also broken. These bonds are broken in the reactants.

After that, a new bond is formed between the carbon atom in ethene and the bromine atom in hydrogen bromide, resulting in the formation of the bromoethane molecule. So, a new C-Br bond is formed in the product.

We can then look up the bond dissociation energies for the bonds involved in the reaction. The C=C bond has an average bond energy of 607 kJ/mol, the H-Br bond has an average bond energy of 366 kJ/mol, and the C-Br bond has an average bond energy of 276 kJ/mol.

Using these bond energies and applying the formula for enthalpy change, we can calculate the approximate enthalpy change for the reaction:

ΔH = (Energy of bonds broken in reactants) - (Energy of bonds formed in products)

ΔH = (607 kJ/mol + 366 kJ/mol) - 276 kJ/mol

ΔH ≈ +697 kJ/mol

Therefore, the enthalpy change for the reaction is approximately +697 kJ/mol.