What is the enthalpy change for the following reaction (the reaction of ethane with hydrogen bromide to form bromoethane) :
CH₂CH₂ + HBr $\rightarrow$ CH₃CH₂Br
Use the following bond energies where appropriate:
C-H = 412 kJ mol^-1; C-C = 348 kJ mol^-1; C=C = 610 kJ mol^-1; H-Br = 365 kJ mol^-1; C-Br = 280 kJ mol^-1.

Question

What is the enthalpy change for the following reaction (the reaction of ethane with hydrogen bromide to form bromoethane) :
CH₂CH₂ + HBr

\rightarrow

CH₃CH₂Br
Use the following bond energies where appropriate:
C-H = 412 kJ mol^-1; C-C = 348 kJ mol^-1; C=C = 610 kJ mol^-1; H-Br = 365 kJ mol^-1; C-Br = 280 kJ mol^-1.

Quizplus · Accepted Answer

The enthalpy change is calculated by subtracting the total bond energies of the products from the total bond energies of the reactants. Breaking the C=C bond (610 kJ/mol) and the H-Br bond (365 kJ/mol) requires energy, while forming two C-H bonds (2 x 412 kJ/mol) and one C-Br bond (280 kJ/mol) releases energy. The net change is -65 kJ/mol.

What Is the Enthalpy Change for the Following Reaction (The →\rightarrow→

What Is the Enthalpy Change for the Following Reaction (The $\rightarrow$