Question

The standard heats of formation for CH4, CHCl3, and HCl are -74.8, -132, -92.0 KJ/mole, respectively. Use this information to calculate the heat of reaction for the following:

CH4+3Cl2=CHCl3+3HCl

Answer 1

$\Delta{H_{reaction}}=\sum{\Delta{H_f({products)}}}-\sum{\Delta{H_f({reactants)}}}=3\times(-92.0)+(-132)-(-74.8)=-333kJ/mol$

Answer: -333 kJ/mol