Question
Given that
∆Hº(Br(g)) = 111.9 kJ/mol
∆Hº(C(g)) = 716.7 kJ/mol
∆Hº(CBr4(g)) = 29.4 kJ/mol
calculate the average molar bond enthalpy of the carbon-bromine bond in a CBr4 molecule.
∆Hº(Br(g)) = 111.9 kJ/mol
∆Hº(C(g)) = 716.7 kJ/mol
∆Hº(CBr4(g)) = 29.4 kJ/mol
calculate the average molar bond enthalpy of the carbon-bromine bond in a CBr4 molecule.
Answers
I believe its products - reactants, but I am not sure. maybe Dr. Bob222 will come along and give you the right answer.
29.4- 828.6=-819.2kj/mol
Exothermic reaction.
29.4- 828.6=-819.2kj/mol
Exothermic reaction.