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.
To calculate the average molar bond enthalpy of the carbon-bromine bond in a CBr4 molecule, we need to use the concept of Hess's law. Hess's law states that the enthalpy change of a reaction is independent of the pathway taken.
Let's break down the formation of CBr4 into two steps:
1. Break the carbon-bromine bonds in Br2(g).
∆H1 = 2 * ∆Hº(Br(g)) = 2 * 111.9 kJ/mol
2. Form the carbon-bromine bonds in CBr4(g).
∆H2 = ∆Hº(CBr4(g)) = 29.4 kJ/mol
Now, based on Hess's law, the overall enthalpy change for the reaction can be calculated by subtracting the energy used to break the bonds from the energy released in forming the bonds:
∆Hº(C(g)) + 2 * ∆Hº(Br(g)) = ∆Hº(CBr4(g))
∆Hº(C(g)) = ∆Hº(CBr4(g)) - 2 * ∆Hº(Br(g))
= 29.4 kJ/mol - 2 * 111.9 kJ/mol
= -194.4 kJ/mol
Therefore, the average molar bond enthalpy of the carbon-bromine bond in a CBr4 molecule is -194.4 kJ/mol. Keep in mind that the negative sign indicates that the process is exothermic.
To calculate the average molar bond enthalpy of the carbon-bromine (C-Br) bond in a CBr4 molecule, we need to consider the enthalpy changes involved in breaking and forming the bonds.
The average molar bond enthalpy can be calculated using the equation:
∆Hº(R-X) = Σ ∆Hº(reactants) - Σ ∆Hº(products)
where ∆Hº(R-X) represents the average molar bond enthalpy of the bond between atom R and atom X, Σ ∆Hº(reactants) represents the sum of the enthalpies of the reactants, and Σ ∆Hº(products) represents the sum of the enthalpies of the products.
In this case, we can use the given enthalpy changes:
∆Hº(Br(g)) = 111.9 kJ/mol
∆Hº(C(g)) = 716.7 kJ/mol
∆Hº(CBr4(g)) = 29.4 kJ/mol
To calculate the average molar bond enthalpy of the C-Br bond, we need to subtract the enthalpy changes of the reactants from the enthalpy change of the product.
∆Hº(C-Br) = ∆Hº(CBr4(g)) - ∆Hº(C(g)) - 4∆Hº(Br(g))
∆Hº(C-Br) = 29.4 kJ/mol - 716.7 kJ/mol - 4(111.9 kJ/mol)
∆Hº(C-Br) = 29.4 kJ/mol - 716.7 kJ/mol - 447.6 kJ/mol
∆Hº(C-Br) = -1135.4 kJ/mol
Therefore, the average molar bond enthalpy of the carbon-bromine (C-Br) bond in a CBr4 molecule is -1135.4 kJ/mol.
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.