Given the following bond dissociation energies, calculated the enthalpy of the following reaction.CH4 + 2F2 → CF4 + 2H2Hint: You might need to draw Lewis Dot Structures for each of these before you can count the bonds on the reactant and product sides.C-F 450 kJ/molC-H 410 kJ/molF-F 158 kJ/molH-H 436 kJ/molGroup of answer choices318 kJ-716 kJ-318 kJ716 kJ
Question
Given the following bond dissociation energies, calculated the enthalpy of the following reaction.CH4 + 2F2 → CF4 + 2H2Hint: You might need to draw Lewis Dot Structures for each of these before you can count the bonds on the reactant and product sides.C-F 450 kJ/molC-H 410 kJ/molF-F 158 kJ/molH-H 436 kJ/molGroup of answer choices318 kJ-716 kJ-318 kJ716 kJ
Solution
To calculate the enthalpy of the reaction, we need to subtract the total energy of the bonds broken from the total energy of the bonds formed.
Step 1: Identify the bonds broken and formed In the reactants, we have 4 C-H bonds in CH4 and 2 F-F bonds in 2F2. In the products, we have 4 C-F bonds in CF4 and 2 H-H bonds in 2H2.
Step 2: Calculate the total energy of the bonds broken 4 C-H bonds * 410 kJ/mol = 1640 kJ/mol 2 F-F bonds * 158 kJ/mol = 316 kJ/mol Total energy of bonds broken = 1640 + 316 = 1956 kJ/mol
Step 3: Calculate the total energy of the bonds formed 4 C-F bonds * 450 kJ/mol = 1800 kJ/mol 2 H-H bonds * 436 kJ/mol = 872 kJ/mol Total energy of bonds formed = 1800 + 872 = 2672 kJ/mol
Step 4: Calculate the enthalpy of the reaction ΔH = Energy of bonds formed - Energy of bonds broken ΔH = 2672 kJ/mol - 1956 kJ/mol = 716 kJ/mol
So, the enthalpy of the reaction is 716 kJ/mol.
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