Calculate the enthalpy change in the reaction of H2 with F2 to produce HF. The H–H and F–F bond energies are 436 and 155 kJ/mol, respectively, and the H–F bond energy is 567 kJ/mol
H2 (g) + F2 (g) --> 2HF (g)
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To calculate the enthalpy change in the reaction of H2 with F2 to produce HF, we will use the concept of bond energies. The enthalpy change can be calculated by subtracting the sum of the bond energies of the reactants from the sum of the bond energies of the products.
Let's start by calculating the bond energy of the reactants:
H2 (g) + F2 (g)
H-H bond energy = 436 kJ/mol
F-F bond energy = 155 kJ/mol
For H2, there are two H-H bonds, so the total bond energy is:
2 x 436 kJ/mol = 872 kJ/mol
For F2, there is one F-F bond, so the total bond energy is:
1 x 155 kJ/mol = 155 kJ/mol
Now let's calculate the bond energy of the products:
2HF (g)
H-F bond energy = 567 kJ/mol
For 2HF, there are two H-F bonds, so the total bond energy is:
2 x 567 kJ/mol = 1134 kJ/mol
Next, subtract the sum of the bond energies of the reactants from the sum of the bond energies of the products:
Enthalpy change = (sum of bond energies of products) - (sum of bond energies of reactants)
Enthalpy change = 1134 kJ/mol - (872 kJ/mol + 155 kJ/mol)
Enthalpy change = 1134 kJ/mol - 1027 kJ/mol
Enthalpy change = 107 kJ/mol
Therefore, the enthalpy change in the reaction of H2 with F2 to produce HF is 107 kJ/mol.