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)
Ooops did not mean to post this twice!
dHrxn = (n*BE reactants) - (n*BE products)
BE = bond energy.
To calculate the enthalpy change in the reaction, we need to calculate the energy required to break the bonds in the reactant molecules and the energy released when the bonds are formed in the product molecules.
Step 1: Calculate the energy required to break the bonds in H2 and F2.
Since there are two H–H bonds and one F–F bond in the reactants, the energy required to break these bonds can be calculated as follows:
Energy required to break H–H bonds = 2 * H–H bond energy
= 2 * 436 kJ/mol
= 872 kJ/mol
Energy required to break F–F bond = 1 * F–F bond energy
= 1 * 155 kJ/mol
= 155 kJ/mol
Step 2: Calculate the energy released when the bonds in HF are formed.
Since there are two H–F bonds formed in the product, the energy released when these bonds are formed can be calculated as:
Energy released when H–F bonds are formed = 2 * H–F bond energy
= 2 * 567 kJ/mol
= 1134 kJ/mol
Step 3: Calculate the enthalpy change of the reaction.
The enthalpy change is given by the difference between the energy required to break the bonds in the reactants and the energy released when the bonds are formed in the products:
Enthalpy change = (Energy required to break bonds in H2 + Energy required to break bonds in F2) - (Energy released when H–F bonds are formed)
= (872 kJ/mol + 155 kJ/mol) - 1134 kJ/mol
= 1027 kJ/mol - 1134 kJ/mol
= -107 kJ/mol
Therefore, the enthalpy change in the reaction of H2 with F2 to produce HF is -107 kJ/mol.
To calculate the enthalpy change in the reaction, we need to consider the difference in the energy of the bonds broken and formed in the reaction. The enthalpy change (ΔH) can be calculated using the following equation:
ΔH = Σ(Bonds Broken) - Σ(Bonds Formed)
Let's calculate it step by step:
1. Bonds broken:
In the reaction, we have one H-H bond and one F-F bond broken.
Bonds broken = (1 mol H-H) × (436 kJ/mol) + (1 mol F-F) × (155 kJ/mol)
2. Bonds formed:
In the reaction, we have two H-F bonds formed.
Bonds formed = (2 mol H-F) × (567 kJ/mol)
3. Enthalpy change:
ΔH = Bonds broken - Bonds formed
Now, let's substitute the values and calculate:
Bonds broken = (1 mol H-H) × (436 kJ/mol) + (1 mol F-F) × (155 kJ/mol)
= 436 kJ/mol + 155 kJ/mol
= 591 kJ/mol
Bonds formed = (2 mol H-F) × (567 kJ/mol)
= 2 × 567 kJ/mol
= 1134 kJ/mol
ΔH = Bonds broken - Bonds formed
ΔH = 591 kJ/mol - 1134 kJ/mol
ΔH = -543 kJ/mol
Therefore, the enthalpy change in the reaction of H2 with F2 to produce HF is -543 kJ/mol. Since the enthalpy change is negative, it indicates that the reaction is exothermic, meaning it releases energy in the form of heat.