Bonds broken: 6 Cl-F and 2 O=O.
Bonds formed: 2 Cl-O and 6 O-F
delta H = (sum reactants)-(sum products) = delta H rxn.
It is unclear is this is the reaction you want.
2ClF3(g) + 2O2(g) Cl2O(g) + 3OF2(g)
I did this:
1530 + 996 - (410 + 570) = 1546.
But the answer was wrong. I don't know where I did wrong.
Thanks!
Bonds formed: 2 Cl-O and 6 O-F
delta H = (sum reactants)-(sum products) = delta H rxn.
It is unclear is this is the reaction you want.
Here's how you can do it step by step:
1. Write down the balanced chemical equation for the reaction:
2ClF3(g) + 2O2(g) → Cl2O(g) + 3OF2(g)
2. Identify all the bonds that break in the reactant molecules:
In 2ClF3(g), you have 6 Cl-F bonds that need to be broken.
In 2O2(g), you have 4 O=O bonds that need to be broken.
3. Look up the bond energies for the bonds you identified. From the "Bond Energy" table, you provided:
Cl-F bond energy = 410 kJ/mol
O=O bond energy = 570 kJ/mol
4. Calculate the energy needed to break all the bonds in the reactants:
Energy needed to break Cl-F bonds = 6 Cl-F bonds * 410 kJ/mol = 2460 kJ/mol
Energy needed to break O=O bonds = 4 O=O bonds * 570 kJ/mol = 2280 kJ/mol
5. Identify all the bonds that form in the product molecules:
In Cl2O(g), you have 1 Cl=O bond that forms.
In 3OF2(g), you have 6 O-F bonds that form.
6. Look up the bond energies for the bonds you identified:
Cl=O bond energy = 630 kJ/mol
O-F bond energy = 710 kJ/mol
7. Calculate the energy released when all the bonds form in the products:
Energy released when Cl=O bond forms = 1 Cl=O bond * 630 kJ/mol = 630 kJ/mol
Energy released when O-F bonds form = 6 O-F bonds * 710 kJ/mol = 4260 kJ/mol
8. Calculate the energy change (ΔHo) for the reaction:
ΔHo = (Energy needed to break bonds in reactants) - (Energy released when bonds form in products)
= (2460 kJ/mol + 2280 kJ/mol) - (630 kJ/mol + 4260 kJ/mol)
= 4740 kJ/mol - 4890 kJ/mol
= -150 kJ/mol
Therefore, the estimated ΔHo for the reaction is -150 kJ/mol (negative indicates an exothermic reaction).