The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
Answer:
One mole of water was produced from this reaction.
Explanation:
According to this question, the following equation is given as follows:
2H2 + O2 → 2H2O
Two (2) moles of hydrogen gas produces two (2) moles of water in this balanced chemical equation.
If 1 mole of hydrogen gas was used, then:
1 × 2/2 moles of water would be produced
1 mole of water would be produced.
Answer:
Percent composition tells you which types of atoms (elements) are present in a molecule and their levels. Percent composition can also tell you about the different elements present in an ionic compound as well.
they all have one thing in common and that its all made up of atoms. When these components are active it creates energy
