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:
8.33 moles CO2 X. 25mol O2. 16mol CO2. = 13.0 moles
Refer to the attachment for answer.
SOME EXTRA INFORMATION:
Bromo is used for Bromine (Br)
Chloro is used for Chlorine(Cl)
'ene' represents double bond between carbon atoms.
'ol' is used for alcohol
HOPE IT IS USEFUL
To determine the amount of a certain element in a compound, we use the ratio of the elements from the compound. We calculate is follows:
45.0 g CCl4 ( 1 mol CCl4 / 153.82 g CCl4 ) ( 1 mol C / 1 mol CCl4 ) ( 12.01 g C / 1 mol C ) = 3.5135 g carbon present
Hope this answers the question. Have a nice day.
