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.
Of what?
Attach a link or pic
Answer: 1.8 x 10^1 or the scientific e notation is: 1.8e1
Answer:
Liquid to Gas
Explanation:
The particles need energy to rise and over come the attractions between them as the liquid gets warmer more particles have sufficient, energy to escape from liquid. eventually even particles in the middle of the liquid form bubbles of gas in the liquid At this point the liquid is boiliing and turning into gas.
