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:
+125.4 KJmol-1
Explanation:
∆H C4H10(g) = -2877.6kJ/mol
∆H C(s)=-393.5kJ/mol
∆H H2(g) = -285.8
∆H reaction= ∆Hproducts - ∆H reactants
∆H reaction= (-2877.6kJ/mol) - [4(-393.5kJ/mol) +5(-285.8)]
∆H reaction= +125.4 KJmol-1
Hi! Well the formula of kinetic energy is Ke = 0.5 x M x V^2
0.5 x 0.5 x 10^2
0.25 x 20
5 J (joule)
Gain enough kinetic energy to get past each other. Ad you heat up a substance, the temperature increases as does the kinetic energy of the particles. At a point the temperature of the substance will stop increasing. The energy is now being used to increase the potential and move the particles further apart.
