Answer:
-125 kJ
Explanation:
You calculate the energy required to break all the bonds in the reactants. Then you subtract the energy to break all the bonds in the products.
H₂C=CH₂ + H₂ ⟶ H₃C-CH₃
Bonds: 4C-H + 1C=C 1H-H 6C-H + 1C-C
D/kJ·mol⁻¹: 413 612 436 413 347
The formula relating ΔHrxn and bond dissociation energies (D) is
ΔHrxn = Σ(Dreactants) – Σ(Dproducts)
(Note: This is an exception to the rule. All other thermochemical reactions are “products – reactants”. With bond energies, it’s “reactants – products”. The reason comes from the way we define bond energies.)
<em>For the reactant</em>s:
Σ(Dreactants) = 4 × 413 + 1 × 612 + 1 × 436 = 2700 kJ
<em>For the products:</em>
Σ(Dproducts) = 6 × 413 + 1 × 347 = 2825 kJ
<em>For the system</em>
:
ΔHrxn = 2700 - 2825 = -125 kJ
There are a lot of ways to increase the solubility of the solute. <span>Increasing the temperature, mixing time and surface area of a solvent increases the solubility of the solute</span>
Potential to kenetic energy
<span>Of all the sub-shells shown ( 1s ,2s ,2p ,3s ,3p ,4s ,3d ,4p ,5s ,4d ,5p ,6s ,4f ,5d) the ones that act as core orbital of HF (Hydrogen Fluoride) is 6s and 5d</span>
