Alpha Particles hope this helps
Answer:
The standard change in free energy for the reaction = - 437.5 kj/mole
Explanation:
The standard change in free energy for the reaction:
4 KClO₃ (s) → 3 KClO₄(s) + KCl(s)
Given that ΔGf(KClO3(s)) = -290.9 kJ/mol;
ΔGf(KClO4(s)) = -300.4 kJ/mol;
ΔGf(KCl(s)) = -409 kJ/mol
According to Hess's law
ΔGr (Free energy change of reaction)= ∑(Product free energy - reactant free energy)
⇒ ΔGr⁰ = {3 x (-300.4) + (-409)} - {3 x (- 290.9)}
= - 901.2 - 409 + 872.7
= - 437.5 kj/mole
Species that have a lone pair of electrons often donate electrons by resonance while substituents that are electron deficient take away electrons by resonance.
<h3>What is resonance?</h3>
The term resonace has to do with the movement of electron pairs in a molecule. Inductive effects has to do with the drawing of electron density towards an atom or bond.
The two effects depends on the nature of a substituent. For instance, species that have a lone pair of electrons often donate electrons by resonance while substituents that are electron deficient take away electrons by resonance.
The question is incomplete hence the exact nature of the substituents can not be determined.
Learn more about resonance: brainly.com/question/23287285?
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Be and F,
E.N of Fluorine = 3.98
E.N of Beryllium = 1.57
________
E.N Difference 2.41 (Ionic Bond)
For H and Cl,
E.N of Chorine = 3.16
E.N of Hydrogen = 2.20
________
E.N Difference 0.96 (Polar Covalent Bond)
For Na and O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic Bond)
For F and F,
E.N of Fluorine = 3.98
E.N of Fluorine = 3.98
________
E.N Difference 0.00 (Non-Polar Covalent Bond)
Result:
A polar covalent bond is formed between Hydrogen and Chlorine atoms.
