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
HNO₃ + H₂S → S + NO + H₂<span>O
Assign Oxidation Number:
L.H.S R.H.S
N in HNO</span>₃ = +5 +2 = N in NO
S in H₂S = -2 0 = S in S
Write Half cell Reactions:
Reduction Reaction:
3e⁻ + HNO₃ → NO -------(1)
Oxidation Reaction:
H₂S → S + 2e⁻ -------(2)
Multiply eq. 1 with 2 and eq. 2 with 3 to balance electrons.
6e⁻ + 2 HNO₃ → 2 NO
3 H₂S → 3 S + 6e⁻
Cancel e⁻s,
______________________________
2 HNO₃ + 3 H₂S → 2 NO + 3 S + H₂O
Balance Oxygen Atoms by multiplying H₂O with 4, Hydrogen atoms will automatically get balance.
2 HNO₃ + 3 H₂S → 2 NO + 3 S + 4H₂O
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