Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
The bond dissociation energy of the Cl - Cl bond is -958 kJ mol^-1.
<h3>What is the dissociation enthalpy?</h3>
Given that;
H-H bond energy = 435 kJ mol^-1
H-Cl bond energy = 431 kJ mol^-1
ΔHfO of HCL(g) = -92kJ mol^-1
Bond dissociation enthalpy of the Cl-Cl bond = x
-92 = 435 + 431 + x
x = -92 - (435 + 431)
x = -958 kJ mol^-1
Learn More about dissociation enthalpy:brainly.com/question/9998007?
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