Answer:
ΔS° = 180.5 J/mol.K
Explanation:
Let's consider the following reaction.
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)
The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.
ΔS° = ∑np × S°p - ∑nr × S°r
where,
ni are the moles of reactants and products
S°i are the standard molar entropies of reactants and products
ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))
ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol
ΔS° = 180.5 J/K
This is the change in the entropy per mole of reaction.
Answer:
H₂(g) + Cl₂(g) → 2HCl(g) + 185kJ
Explanation:
In a chemical reaction, enthalpy of reaction ΔH is a thermodynamic constant that gives information if the reaction is exothermic (Produce heat if reacts) or endothermic (Consume heat if reacts).
In the reaction:
H₂(g) + Cl₂(g) → 2HCl(g) ΔH = -185kJ
As ΔH <0, the reaction is exothermic, that means, <em>produce heat</em>, writing a balanced thermochemical equation:
<em>H₂(g) + Cl₂(g) → 2HCl(g) + 185kJ</em>
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The enthalpy is as a product beacause an exothermic reaction produces heat.
I hope it helps!
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<span>Conductor, and there you go, i hope this helped but if its wrong, i am extremly sorry</span>
And the significant amount of volume can be differed by its solitude
