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
Answer:
The answer to your question is 7.64 atm
Explanation:
Data
Volume 1 = V1 = 2.1 l
moles = 4.65
Pressure 1 = P1 = 3.82 atm
Volume 2 = Volume 1/2
Pressure 2 = ?
Process
1.- Calculate the new volume
Volume 2 = 2.10/2
Volume 2 = 1.05 l
2.- Use Boyle's law to find the Pressure
P1V1 = P2V2
-Solve for P2
P2 = P1V1 / V2
-Substitution
P2 = (3.82)(2.1) / 1.05
-Simplification
P2 = 8.022/1.05
-Result
P2 = 7.64 atm
