Answer: Yes it is. Rust would be considered a chemical reaction because it changes the chemical makeup of the metal.
Explanation:
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: 3.00
Explanation: got the answer from my teachers notes
Answer:
T2 = 94.6 C
Explanation:
Use Clausius-Clayperyon equation.
ln P1/P2 = ∆Hvap/R (1/T2 - 1/T1) where R = 8.314 J/mol-K and T is in degrees K
P1 = 760 mmHg
P2 = 630 mmHg
T1 = 373 K
T2 = ?
∆Hvap = 40.7 kJ/mole
R = 0.008314 kJ/mole-K (NOTE: change R to units of kJ)
Plug in and solve for T2
ln 760 mmHg/630 mmHg = 40.7 kJ/mole (1/T2 - 1/373K)
T2 = 367.74 K = 94.6 C
