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:
2666.7 hours
Explanation:
The key to solve this problem is that we are given the propane gas consumed in one hour by giving us the information of the volume consumed at 1 atm, 298 K (25 +273). Using the gas law we can calculate the rate of consumption of propane per hour, and from here we can calculate its mass and converting it to gallons and finally diving the 400 gallos by this number.
PV = nRT ∴ n = PV/RT
n = 1 atm x 165 L/ (0.08206 Latm/kmol x 298 K ) = 6.75 mol propane
Mass propane :
6.75 mol x 44 g/mol = 296.88 g
convert this to Kg:
296.88 g/ 1000 g/Kg = 0.30 Kg
calculate the volume in liters this represents by dividing by the density:
0.30 Kg / 0.5077 Kg/L = 0.59 L
changing this to gallons
0.59 L x 1 gallon/3.785 L = 0.15 gallon
and finally calculate how many hours the 400 gallons propane tank will deliver
400 gallon/ 0.15 gallon/hr = 2666.7 hr
Uh.....Idk i could hardly read it sorry
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