Consider the reaction: 2NH3 (9) + 3N2O(g)→ 4N2(g) + 3H₂O(g) Using standard thermodynamic data at 298 K, calculate the entropy change for the surroundings when 1.54 moles of NH3 (g) react at standard conditions. A.S⁰ Substance AH (kJ/mol) NH3 (9) -46.1 N₂O(g) 82.1 N₂ (9) 0.0 H₂O(g) -241.8 surroundings J/K

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Chapter15: Energy And Chemical Change
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Problem 95A
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Consider the reaction:
2NH3 (9) + 3N2O(g) → 4N2 (g) + 3H₂O(g)
Using standard thermodynamic data at 298 K, calculate the entropy change for the surroundings when 1.54 moles of NH3(g) react at
standard conditions.
AS⁰
Substance AH (kJ/mol)
NH3 (9)
N₂O(g)
N₂ (9)
H₂O(g)
surroundings
-46.1
82.1
0.0
-241.8
J/K
Transcribed Image Text:Consider the reaction: 2NH3 (9) + 3N2O(g) → 4N2 (g) + 3H₂O(g) Using standard thermodynamic data at 298 K, calculate the entropy change for the surroundings when 1.54 moles of NH3(g) react at standard conditions. AS⁰ Substance AH (kJ/mol) NH3 (9) N₂O(g) N₂ (9) H₂O(g) surroundings -46.1 82.1 0.0 -241.8 J/K
Hint: No calculations are required.
For the reaction
6CO₂(g) + 6H₂O(l)→→→C6H₁2O6 + 60₂(g)
AH° = 2.80×10³ kJ and AS° = -259 J/K
At standard conditions, this reaction would be product favored
at relatively high temperatures.
at no temperature.
at all temperatures.
Oat relatively low temperatures.
Transcribed Image Text:Hint: No calculations are required. For the reaction 6CO₂(g) + 6H₂O(l)→→→C6H₁2O6 + 60₂(g) AH° = 2.80×10³ kJ and AS° = -259 J/K At standard conditions, this reaction would be product favored at relatively high temperatures. at no temperature. at all temperatures. Oat relatively low temperatures.
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