Consider the reaction below at constant pressure. Use the information provided to determine the value of ASsurr at 398 K, and predict whether or not this reaction will be spontaneous at this temperature. 4NH3(g) + 302(g) → 2N2(g) + 6H20(g) AH°rxn = -1267 kJ ASsurr = +12.67 kJ/K, reaction is spontaneous ASsurr = +3.18 kJ/K, reaction is spontaneous ASsurr = -12.67 kJ/K, reaction is spontaneous O ASsurr = +3.18 kJ/K, it is not possible to predict the spontaneity of this reaction without more information.

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Chapter10: Entropy And The Second Law Of Thermodynamics
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Consider the reaction below at constant pressure.
Use the information provided to determine the value of ASsurr at 398 K, and predict whether or not this
reaction will be spontaneous at this temperature.
4NH3(g) + 302(g) → 2N2(g) + 6H20(g)
AH°rxn = -1267 kJ
%3D
ASsurr = +12.67 kJ/K, reaction is spontaneous
ASsurr = +3.18 kJ/K, reaction is spontaneous
ASsurr = -12.67 kJ/K, reaction is spontaneous
ASsurr = +3.18 kJ/K, it is not possible to predict the spontaneity of this reaction without more information.
Transcribed Image Text:Consider the reaction below at constant pressure. Use the information provided to determine the value of ASsurr at 398 K, and predict whether or not this reaction will be spontaneous at this temperature. 4NH3(g) + 302(g) → 2N2(g) + 6H20(g) AH°rxn = -1267 kJ %3D ASsurr = +12.67 kJ/K, reaction is spontaneous ASsurr = +3.18 kJ/K, reaction is spontaneous ASsurr = -12.67 kJ/K, reaction is spontaneous ASsurr = +3.18 kJ/K, it is not possible to predict the spontaneity of this reaction without more information.
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