4C3 ZgN3O9(1)6N2(g) +12CO2(g) +10H20(g) + O2(g) ,y8tem =-5680kJ Part A-Predictin

Question

4C3 ZgN3O9(1)6N2(g) +12CO2(g) +10H20(g) + O2(g) ,y8tem =-5680kJ Part A-Predicting the entropy change for the system, 5,ystem» from the balanced equation Predicting the entropy change for the system, AS em, from the balanced equation: Positive Is ASsystem positive or negative? Is this entropy change a driving force for spontaneity? Negative Yes Predicting how the enthalpy change for the system, H system, affects ASsurr No Based on the sign of AHsystem, is this reaction endothermic or exothermic? Does the temperature of the surroundings increase or decrease? Is ASsurroundings positive or negative? Is this enthalpy change a driving force for spontaneity? Endothermic Exothermic Increase Decrease Consider the signs ofas you determined above and that Suniverse- What is the sign of ASunvrs for this reaction? Reset Help Submit My Answers Give Up

Explanation / Answer

Part A :

Positive : delta S system is positive , as there are more number of gaseous species in products side.

Yes : The increase in entropy of the system is always favourable.

Exothermic : Since the enthalpy change is negative , it indicates that the heat is given off by the reaction.

Increase : The reaction releases heat energy , so the temperature of the surroundings increase.

Negative : Since the entropy change associated with the reaction is positive , it will be negative with respect to the surroundings.

Yes : The negative enthalpy change causes the reaction to be spontaneous.

positive : All the spontaneous reactions proceed through increase in entropy of the universe.

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