For the reaction NH,NO3(aq) → N½O(g) + 2 H,0(1) AG° =-177.8 kJ and AH° = -149.6 kJ at 282 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 282 K. The entropy change for the reaction of 2.03 moles of NH,NO3(aq) at this temperature would be J/K.

Consider the reaction 2NaOH(aq) + H2(g)—2Na(s) + 2H2O(l) for which AH° = 368.6 kJ and AS° = 15.30 J/K at 298.15 K. (1) Calculate the entropy change of the UNIVERSE when 1.759 moles of NaOH(aq) react under standard conditions at 298.15 K. ASuniverse= J/K (2) Is this reaction reactant or product favored under standard conditions? (3) If the reaction is product favored, is it enthalpy favored, entropy favored, or favored by both enthalpy and entropy? If the reaction is reactant favored, choose 'reactant favored'. Submit Answer

What is the change in entropy, AS, for the following reaction? Use Table G in your book for values of standard entropies. Your answer should be in J/K mol. CaCO3(s) + 2 H'(aq) → → Ca² (aq) + H₂O(l) + CO2(g) O 264.9 J/K mol O 118.6 J/K mol 394.3 J/K mol 137.7 J/K mol 2

For the reaction Fe3O4 (s) + 4H2(g) → 3Fe(s) + 4H₂O(g) AH° = 151 kJ and AS° = 169 J/K The equilibrium constant for this reaction at 342.0 K is Assume that AH° and AS are independent of temperature.

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Using H° data, calculate the standard enthalpy change for the reaction at 298K. 2 NH3(g) HNO3(aq) --> NO2(g) 2 H20(1) AH° (KJ/mol) -46.19 -206.6 33.84 -285.83 (Note: the table AH values are located directly below the corresponding compound, this is the number found in the Appendix)

For the reactionNH4NO3(aq)  N2O(g) + 2 H2O(l)G° = -175.3 kJ and H° = -149.6 kJ at 257 K and 1 atm.This reaction is (reactant, product)  favored under standard conditions at 257 K.The entropy change for the reaction of 1.74 moles of NH4NO3(aq) at this temperature would be  J/K.