Question

Predict the sign and calculate ASⓇ for a reaction. Consider the reaction 2NH3(g) + 3N2O(g) +4N2(g) + 3H2O(g) Based upon the stoichiometry of the reaction the sign of ASørxn should be ' Using standard thermodynamic data (in the Chemistry References), calculate A Sºrxn at 25°C. AS rx = J/Komol

Answer 1

creases. a 2NH₃(g) + 3N2018) — 442 (9) + 3H2018) Based on the storniometery of the reaction, there are s motes of most gaseous particles on the reactant side and 7 moles of gaseous particles on the product side. As the number of gaseous particles the per product side, the degree of randomnessie, entro Hence, As reaction should be positive. from Standard thermodynamic data, at 25°C, S (NH3 (91) = 192.45 J K t mol-1 S (N2O) = 219.85 J kimoll 5 (42) = 191.61 J K 1 molt S (H₂018 = 188.83 J K 1 molt As reaction = { s° C foodueks) - Sºl Reactants) = [4 8(a) + 3 sº1+30) - [ 28°(NH3) + 3 5°(N>0)) - 41191.61) + 3(18883) - ( 24192.45.)+3(2.19.85)] JK 'moto - 766-44 +566.49 - [384.9+ 659.55]J K1 mol4 = [1332.93 - ( 1044.45)]JK-Imol 7 As reach 288.48 Ik motll 4 NH3(g) + 502 (9) UNO(g) + 6H2019) As reaction - 180.55/k from standard thermodynamic data, AH (HO) = 90.25 kJ molt ALHO (H2O(g) = -241.82 kJ molt AHO (02 (9) = 0 kJ molt HO NH₂) = -46.11 ki mol 1

Hreaction = 4 of Hº (ND) + 6 Ag*°(490)- [5 4449/03) +44p4°(NH3) = 4190.25) + 6 (-241.81)-10+41-46 11)) ki ima l = 361 - 1450.86 + 184.44 KJ molt = - 905.42 kJ molt AS AH reaction is - 905.42 kJ moll this means this amount of energy is absorbed by the surroundings (Allurr a is surroundings = AH = + 905.42 kimoes kumolt 1 =7905.42 As surroundings 298 K = 905.42 x 1000 J k l mol-1 298 - 3038.322 J kimalt As system = 180.5JKT ginen) As universe a AS system + As surrounding = 180.5 30 38.322 = 3218-822 JK-