Question

3. (a) Use the data given below and calculate ∆Ho, ∆So, ∆Go, and Kp at 25° C for the reaction: 4 NO (g) → 2 N2O (g) + O2 (g)

(b) Calculate ∆G for the reaction at 250 °C.

(c) At what temperature (°C) is ∆G equal to zero? In what temperature range is this reaction product favored?

Compound	∆ Ho , kJ/mol	S o , J/mol• K
NO (g)	90.9	210.76
N2O (g)	82.05	219.85
O2 (g)	0	205.14

Answer 1

4NO(g) 2N2O(g) + O2(g) DHO KJ/mol for Nog = 90.9 kJ/mol. Dho kJ/ and for N₂O(g = 82.05 kJ/mol AHO kJ/mol det for 22 (9) = 0 kJ/mol calculate Al for reaction which is Hess daw. AH Reaction - Diproducts - All Reactants, fiest As Reaction = 2 XA1 f (N2O) + suf (02) - 4 (BHNO) = 2x82.05 + 0-4/90.9) kJ/mol - 164-170-1963.6 ks/m of - - 199.5 kJ/mol = 2500 + 2739 - 523 K | AS Reaction = Asproduct - As Reactants - $ 0,2 + (No)- vo) = 205-14+ 2x219.85 - 4(210) Imolok = 205.14 + 439.7- 840 20604.84-840-195 smotiks ] Da DH-TAS selleleggese ragazze x AG= ? AH = – 199.5kJ/mol Scandith - 195 umol = + 0.195 kJ mott s Scanner – 195 Imol k ollonioules - 1000 Joules CamScanner

AGE AH-TAS" DGS - 199.5 - 523% (-0.195) OG": -199.5+ 101.985 09° = -99.515 Kill o Soules mots calculate 84 for 256 25+273 298 Kelvin To calculate use Gibbs free Energy equation which is Du = DHTAS helgy equation which is DH -19.5 ksmolt @gler a ceae -0.195 kJ 09": -199.5 – 298* (4195) DG = -119.5 + 58:11 09'- - 141.39 Killo Joule moet OS 2-0.195 kJ mulut (6) At what temperature (°C) u AG equals to hers ? s9 = sn -TOS 03-199.5 - T(-0-195) 03-199.5+ Tx0.195 +199.5 = TX 0.195 199.5 = Temp 0:195 Temp = 1023.07 kelvin ca Scanned with e CamScanner

for condition to be montaneous SG is negative the process ń spontaneous - - O → ou u hero, the system i equilibrium The process does not occur Auntve the process does not occur in the forward reacton , It may occus as the back word direction. AG at 523K (250 c) = -97 5 15 killo Joules not DG at 298K (25°c) = -141-39 kellojoules mol AG at At 1023.07 kelvin o kello Youles on of we can notice a4 at 523 k u greater than saat 298k So at A4 for reaction favoured will be 5234 (250°C) rules mots CS Scanned with CamScanner

Calculate kp at 25°c use the formulla sa = -2-303 RT log ke DG3 -141-39 Killolone mot" -141.39 = -2-303 X 8.314 x 100% -141-39 = – 2-303 x 8-314 X 10*x 298 loghe +14139 z 72-303¥ 8-314x10-32 ge = log hp 141.39 5705,84 X10- 3 141:39 x 1000 = log ke 5705.84 24. 77 = log kp kp- antilog (a Apr mantersa characteristics 60QQ00011 .7798 24 = 6028 = = 10 x 10 = 6:022x 10 27 Aus Icdscanned with Cthank genner