Question

The equilibrium constant for the following reaction

$Ag_{2}O(s)\rightarrow 2Ag(s)+\frac{1}{2}O_{2}(g)$

can be expressed by the empirical formula

$\ln K_{P}=0.9692+\frac{5612.7K}{T}-\frac{2.0953\times 10^6K^2}{T^2}$

in the interval 445 K < T < 460 K. Use this expression to calculate ∆_rH◦ and ∆_rS ◦ at 450 K and derive an equation for ∆_fH◦ of Ag₂O (s) in this temperature range. Now use the following heat capacity data:

$C^{^{\circ}}_{P}[O_{2}(g)]/R=3.27+(5.03\times 10^{-4}K^{-1})T$

$C^{^{\circ}}_{P}[Ag(s)]/R=2.82+(7.55\times 10^{-4}K^{-1})T$

$C^{^{\circ}}_{P}[Ag_{2}O(s)]/R=6.98+(4.48\times 10^{-3}K^{-1})T$

to calculate ∆_rH◦ , ∆_rS◦ , and ∆_rG◦ at 298 K. Would Ag₂O be automatically decomposed when exposed to the atmosphere (the partial pressure of O₂ in the atmosphere is about 0.2 atm)?

Answer 1

Pas. .. Aggo (s) + 2Ag(s) + £ Q2 Ca) Given thals In Kp = 0.9692 + 5612.7K _ 2.0953x109K T temperature range, 445K<<< 460k #AH at 450k According to Vant-Holf en - Thef 92- R2 o 6 8. AH = R* (ankee ERE ((0.09692+ SP.K. 20 ) | 5 6 12.7K 2.0953 NOK - R 22 aces 28 2.095 3x10 56 12.7. = R 6 AH T . :73 Putting the value T. = 450k, we get- Ayo = 8.314 x (250) * (- SOP at an 45o5} [ 5617.7 2620953x1067 = 30. 76x10 J/mol. . at 450k AH = 30.76 kJ/mol #Arsº at 450K As per defination, - Ans - Ato - AGO DT / Since, Dr H t RT linke OPG =- RT lnke Os, ApS = os, Aps = Our Anso = App +R linke

og Afisa - 30.76 kJ/mal 4 50K + RX 0.09692 + 5612.7 T - 2.0953710 (450) Os, Ars' = 0.068 kJ[x]mal + 8.314 ( 212234 ) J.x'mal' = 0.068 kJ/x/mad + 18.477 J | at mal = 0.068 kJ /k/muel + 0.018 KJ|K)mal -0.086 kJ/k/mal ons's 0.086 K5/x/mall at 450K. # De M Ag20 (3) ► 2 Ag(s) + Ź Q (9) OLAR, DpH = Ap Hoxca) + 204 HAqcs) * – AMÂg,0 (s) os, Aph Ag 20 (s) - OC) =!. Dy Hº - Ž Af Meg () . - 2 AG HÅg(s) > Required expression. # Dr H at 298 K According to Kirchhoff equation, Dr H = Arkin +"Sociedt biene, se pare nts red Produ

Now, Ti = 450k, Am Husok = 30.46 K3/med. T2 = 298k ACP = 2*(CP) Agco + 5* ().gro) - (CP) Agora) on, AC = T2+{282 + (7.85x104) 7} +$ {3.97 +(5038197} -{5198+ (448X10%)] TR Os, Aco = {0.295 – 2.7105610°1} R Nowy An Masok = Arriusok + sapat 09, Dp Hagex = 30.76 ks/male + [fo.298 – 2,7195krőt Jedi Os, Dep H 2 SOK = 30.76K5]mel + RX 0.295* (Tx - 7.) - 2.7185*o**** (at) - 2.7180.5 x Ti Tz 0.7185*10 TS 50-29SX - 30.12 - 2.718 5X10 XR X 09, Dp HZ9SK = 30.76 kJ/mol + 8.314 80-295 (298–450) J/mal 1299 - 4să IJ/mol 03 - 2.7185410 x8.314 x 1 Os, APH298K = 30.76 kJ/mol & -0.3728 kJ/mal + 1.2849 kJ/mol. Ar H290r = 31.6721 kJ/mol

K T T D # Ange at 298k APG y = – RT In Kp e branz-RT (0,9692 + 5612. * - 3:09.534164) putting the value, T = 298k , we get, Amci pe = - 3.3046 28 x (0:9692 + stan - Os, Arcinox = + 9.39 x 103 Jlmal. 562-7 2-0 2r 3 x - 298 (3989 . - 9.39 kJ/mol :. Apci = 9-39x3med at 298K # Apsº at 298k Dmit braai by defination, Arsa - T Or, An Spok - Dr H298k -An Gork 16T (31.6721 -9.39] kJ/k/mal 2.98 = 0.075 kJ/k/mal. . Carsº= 0.075 K 5\x/mal at 298 K # Since the reaction gibbs free energy at room temperature , i.e, at 298k is positive (+9.39 kJ/mal thus it will not decompose automatically.