Question

Predict the signs, comment and then calculate the enthalpy and entropy changes for the reaction of combustion ofmethane at 500 K. Will the reaction be spontaneous at 500 K?

CH4(g) + 2 O2(g) → CO2(g) + 2H2O(g)

We give: The enthalpies of formation of the products at 298.15 K in kJ/mol are:

∆Hf(CO2) = - 393.51; ∆Hf(H2O) = - 241.82; ∆Hf(CH4) = -74.81.

The heat capacities at constant pressure in J.mol-1.K-1:

Cp(CO2) = 37.11; Cp(H2O) = 33.58; Cp(CH4) = 35.31; Cp(O2) = 29.36.

The standard molar entropies of the compounds at 298.15oC in J.mol-1.K-1:

So(CO2) = 213.74; So(H2O) = 188.83; So(CH4) = 186.26; So(O2) = 205.14.

Answer 1

solution - -* *-* - case- 0 Criven- Recection - -*-*- CH4 cy: +20213) - 60947) +242017) -O enthalpy of formation - at Temperature = 298.150k AH Amp (C04) = - 393-51 le , 44p [%50) = - 241-82 kg AM (CH4) = -74.81 page, logig) zo mol te - Heat of Reaction calculated as a at Temperaturer T=298.151 *** = (3 ang product (anf Jeracione (E Hi product = 1x AHýl Coggi) + 2x 44° C 450 g) = 1-393-51 +-241.82 mol mo = -877.15 KJ more (EAHReuchende AHCehq) +28 AH (02) = -79,816x9 +0 mo --74.81KO mee

Since - a A to B cctd D A4 for Reaction calculated ar- AH° = (cx amic c) + dx A49 (01)-(ux agºca) +6& 403°(B) ) Apply in the above calculation, 30 — put the value of (E AK") produe et (EA MOJ Remetean - pu AHO 1-877115 198 -1-74.81.145 mol امی AH° = -877,15 105.7 74.81 43 mol at G; 809 g (298115) 4 195 mol - 0 at Fanporakexe at temperature T= 298.15K coge- _ eriven- Cpl Coa) = 37.11 I , Co (490) = 33.58 molek en nad Gol CHY) = 35.31 I , 6 (0) = 29. 36 I malik Alpmix) of the Reaction calculated as a Imixa Lowhere Pis stoichiometrie coetticida from Reaction - tve for product and are for Pice M4.) = a), Dircondai Reactante QpO2) ='62, pe 420) = 2 {p;= for component i

(A Cp Imix = -1x cpécha) - 2x Cploa) +1x gecom) + 9x Cp (420) - out the value of Colchas, Cocogs, collo & G (Ago) A (mix) = -lx 35.31 3 - 2x29.36 - 35:31 -2%29.36 3 molok t (AC) mix malik + 1x37.11 I +9 x 33.583 Molk malal = 10.24 3 molok > Cniven - AH ay initieel Tempore teeme - T= 298.15K final Temperature - To = 500k Reaction at at a 500k - Heat of calculated as a AH = AH0+ ☺ T here (AC) mix a constant AH = AH° +6C Smix mix old Sincea 7 erarti) AH = AH° + (4 %) mixa (Tam Tv ) + put the value By AH°, (ACP) mix, , T, & to in this Reaction

Since - АН°. - go2, 24 | 5 mol 185nee- 1 1 1 3 = to'y Ан° - 802.34 хто 1 mul дн". — 302 3до з. ир Ана- 802340 mol (Ace mix - 10.24 1 - molok Т) - 2 98/ Т. - Sook ДН – – go 2340 ја то + 10• 24 3 х (foo- 298js) К тек AH - - Во 0113, 056 3 moe I Coge (3) — + — + — criven - Revelion - ена 1) + 20) — 02 (1) + 2ң,0 ty) Сојуел — 8° с со = а13. 14 тоен л , s°ін;0) = 1 88 -8 3 3 mol.k °ін, 39 сен, ) = 186 .16 т мое. К Т ° (02) 52os.14 3 mulik a entropy change of Reaction - calculated as a | 45° = ([ 5° product - ES Plexeteinke

Asº = { sºcc0g)+2x 8°C 460]- 's® ( Cha) +2x 5°009) ] -> put the verdure ay 890., Saxo), Scenes (int 5%,,— Asº = [ 213.793 + 2x188.33 I тi.k 7- 71861293 +28205.1991 moeck moook ASO - -6.12 I molik makit > As ont S entropy cucinge cito Temperature (T=500k) calculated Oise AS - A ( Amoxild or os AS = 48° +leep mig. J * 4,0T sincer App in 34а ч Фоvе tuzhion — | as = 28°44) is hy - put the value of Aso (AC) mix , T, and to in the above equation, O

AS = - 6.12 - molik +10.24 I x In 1 500 molek AS = -0.826 3 Molok ) at Tembera deere - T= 500k -14 8- Cribbe- free energy Related to enthalpy & entropy cra AS AH-TAS - put the voluе d AH, AS AT since - AH = -800273.056,5 1 mol ASE-0.896 I molok T = 500k AC: -8002 73.056 I - 500K X-0.826 I molek 5-799860.056 I s mo Aleave at this temperaturer so Reaction is spontaneoux- Sincer Acn=ave Alla the sponteneous non-spoon teneur