Question

Be sure to answer all parts. Report problem Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presence of superheated steam. Guided Solution (a) Given these data, find A , ?G , and AS at 298 K: rxn rxn 0 ?? AG (kJ/mol) (kJ/o) J/mol K) Ethylbenzene, CgHs CH2CH3 12.5 119.7255 Styrene, C,H5CH-CH2 103.8 202.5 238 0 AG rxn rxn rxn J/K (b) At what temperature is the reaction spontaneous? (c) What are AGand K at 600.°C? rxn AG rxn kJ/mol x 10 Enter your answer in scientific notation.

Answer 1

The catalytic dehydbogenation of ethylbenzene is given as High Temp CH₂ - CH₂ CH2 Ch CH₂CH=CH₂ + H (9) @ To find onen Fromulo Dytm - emmat pies Sum of Sum of enthalples of reactante of producte OR Aliran = {DH ( Products) - {arif (Reactants) stiren - AHF (CMSCH=-1) + 04°f(460)) – AHF (Stig wonen) Given Alf (Cath 5 cr CH2) = -12.5 kJ/mol DH f (CM CH=CH2) = 103.8 kJ/mol Remember DHF (4.68)] = 0 kg Put these vahen in equ Horn = 310308 +0-(-12:5) = 103.8 + 12.5 = 116.3 kg (ahorn = 116.3 k]] To find allen su on = { AGit Bordecto - Esmir (rectant) sapotn = soif (cally Cruz.) +004(4) - ong (4444) Given Dif (Shy CH Caly) = 119.7 kJ / mot Doof (GHg CH = CH2) = 202.5 kJ/mol Remember of (H₂(3)] = 0 107/mor Put all these values in ean ③ Dronn = 202.5 to 119.7 = 82.8 kJ/mol Darn - 82.8 kJ

n Reactants To find osórn oson = 55° C Products ) - {s' Peacrendes osorn = 5° (CMICH=CH.) +5°CH) – 5°Ccous couch) Given so( CH Cheb) = 255 J/molk s (CHI CH=CH2) = 238 ] mork Remember 3 (H₂) = 131 / mork Put all these values in above een Asoon = 238 + 131-255 = 114 J/mor.k A Soren= 1145/6

(6) At what Temp is the oln spontaneous ? for this we have to calculate the temperation at which so is zero we know that AG = AH - TASI For Dor=0 DHO - TA5° = 0 - OR ay = Tas ~0 we know Driti = 116.3 kJ s Alreadly Calculated - Doro= 114 J l l in part ① Put in above eq" 116.3 kJ = TX 1143/k r convert ky to ]] 1 kg = 1000 ] 116.3 X 1000 y = Tx 1143/ T= 116.3 x locoy - 1020.2k 114 XI To convert it to oc, Bubtract 213k from this 1020:2 - 273 = 747.2°C or 7476(approx) We know that 190 - AH - TASCO han = 116.3 kJ Asan = 1147 = 114 kJ/K = 0.114 kJ/4 T= 600°c = 600+2736 = 873k substitute all these values in eqr ② 100 AG 16.3 - 873 X 0.114 soon - 116.3 - 99.522 = 16.79 kJ/mol oorn = 16.79 145/no/

porn = 16.79 145/mol To finak Relation b/w DG & equilibrium constant (k) is given by Gan = -RT ln K OR AGO = -2.303 RT logk-① scrigen = 16.77 ks/moe = 16.77X1000 mcp = 167703/mu use R= 8.314 lk. mot T=8136 Put in above ean 16770 = - 2.303 X 8.314 X 873 lock 16770 = -16715.455 lojk losk = 16770 =- 1.003 16715.455 OR K = Antilos (-1.003) fculator of table Antilog (-1.033) = 0.099 K = 0.099 = 9.9x10² or - 2 You can also use Cappoox) 10x102