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5. Calculate deltaG under the following conditions.

AH+ (kJ/mol) 30.71 Sº (1/K*mol) 245.3 AG+ (kJ/mol) 3.14 52.30 219.4 68.11 -235.1 282.7 - 168.5 0 222.96 0 0 202.7 0 130.58 0

a.C2H5OH (g) --> C2H4 (g) + H2O (g) Pressure of all gasses=0.200 atm, T=200oC

b. C2H5OH (g) --> C2H4 (g) + H2O (g) Pressure of all gasses=0.200 atm, T=400oC

c.PCl3 (g) + 3 H2 (g) --> PH3 (g) + 3 HCl (g) PPH3=3.00 atm PPCl3=1.00 atm PH2=0.250 atm PHCl=0.250 atm T=298 K

d. PCl3 (g) + 3 H2 (g) --> PH3 (g) + 3 HCl (g) PPH3=3.00 atm PPCl3=1.00 atm PH2=0.250 atm PHCl=0.250 atm T=400 K

e. PCl3 (g) + 3 H2 (g) --> PH3 (g) + 3 HCl (g) PPH3=2.50 atm PPCl3=1.50 atm PH2=0.250 atm PHCl=0.250 atm T=400 K

Compare your answer to parts a and b to your answer to parts c and d. Is the effect of the temperature change the same? If not, why would this happen

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Answer #1

Part (a) C2H5OH (2) - C2 Hu (2) + H2O (2) AH - 235.1 52.30 -241.82 KJ mole Sº 282.7 219.4 188.83 J Molcok AH {(AHp Product {(AGO RT dn(Q) R= 8.314 J/molik Where - 13814.52 + (8.314 x 473115) ln (0.2) Дл - 13814.52 - 6331.157 AG - 20145.677 Ilmole -20part (ol PU3 (21 + 3 H₂ (9) 7 PH3 (2) + 3HO (2) DHS (2) - 288.07 5.4 - 92.3 S° 311.7 130.58 210.2 186.69 molik AHO { (DHål PrDh= AGO + RT ln (16) - 3345.34 + 8.314 x 298 x ln (3) 623.4 J/mole AU 6234 kJ/mole Part (d) only Temp. change T= look AH - TA+ 3 Hlll Past (el RAN: PU3(g) + 3 H₂ (7) PH3 (9) Since RXN is same as part (C), (d) than AH = 16570 Ilmole ago = 66.83 I mole

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