a) What is G rxn o (in kJ) at 2715 K for the following reaction? 2POCl 3 (g) 2PC
ID: 962571 • Letter: A
Question
a) What is Grxno (in kJ) at 2715 K for the following reaction?
2POCl3(g) 2PCl3(g) + O2(g)
POCl3(g): Hfo = -592.7 kJ/mol and Sº = 324.6 J/K mol)
PCl3(g): Hfo = -287.0 kJ/mol and Sº = 311.7 J/K mol)
O2(g): Hfo = ? kJ/mol and Sº = 205.0 J/K mol)
b) At what temperature (IN KELVIN) does the above reaction become spontaneous?
c) What is Grxno (in kJ) at 488 K for the following reaction?
PbO(g) + CO2(g) PbCO3(s)
PbO: Hfo = -219.0 kJ/mol and So = 66.5 J/K mol)
PbCO3(s): Hfo = -699.1 kJ/mol and So = 131.0 J/K mol)
CO2: Hfo = -393.5 kJ/mol and So = 213.6 J/K mol)
d) At what temperautre (in K) does the above reaction become spontaneous?
Explanation / Answer
a)
dGrxn = Hrxn - T*Srxn
Hrxn = 2*-287.0 + 0 - -592.7 = 18.7 kJ/mol
Srxn = 2*311.7 + 205.0 - (2*324.6 ) =179.2
G = 18700 - 298*179.2 = - 34701.6 J/mol = -34.7 kJ/mol
b)
G = H - TS
H - TS < 0
H < TS
H/S < T
18700/179.2 < T
T > 104.352 K
c)
Grxn = H - TS = 18700 - 488*179.2 = -68749.6 J/mol = 68.749 kJ/mol
d)
G < 0
H - TS <0
H/S < T
H = (-699.1 ) - (-393.5 + -219.0 ) = -149.1
S = 131.0 - (213.6 +66.5 ) = -86.6
H/S < T
149100 / 86.6 < T
T > 1721.7K
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