Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

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

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote