CO2 (g) + CCl4 (g) <--> 2 COCl2 (g) Use the data from the tables to calculate de

Question

CO2 (g) + CCl4 (g) <--> 2 COCl2 (g)
Use the data from the tables to calculate delta G for this reaction under the following conditions:
a) standard conditions
b) P CO2 = 0.112 atm; P CCl4 = 0.174 atm; P COCl2 = 0.744 atm
c) at equilibrium (hint: you should not have to do any calculations here)
d) what is the equilibrium constant for the reaction at 25°C? 3. (6 pts) Consider the following reaction: under the following conditions: a) standard conditions b) Pco2 -0.112 atm; Pccu-0.174 atm; PcoC-0.744 atm co2 0.1 e) at equilibrium (hint: you should not have to do any calculations here) d) What is the equilibrium constant for the reaction at 25°C?

Explanation / Answer

a)

using these dGf's
CO2 (g)= -394.4 kJ/mol
CCl4 (g)= -60.63 kJ/mol
COCl2 (g)= -206.8 kJ/mol         you will have to re-do these calcs if your text has different dGf's
we first find the statndard dG for the reaction

dGo = dGf products - dGf reactants
dGo = [(2) (-206.80] - [-394.4 & -60.63]
dGo = -413.6 - (-455.03)

dGo = + 42.43 kJ

b)

now we use
Grxn = G°rxn + RTlnQ

dG = + 42.43 kJ + (0.008314) (298) ln ( [COCl2]^2 / [CO2] [CCl4]


dG = + 42.43 kJ + (0.008314) (298) ln ( [0.744]^2 / [0.112] [0.174]

dG = + 42.43 kJ + (0.008314) (298) ln ( 28.40)

dG = + 42.43 kJ + (2.4776) (3.3465)

dG = + 42.43 kJ + 8.29

dG = 50.72 kJ

c)   at equilibrium dG = 0

d)

Kp = PCOCl2/PCO*PCl2 = 0.744 / (0.174*0.112) = 38.177

Kp = 38.177

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