Question

CO2(g) + CCl4(g) => 2 COCl2(g) Calculate ∆G for this reaction at 25 oC under these conditions:

PCO2 = 0.112 atm

PCCl4 = 0.174 atm

PCOCl2 = 0.744 atm

Answer 1

Answer G = 55.2 kJ/mol

the given reaction is

CO2(g)   +   CCl4(g)   ⇌   2COCl2(g)

We know that

ΔG_rxn   = ΔG⁰_rxn   + R T lnQ

it is given that
CO2 (g)= -394.4 kJ/mol
CCl4 (g)= -62.3 kJ/mol
COCl2 (g)= -204.9 kJ/mol

you will have to re-do these calculations if your text has different dGf's

we first find the standard dG for the reaction

Go = dG_f [products – reactants]

Go = [ 2 (-204.9 kJ/mol)] - [(-394.4 kJ/mol) + (- 62.3 kJ/mol)]

Go = - 409.8 kJ/mol - (-456.7 kJ/mol)

Go = - 409.8 kJ/mol + 456.7 kJ/mol

Go = + 46.9 kJ/mol

put all theses values in the below equation

G_rxn   = G⁰_rxn   + R T lnQ

G = + 46.9 kJ/mol + (0.008314 kJ/mol K) (298 K) ln ( [COCl₂]² / { [CO₂] [CCl₄] } )

G = + 46.9 kJ/mol + (0.008314 kJ/mol K) (298 K) ln ( [0.744]² / { [0.112] [0.174] } )

G = + 46.9 kJ/mol + (0.008314 kJ/mol K) (298 K) ln ( 28.4)

G = + 46.9 kJ/mol + (2.4776 kJ/mol) (3.346)

G = + 46.9 kJ/mol + 8.29 kJ/mol

G = 55.19 kJ/mol

G = 55.2 kJ/mol

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