Question

Consider the following reaction:

CO2(g)+CCl4(g)⇌2COCl2(g)

Calculate ΔG for this reaction at25 ∘C under these conditions:

PCO2= 0.140 atm

PCCl4 = 0.170 atm

PCOCl2= o.740 atm

ΔG∘f for CO2(g) is −394.4kJ/mol, ΔG∘f for CCl4(g) is −62.3kJ/mol, and ΔG∘f for COCl2(g) is −204.9kJ/mol.

Express the energy change in kilojoules per mole to one decimal place.

Answer 1

ANS)

from above data that

CO2(g)   +   CCl4(g)   ⇌   2COCl2(g)

We know that

ΔG_rxn   = ΔG⁰_rxn   + R T lnQ

Using these dGf's
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

Now we use

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 ( [0.740]² / { [0.140] [0.170] } )

G = + 46.9 kJ/mol + (0.008314 kJ/mol K) (298 K) ln ( 23.01)

G = + 46.9 kJ/mol + (2.4776 kJ/mol) (3.136)

G = + 46.9 kJ/mol + 7.77 kJ/mol

G = 54.67 kJ/mol

G = 54.7 kJ/mol