Question

Consider the following reaction:

CO2(g)+CCl4(g)⇌2COCl2(g)

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

PCO2=0.100atm

PCCl4=0.160atm

PCOCl2=0.760atm

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

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

$\Delta$ Go = dG_f [products – reactants]

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

$\Delta$ Go = - 409.8 kJ/mol - (-456.7 kJ/mol)

$\Delta$ Go = - 409.8 kJ/mol + 456.7 kJ/mol

$\Delta$ Go = + 46.9 kJ/mol

Now we use

$\Delta$ G_rxn   = $\Delta$ G⁰_rxn   + R T lnQ

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

$\Delta$ G = + 46.9 kJ/mol + (0.008314 kJ/mol K) (298 K) ln ( [0.760]² / { [0.100] [0.160] } )

$\Delta$ G = + 46.9 kJ/mol + (0.008314 kJ/mol K) (298 K) ln ( 36.10)

$\Delta$ G = + 46.9 kJ/mol + (2.4776 kJ/mol) (3.586)

$\Delta$ G = + 46.9 kJ/mol + 8.88 kJ/mol

$\Delta$ G = 55.78 kJ/mol

$\Delta$ G = 55.8 kJ/mol