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Consider the following reaction:

CO2(g)+CCl4(g)⇌2COCl2(g).
Calculate ΔG for this reaction at 25 ∘C under the following conditions.

PCO2= 0.140 atm
PCCl4= 0.160 atm
PCOCl2= 0.745 atm

User Bensstats
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1 Answer

6 votes

Answer: -7956 J

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
K_(eq)


CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

The expression for
K_(eq) is written as:


K_(eq)=((p_(COCl_2))^2)/(p_(CO_2)* p_(CCl_4)^1)


K_(eq)=([0.745]^2)/(0.140* 0.160)


K_(eq)=24.8

The Gibbs free energy is related to equilibrium constant by following relation:


\Delta G=-2.303RTlog K

R = gas constant = 8.314 J/Kmol

T = temperature in kelvin =
25^0C=25+273=298K

K = equilibrium constant


\Delta G=-2.303RTlog K


\Delta G=-2.303* 8.314* 298* log(24.8)


\Delta G=-7956J

Thus ΔG for this reaction at 25 ∘C is -7956 J

User ViviG
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