Question

A reaction Alaq) + Blaq) = Claq) has a standard free-energy change of -3.44 kJ/mol at 25°C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively? Number Number Number [A]= M [B] = [c]= 0 How would your answers above change if the reaction had a standard free-energy change of +3.44 kJ/mol? O O All concentrations would be lower. All concentrations would be higher. There would be less A and B but more C. There would be no change to the answers. There would be more A and B but less C. O

Answer 1

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Dear student, The reaction provided u e- Alaq) + Blag) = c lag) initial 0.30 0.4o o l at 10.30-x) (0.40 x x equilibrium Given AG = -3.44 KJ/mol Tz 95°C = 298 K AG ² – RT en keq, enkeq - AGO RT = - (-3.44 Kg/mol) 8,314 Jkt malt 998k en keg - 3.44 x1000 J/mol 8.3145 ktma- x 298 k Keg a 4.0L keg = [c CAI [B] 4.01 = x 10.30-2) (0.40-x) 4.01 = x 0.18-0.30x -0.you +22 4.01= x 0,12 -0.702 +22 10.4812 - 2.807x + 4.010²= x 0.0122 - 3.807 x + 0.481920 P.To

solving quadratic equation x= 0.798 (not possible) az 0.1501 W azo.799 not possible becaus we started with 0.30M and 0.40M: theus x=0,1501] Calea Aleq = 0.30-0.1501 ~ 0.1499 M deq 20.40 -0. Isolo .2499 M Jeg = 0.150M When AG = +3.44 kJ/mol (then reaction. will be non-spontaneous). This implies reaction is not favourable at standard conditions, thus we end up with more of deactants A and B and less product C. Best wishes!!