Question

I ACCIDENTALLY ASKED THE SAME QUESTION TWICE. PLEASE DO NOT ANSWER

A reaction A(aq) +B(aq)C(aq) has a standard free-energy change of-3.38 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? [A] м [B М [C М How would your answers change if the reaction had a standard free-energy change of +3.38 kJ/mol? There would be no change to the answers. All concentrations would be lower. There would be less A and B but more C. There would be more A andB but less C. O All concentrations would be higher.

Answer 1

= -3.38 kJ/mol = -3.38x 103 J/mol Standard free energy change of the reaction,AG Gas constant, R= 8.314 JK-mol Temperature, T=25° C273 298 K Standard free energy change of the reaction, AG° = - RTln K AG In K RT -3.38x103 J/mol = 1.364 8.314 JKmol1x 298 K Equilibrium constant, K= e1364 - 3.91 C(aa) Equilibrium equation: A(aq) B (a) Equilibrium constant, 3.9 ICE table s A В C Initi al (M) Change (M 0.30 0.40 x Equilibrium (M): 0.40 x 0.30 x [C] AB Equilibrium constant, K 3.91 (0.30-x)(0.40-x) x0.15 According to the ICE table, Equilibrium concentration of A,[A] (0.30-x) M (0.30-0.15) M 0.15 M Equilibrium concentration of B,[B (0.40-x)M(0.40-0.15) M- 0.25 M Equilibrium concentration of C,[C] (x) M 0.15 M = When the value of AG s negative (-3.38 kJ/mol), the reacti on is spontaneous which means m ore and more am ount of 'C' will be produced When the value of AGs positive(+3.38 kJ/mol), the reaction is nonspontaneous which means very less amount of 'C" will be produced. Thus, there would be more 'A' and B'but less 'C when the std free energy is changed from -3.38 kJ/mol to +3.38 kJ/mol 1