Question

please help me find the correct answer to this problem

Free Energy of Dilution Calculate AG for the dilution of aqueous HCl from 1.98 M to 0.260 M at 25°C. 5029.47 mol The process you are considering is: H+ (0) + Cl(aq)H+ (aq) + Cl(aq) (for which AGº=0 i.e. standard states all round--both sides are the same). Now use AG=AGº + RTIN(Q). (Use the actual concentrations to determine Q). Does this agree with what you intuitively know about the spontaneity of diffusion?

Answer 1

oh= on art ind a & product concentration reactand concentration final initial concentration concentration - 0.26 (given in question logg 0.1313 - an ao (given in question) 9=298< (213 +25) (given in question) R=80314 (constant on stand then a h=154 Rilna oh=0+ 8.314x298 in 8.1313 on = of 8.348 298 x (-2.030) = -5030.14 An - (os=-503014

In 001313 - 200302 from Or log table net 0530+6299* *.319) * • (- 200302) = - 5030. 1413 AG = -5030.1413

oh= on +RT ind Q = product concentration reactant concentration final initial concendration conconvention In question 0.26 (given in question) given logg 0.1313 - an ao (given in question) T= 298 € (273+25) (given in question) R= 80314 (constand) 4he A6 Aut of the oh=0+ 8.314x298 in 0.1313 = Of q.3kx 298 × (-2.030)