Why Do We Square Molarities in Free Energy Calculations for HCl Dilution?

[salman213]

Calculate ΔG for the dilution of aqueous HCl from 0.89 M to 0.253 M at 25°C.

THis was a question I had to do some time earlier for one of my assignments. I got some help from a classmate and he said THE FOLLOWING:

"H+ and Cl- are equimolar so you have to put them to the
power of two...

so the equation reads
Deltag = RTln Q
delta g = 8.314472J/K/mol*298.15K * 1kJ/1000J *ln (second
molarity^2/first molarity^2)"





WHEN he says EQUIMOLAR how do i find this out like i have no idea how he figured that out. If i did this question i would just do products over reactions to find the value of Q.

But he says put them to power of 2. WHY?? Can someone explain?

 

[staf9]

HCl is a strong acid and therefore completely disassociates in water. Therefore, whatever the molarity of HCl = molarity of H+ = molarity of Cl-

Hence equimolar