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

 

[ravenprp]

Equimolar means that the concentrations of both H+ and Cl- ions are equal in the solution. In this case, the initial concentration of HCl is 0.89 M, which means that the concentration of H+ ions is also 0.89 M and the concentration of Cl- ions is also 0.89 M. When the solution is diluted to a final concentration of 0.253 M, the concentration of H+ ions and Cl- ions will also be equal, as they are in an equimolar ratio in HCl.

The reason why we need to raise the molarities to the power of 2 is because of the stoichiometry of the reaction. The dilution of HCl can be represented by the following equation:

HCl (aq) ⇌ H+ (aq) + Cl- (aq)

According to the stoichiometry of this reaction, for every 1 mole of HCl that is diluted, 1 mole of H+ ions and 1 mole of Cl- ions are formed. Therefore, when calculating the value of Q, we need to take into account the concentrations of both H+ and Cl- ions, which is why we raise the molarities to the power of 2.

I hope this explanation helps to clarify why we need to consider the equimolar concentrations and raise the molarities to the power of 2 when calculating the free energy of dilution for HCl.