So Ag+ + Cl- ---> AgCl and delta H is -65.5 kJ. I'm supposed to calculate delta H when one mole of AgCl dissolved in water, so 1 mol AgCl *(65.5 kJ/1mol Agcl) right? But the answer is 65.5 kJ, and why is it positive, not negative, is my question.

Ag^+ + Cl^- ==> AgCl dH= -65.5

so AgCl ==> Ag^+ + Cl^-, the reverse of what is -65.5 gives you +65.5 kJ/mol.

I think it's becos it's a formation reaction enthalpy is always: -deltaH:-1*-65.5=65.5KJ

Yes, it is a dHo (formation) which is the reverse of what you want.

To calculate the delta H when one mole of AgCl dissolves in water, you are correct in using the equation:

1 mol AgCl * (65.5 kJ / 1 mol AgCl)

However, in this case, the delta H you calculated is the magnitude of the value because it represents the energy released when one mole of AgCl is formed. Since the reaction you provided shows the formation of AgCl (solid) from Ag+ and Cl- ions (in solution), a negative delta H value (-65.5 kJ) indicates that the reaction is exothermic. This means that energy is released in the form of heat during the reaction.

When AgCl dissolves in water, the process is referred to as dissolution or hydration. The delta H for this process is the opposite of the delta H for the formation reaction. Therefore, the delta H for the dissolution of AgCl in water is +65.5 kJ.

This positive value indicates that the dissolution process is endothermic, meaning it requires energy input. Energy from the surroundings is absorbed during the dissolution, resulting in an increase in the system's energy.

In summary, the formation of AgCl from Ag+ and Cl- has a negative delta H (-65.5 kJ) because it is exothermic, while the dissolution of AgCl in water has a positive delta H (+65.5 kJ) because it is endothermic.