Calculate Enthalpy of Ionization of Weak Acid: ΔH3

[BayernBlues]

Homework Statement

Find the enthalpy of ionization of the weak acid, ΔH3

H3O+ + OH- <-> 2H2O....ΔH1
HA + OH- + H2O <-> A- + 2H2O...ΔH2
HA + H2O <-> H3O+ A-...ΔH3

The Attempt at a Solution

I won't show it here but I've found ΔH1 to be -52.33 kJ/mol and ΔH2 to be -5.21 kJ/mol. I don't know how to get the third enthalpy though, I know you'd use Hess' law but I don't really understand what to do here.

 

[Borek]

Obviusly you have to start with the second reaction, as it has HA and A^- in correct places. Then use the first reaction to get rid of OH^- on the LHS and H₂O on the RHS.

 

[Blueshift5]

To calculate the enthalpy of ionization of the weak acid, ΔH3, we can use Hess' law. This law states that the enthalpy change for a chemical reaction is the same regardless of the route taken from reactants to products. In this case, we can use the enthalpies of the two given reactions, ΔH1 and ΔH2, and combine them in a way that will give us the enthalpy of the third reaction, ΔH3.

To do this, we can use the following equation:

ΔH3 = ΔH1 + ΔH2

Since ΔH1 and ΔH2 are both negative, we can add them together to get the final enthalpy, ΔH3. This will give us the enthalpy of ionization of the weak acid. It is important to note that we are assuming that the enthalpy of the third reaction, ΔH3, is also negative. If it is positive, this would indicate an endothermic reaction and our calculation would be different.

In summary, to calculate the enthalpy of ionization of the weak acid, we can use Hess' law and add the enthalpies of the two given reactions, ΔH1 and ΔH2. This will give us the final enthalpy, ΔH3, which represents the enthalpy change for the third reaction.