[jinjo13]

For the reaction between aqueous calcium hydroxide and aqueous hydrochloric acid, which is Ca(OH)2(aq) + 2HCl(aq) --> CaCl2(s) + 2H2O(l), I have come up with some conclusions and data.  Could the forum members here please help me check some of these results, looking at this excerpt from my project?

"The conclusion that this reaction is exothermic is further supported in calculations of enthalpy (the values of which are in kJ/mol, and express heat):
 Ca(OH)2(aq) + 2HCl(aq) --> CaCl2(s) + 2H2O(l)
           -1002.82 + 17.46 > -795.8 - 285.83
            -985.3 > -1081.6
Because the sum of the enthalpies on the reactants side is greater than that of the products, the reaction produces heat, making it exothermic.  (In the actual experiment, the beaker in which the reaction takes place will probably be hotter after the reaction than before.)

The reaction can now be written as follows:                          
Ca(OH)2(aq) + 2HCl(aq) -  (delta)H --> CaCl2(s) + 2H2O(l)

More specifically, using the enthalpy values found earlier, the reaction can be written as such:  
Ca(OH)2(aq) + 2HCl(aq) --> CaCl2(s) + 2H2O(l) + 123.3kJ/mol"

The main confusion here is with the enthalpy value of aqueous hydrochloric acid; to find it, I used the difference between its enthalpy of dissolution, which is -74.84, and its "standard enthalpy change of formation" as a gas, which is -92.3. (This difference is +17.46).  Is this the correct way to do this? For all of the other reactants and products, I simply used standard enthalpy change of formation.

[igloo5080]

The progression is from 0.5H₂(g) + 0.5Cl₂(g) HCl(g)   :delta:H = -92.3 kJ mole^-1 (according to you)

and then from  HCl(g) HCl(aq)   :delta:H = -74.84 kJ mol^-1 (once again, according to you)

So you should be ADDING these energies, not subtracting them, giving -167.14 kJ mol^-1

[jinjo13]

So if the enthalpy of aqueous HCl is -167.14kJ/mol, then the reaction (as shown below) is endothermic, and I can expect the beaker after the experiment to be colder?

-1169.96 < -1081.6

 :delta:H =-1081.6-(-1169.96)=88.36

Ca(OH)_2(aq) + 2HCl_(aq) +  88.36kJ/mol--> CaCl_2(s) + 2H₂O_(l)

To clarify, I got the standard enthalpy change of formation of HCl as a gas from here: http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table)

And I got its enthalpy of dissolution here: http://en.wikipedia.org/wiki/Enthalpy_change_of_solution

To make sure that I am correctly doing this, I have a question for you, since you seem pretty knowledgeable about this. Am I even allowed to do this?

(standard enthalpy change of formation of HCl as a gas) + (HCl enthalpy of dissolution) = (enthalpy of aqueous HCl)

[igloo5080]

This reaction is certainly exothermic (as you suspect), but there are one or two issues which you need to clear up:

(1) Your equation has Ca(OH)₂(aq) on the left-hand side and CaCl₂(s) on the right-hand side - it should be CaCl₂(aq) on the right-hand side

(2) I haven't my data-book at hand, but you appear to be using enthalpy changes of formation for these two calcium compounds in the solid form, whereas you need to look up values for the aqueous forms, or, at any rate, deduce them, just as you did for HCl(aq)

(3)  You haven't doubled the enthalpy changes of formation for H₂O(l) and HCl(aq); after all, there are two moles of each involved in the equation

Finally, yes, the enthalpy change of formation of HCl(aq) can be found by adding together the value for HCl(g) PLUS the enthalpy change of dissolution


One final point:  You would save yourself a great deal of 'strife' by realising that this reaction is an ionic one taking place in solution, and that the effective process taking place is:

2OH^-(aq)  +  2H⁺(aq)      2H₂O(l)

The calculation then is very easy: just look up the enthalpy changes of formation of OH^-(aq), H⁺(aq)  and H₂O(l) and carry out some very easy calculations!

[jinjo13]

Thank you very much for pointing out these numerous errors! I feel I have a better understanding of this subject now. Here are my new calculations:

-1002.82 + 2(-167.14) > -877.30 + 2(-285.83)

-1337.10 > -1448.96

 :delta:H = -111.86kJ/mol

Ca(OH)_2(aq) + 2HCl_(aq) --> CaCl_2(aq) + 2H₂O_(l) + 111.86kJ/mol

[igloo5080]

And what's more you now finish up with the correct answer, because H⁺(aq) + OH^-(aq)  H₂O(l) is well known to have an enthalpy change of reaction of about -56 kJ mol^-1, which of course is equivalent to yours, because yours is 'doubled up'.

I'm glad you understand better now!!