Question

Question #556b2

Answer 1

Here's what I got.

Explanation:

You can check out the standard enthalpies and entropies of formation for these two ions here

http://nshs-science.net/chemistry/common/pdf/R-standard_enthalpy_of_formation.pdf

and here

http://www.mrbigler.com/misc/energy-of-formation.PDF

Now, here's how you could determine the standard enthalpy of formation for those two ions. The same principle applies to the standard entropy of formation as well.

Let's say that you're looking at the solvation of ammonium nitrate, `"NH"_4"NO"_3`, which dissociates completely in aqueous solution to give ammonium cations and nitrate anions.

`"NH"_4"NO"_text(3(s]) stackrel(color(red)("H"_2"O") color(white)(xx))(->) "NH"_text(4(aq])^(+) + "NO"_text(3(aq])^(-)`

At this point, you can use the standard enthalpy change of solution, `"DeltaH"_"sol"^@`, for ammonium nitrate, which is lested as being equal to

`DeltaH_"sol"^@ = +"25.69 kJ/mol"`

http://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/enthalpies%20of%20solution%20of%20electrolytes.pdf

and the standard enthalpy of formation, `DeltaH_"f"^@`, for ammonium nitrate, which is equal to `-"365.1 kJ/mol"`, to determine the standard enthalpy of formation for the pair of ions `"NH"_4^(+)` and `"NO"_3^(-)`, in aqueous solution.

As you know, the standard enthalpy of solution can be written as

`color(blue)(DeltaH_"sol"^@ = sum (n xx DeltaH_"f products"^@) - sum( m xx DeltaH_"f reactants"^@))`

In this case, you would have

`+"25.69 kJ/mol" = [DeltaH_text(f)^@("NH"_text(4)^(+)) + DeltaH_text(f)^@("NO"_3^(-))] - (-"365.1 kJ/mol")`

The standard enthalpy of formation for the pair of ions will be

`[DeltaH_text(f)^@("NH"_text(4)^(+)) + DeltaH_text(f)^@("NO"_3^(-))] = 25.69 - 365.1 = -"339.41 kJ/mol"`

Now you can use the same technique to determine the standard enthalpy of formation of the nitrate anion by using the standard enthalpy of solution of nitric acid, `"HNO"_3`.

`"HNO"_text(3(aq]) -> "H"_text((aq])^(+) + "NO"_text(3(aq])^(-)`

The trick here is to use the fact that the standard enthalpy of formation of the hydrogen ion, `"H"^(+)`, in aqueous solution is equal to `"0 kJ/mol"`.

You will thus have

`overbrace(-"33.28 kJ/mol")^(color(blue)("std. enthalpy of solution")) = [overbrace(DeltaH_text(f)^@("H"^(+)))^(color(red)(=0)) + DeltaH_text(f)^@("NO"_3^(-))] - overbrace((-"173.2 kJ/mol"))^(color(blue)("std. enthalpy of formation"))`

This means that the standard enthalpy of formation for the nitrate ion in aqueous solution will be

`DeltaH_"f"^@("NO"_3^(-)) = -33.28-173.2 = color(green)(-"204.48 kJ/mol")`

The listed value is `-"206.6 kJ/mol" ->` close enough.

Use this value to find the standard enthalpy of formation in aqueous solution for the ammonium ion

`DeltaH_text(f)^@("NH"_text(4)^(+)) + (-"204.48 kJ/mol") = -"339.41 kJ/mol"`

`DeltaH_"f"^@("NH"_4^(+)) = color(green)(-"134.9 kJ/mol")`

The listed value is `-"132.8 kJ/mol" ->` close enough