the standard enthalpy is

Question

A scientist measures the standard enthalpy change for the following reaction to be -137.6 kJ:

NH4NO3(aq) N2O(g) + 2 H2O(l)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of NH4NO3(aq) is

kJ/mol.

Answer 1

NH₄NO₃(l) ==> N₂O(g) + 2H₂O(l) ... balanced equation

∑∆H_fproducts - ∑∆H_freactants = ∆H_freaction

From a table of standard enthalpy of formation, I obtained the following values:

N₂O(g) = 82.05 kJ/mol

H₂O(l) = -285.8 kJ/mol

Substituting in the above equation, we have...

(82.05 + 2x-285.8) - (∆H_f NH₄NO₃) = -137.6

-489.55 + 137.6 = ∆Hf

∆Hf = -351.95 kJ/mol

Answer 2

I am taking this to be the decomposition of ammonium nitrate into dinitrogen oxide and water (I would expect the ammonium nitrate to be solid and the two products to be gases, normally) . I'd also expect ammonia as a product.

Ammonium nitrate liquid would require something other that the normal ΔH_f, for the ionic solid, also not readily available in two texts that I checked. These can be googled online.

The equation is not balanced and must be before any calculation can occur.

If you really are starting with the liquid... (AN is ammonium nitrate)

ΔH_{f, AN(l)} = ΔH_{f, AN(s)} + ΔH_fusion (This assumes that enthalpies don't change with T)

You need to make sure you have the right equation and information to get a more thorough answer. Take care.