When 22.2 g of ammonium nitrate is dissolved in 200 g of water, the temperature decreases from 20ºC to 12.5ºC.

Question

When 22.2 g of ammonium nitrate is dissolved in 200 g of water, the temperature decreases from 20ºC to 12.5ºC.

What is the enthalpy of solution of ammonium nitrate? Given, specific heat of solution is 4.20 J/g.K.

Answer 1

Let's first find the heat (enthalpy) of solution for the 22.2 g of ammonium nitrate.

q = mC∆T

q = heat = ?

m = mass = 200 g + 22.2 g = 222.2 g

C = specific heat = 4.20 J/gº

∆T = change in temperature = 20º - 12.5º = 7.5º

q = (222.2 g)(4.20 J/gº)(7.5º)

q = 6999 J

Since the temperature decreased, this tells us that heat is being lost from the surroundings and gained by the system. This makes the dissolution of ammonium nitrate ENDOTHERMIC and the sign is positive.

Now, to find the ∆Hsoln NH4NO3, we divide the heat by the moles as follows:

molar mass NH4NO3 = 80.05 g/mole

moles NH4NO3 used = 22.2 g x 1 mol/80.05 g = 0.2773 moles

∆Hsoln = 6999 J/0.2773 moles = 25,399 J/mole = 25,400 J/mole = 25.4 kJ/mol (3 sig. figs.)