Question

Ammonium nitrate dissolves in water to form ammonium ions and nitrate ions. You decide to determine the enthalpy of this dissociate since it is a type of reaction. ?

You take 8.00 g of ammonium nitrate and dissolve it in a calorimeter that contains 1.000 L. The temperature of the water decreases from 21.03 ˚C to 20.39 ˚C.

1. What was the heat transfer of the water?
2. What was the heat transfer of the reaction (qrxn)?
3. How many moles of ammonium nitrate were used?
4. What was the enthalpy of this dissociation reaction (Δ H r x n )?

Answer 1

`(Delta H_r)/n= 26.3 (kJ)/(mol)`

Explanation:

ammonium nitrate (MM= 80.0 g/mol) `NH_4NO_3` has an endotermic solubilization.
You have `(8g)/(80 g/(mol)) =0,1 mol` of it
Neglecting the costant of calorimeter, and considering that for water 1L = 1kg, you have:
`Q= m xx cp xx Delta T= 1 kg xx 4,186 (KJ)/(Kg) xx (21.03-20,39)°C = 2,63 kJ`
the heat of dissociation is
`(Delta H_r)/n=( 2,63 kJ)/(0.1 mol)= 26.3 (kJ)/(mol)`