ΔH= +32.38 kJ/mol
Assuming 1.62 g is the mass of Ammonium nitrate (NH₄NO₃)
we can find the number of moles of NH₄NO₃
Number of moles = Mass ÷ Molar mass
Molar mass of NH₄NO₃ = 80.05 g/mol
Therefore;
Moles of NH₄NO₃ = 1.62 g ÷ 80.05 g/mol
= 0.0202 moles
Initial temperature = 25.44°C
Final temperature = 23.33°C
Therefore, Change in temperature = 23.33°C - 25.44°C
= -2.11°C
Since, the change in temperature is negative then the reaction is endothermic which means heat was absorbed from the surrounding.
Mass of water = 74.1 g
We can calculate the heat absorbed from the surroundings;
Quantity heat = Mass × specific heat capacity × change in temperature
= 74.1 g × 4.18 J/°C·g × -2.11°C
= -653.547 Joules (1000 joules = 1 kJ)
= -0.654 kJ
But, the heat absorbed from the surroundings is equivalent to the heat of the reaction.
Therefore; Q(reaction) = +0.654 kJ
Required to calculate the Enthalpy of dissolution in kJ/mol.
Enthalpy for 0.0202 moles of NH₄NO₃ is +0.654 kJ
Therefore, for 1 mole
ΔH = +0.654 kJ ÷ 0.0202 mol
= + 32.38 kJ/mol
The value of enthalpy change is positive since the reaction is endothermic.
