In a coffee-cup calorimeter, 1.62 g is mixed with 74.1 g water at an initial temperature of 25.44°c. after dissolution of the salt, the final temperature of the calorimeter contents is 23.33°c. assuming the solution has a heat capacity of 4.18 j/°c·g and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of in units of kj/mol.

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.

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