Question

In a coffee cup calorimeter, 1.60 g of $N{H}_{4}N{O}_3$is mixed with 75.0g of water at an initial temperature of $25.00°C$. After dissolution of the salt, the final temperature of the calorimeter contents is $23.34°C$. Assuming the solution has a heat capacity of $4.18J°/Cg$and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of $N{H}_{4}N{O}_3$in units of $kJ/mol.$

Short answer

The enthalpy change for the dissolution of $N{H}_{4}N{O}_3$is $2656.575kJ/mol$.

Lattice energy is the energy change upon formation of one mole of crystalline ionic compound from its constituent ions.

Enthalpy of hydration is the amount of energy released on dilution of one mole of gaseous ions.

Enthalpy change is the difference between the total reactants and the product molar enthalpy.

Step-by-step solution

Explanation

$\begin{array}{rcl}Massofsolution& =& 1.6g+75.0g=76.6g\\ Fallintemperature& =& 25-23.34°C=1.66°C\\ HeatAbsorbed& =& Mass\times Specificheat\times Temperaturefall\\ & =& 76.6\times 1.66\times 4.18\\ & =& 531.5J=0.5315kJ\\ & & \end{array}$

Molecular weight of $N{H}_{4}N{O}_3=80g$

Heat absorbed by $80g=26.575kJ$

Enthalpy change$=+26.575kJ$$=+26.575kJ$

For dissolution of NH4NO3:

Enthalpy of hydration = 2630 kJ /mol

$\begin{array}{rcl}LatticeEnergy& =& Enthalpychange+Enthalpyofhydration\\ & =& 2630+26.575\\ & =& 2656.575kJ/mol\\ & & \end{array}$