Question

commercial cold packs consist of solid NH4NO3 and water. In a coffee-cup calorimeter, 5.60g NH4NO3 is dissolved in 100g of water at 22.0C; the temperature falls to 17.9C. Assuming that the specific heat capacity of the solution is 4.18 J/(g*K), calculate the enthalpy of dissolution of NH4NO3, in kJ/mol.

Answer 1

Answer:

-1.37 kJ/mol

Explanation:

The expression for the calculation of the enthalpy of dissolution of [tex[NH_4NO_3[/tex] is shown below as:-

$\Delta H=m\times C\times \Delta T$

Where,  

$\Delta H$  is the enthalpy of dissolution of [tex[NH_4NO_3[/tex]

m is the mass

C is the specific heat capacity

$\Delta T$  is the temperature change

Thus, given that:-

Mass of ammonium nitrate = 5.60 g

Specific heat = 4.18 J/g°C

$\Delta T=17.9-22.0\ ^0C=-4.1\ ^0C$

So,  

$\Delta H=-1.25\times 4.18\times 3.9\ J=-95.9728\ J$

Negative sign signifies loss of heat.  

Also, 1 J = 0.001 kJ

So,  

$\Delta H=-0.096\ kJ$

Also,

Molar mass of [tex[NH_4NO_3[/tex] = 80.043 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{5.60\ g}{80.043 \ g/mol}$

$Moles= 0.06996\ mol$

Thus, $\Delta H=-\frac{0.096}{0.06996}\ kJ/mol=-1.37\ kJ/mol$