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

-1.37 kJ/mol

Step-by-step explanation:

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

`\Delta H=m* C* \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* 4.18* 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 = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (5.60\ g)/(80.043 \ g/mol)`

`Moles= 0.06996\ mol`

Thus,
`\Delta H=-(0.096)/(0.06996)\ kJ/mol=-1.37\ kJ/mol`