Chemistry: 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.

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Commercial cold packs consist of solid nh4no3 and

khushailawadip9dkn2

12.01.2022

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

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