Question

A "coffee-cup" calorimetry experiment is run for the dissolution of 3.20 g of aluminum nitrate placed into 103.2 mL of water. The temperature of the solution is initially at 23.2 oC. After the reaction takes place, the temperature of the solution is 17.7 oC. How much heat was absorbed or lost by the surroundings? Use 4.184 J/goC for the specific heat of the solution. Put your answer in units of kJ and make sure the sign is correct. What would be the enthalpy for the dissolution reaction of one mole of aluminum nitrate? Put your answer in kJ/mol and watch the sign for the enthalpy.

Answer 1

Answer: Enthalpy for the dissolution reaction of one mole of aluminum nitrate is -158 kJ/mol.

Explanation:

It is known that the density of water is 1 g/mL. So, mass of water will be calculated as follows.

         Mass = volume × density

                   = $103.2 ml \times 1 g/ml$

                   = 103.2 g

Specific heat of water is 4.18 $J/g^{o}C$.

Now, we will calculate the enthalpy of solution as follows.

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

                     = $103.2 g \times 4.18 J/g^{o}C \times (23.2 - 17.7)$

                     = 2372.5 J

As,    1 J = $10^{-3} kJ$. So, 2372.5 J will be converted into kJ as follows.

           $\Delta H$ = 2.37 kJ    

Molar mass of $Al(NO_{3})_{3}$ = 213 g/mol

Hence, moles of $Al(NO_{3})_{3}$ will be calculated as follows.

         Moles of $Al(NO_{3})_{3}$ = $\frac{3.20}{213}$

                           = 0.015

Therefore, enthalpy for the dissolution will be calculated as follows.

           $\Delta H = \frac{-2.37}{0.015}$

                        = -158 kJ/mol

Thus, we can conclude that enthalpy for the dissolution reaction of one mole of aluminum nitrate is -158 kJ/mol.