Question

A coffee cup calorimeter initially contains 125g of water at 24.2oC. 10.5g of potassium bromide also at 24.2oC is added to the water. After the KBr dissolves the final temperature is 21.1oC. Calculate the enthalpy change for dissolving the salt in J/g and kJ/mol. Assume specific heat of solution is 4.18J/goC.

Answer 1

Explanation:

The given data is as follows.

     Total mass of KBr solution (m) = 125 g + 10.5 g

                                                        = 135.5 g

   Change in temperature (dT) = 24.2 - 21.1 = $3.1^{o}C$

         Specific heat (C) = 4.184 $J/g^{o}C$

Hence, heat absorbed will be calculated as follows.

                  q = mC dT

                     = $135.5 g \times 4.184 J/g^{o}C \times 3.1^{o}C$

                     = 1757.5 J

Therefore, enthalpy change per gram of salt will be calculated as follows.

                   $\frac{1757.5 J}{135.5 g}$

                      = 12.97 J/g

So, no. of moles present in 10.5 g of KBr are calculated as follows.

      No. of moles = $\frac{mass}{\text{molar mass}}$

                            = $\frac{10.5 g}{119 g/mol}$

                            = 0.0882 mol

Change in enthalpy will be calculated as follows.

           $\frac{1757.5 kJ}{1000 \times 0.0882 mol}$

               = 19.93 kJ/mol

Thus, we can conclude that the enthalpy change for dissolving the given salt is 19.93 kJ/mol and 12.97 J/g.