Question

A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 107 g of solution. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 ∘C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water.

Answer 1

Explanation:

The given data is as follows.

             mass (m) = 107 g,      

    $\Delta T$ = $(24.70 - 21.00)^{o}C = 3.7^{o}C$

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

Since, the relation between heat energy, mass and temperature change is as follows.

                     q = $mC \Delta T$

Hence, putting the given values into the above formula to calculate the heat energy as follows.

                     q = $mC \Delta T$

                        = $107 g \times 4.18 J/g^{o}C \times 3.7^{o}C$

                        = 1654.86 J

Therefore, calculate the enthalpy of this reaction for 2.00 mol of a compound as follows.

                     $\frac{1654.86 J}{2.00 mol}$              

                  = 827.43 J/mol

or,               = $827.43 \times \frac{1 kJ}{1000 J}$       (as 1 kJ = 1000 J)

                  = 0.827 kJ/mol

Therefore, we can conclude that enthalpy of this reaction is 0.827 kJ/mol.