Question

14.00 g of anhydrous sodium acetate (82.03 g/mol) was dissolved in 100.00 g of water, the heat capacity of the reaction mixture was 4.045 J/g K. The change in temperature was increased by 6.40 °C.
1. is this an exothermic or a.m. endothermic reaction?
2. the heat transferred (qsys), was?
3. the number of moles of anhydrous sodium acetate dissolved is?
4. the (delta H) of dissolution for anhydrous sodium acetate is?

Answer 1

Ans. It is an exothermic reaction because heat is being released as indicated by increase in temperature of the solution.

2. Using q = m x c x dT       

Where, q= heat change,                m= mass in gram,

c= specific heat                                 dT = change in temperature

Given, c = 1.045 J/gK

            dT = 6.40⁰C = 6.40 K                           [Note: change in temperature in ⁰C = K scale]

            total mass of the system = 14.00 g sodium acetate + 100 gram water. Because the temperature increase is for the solution, not for ‘water only’.

Or, q = 114 g x 1.045 J/gK x 6.40K = 762.432 J

3. No. of moles of anhydrous sodium acetate dissolved (assuming 100% dissolution)

                        = mass of Na-acetate/ molecular mass

                        = 14.00 gram / 82.03 gram per mol = 0.170669 moles

4. dH of dissolution = heat released per mole Na-acetate

We have, q = 762.432 J per 0.170699 moles.

So, heat released per mol = q / number of moles

                                                = 762.432 J / 0.170699 moles

                                                = 4467.31 J mol^-1 = 4.467 kJ mol^-1

Thus, dH dissolution of anhydrous Na-acetate = - 4.467 kJ mol^-1