1. Determining the enthalpy of a reaction

2. Determine the enthalpy of this reaction: Mg(OH) 2 (s) + H 2 SO 4 (aq) → MgSO 4 (aq) + 2H 2 O(l) Method Measure out 50 mL of 1 mol L -1 sulfuric acid (an excess) into a styrofoam cup. Weigh accurately about 1.5 g of Mg(OH) 2 powder. Measure the temperature of the sulfuric acid. Quickly add the Mg(OH) 2 powder, stir and measure the maximum temperature of the reaction mix.

3. Measure 50 mL of about 1 mol L -1 sulfuric acid. Pour the acid into a styrofoam cup (two stacked together will give even better insulation).

4. Weigh accurately about 1.5 g of Mg(OH) 2 powder.

5. Measure the initial temperature of the sulfuric acid solution. T initial = 16.5 °C

6. Add the Mg(OH) 2 to the acid.

7. Stir well and watch the temperature. All the powder should dissolve since we have excess acid. Measure the maximum temperature change.

8. T final = 28.0 °C

9. Calculate: The amount, in moles, of Mg(OH) 2 reacting. The total energy released during the reaction. The energy change per mole of Mg(OH) 2 reacting.

10. The energy released by the reaction has made the temperature of the solution rise by 28.0 °C – 16.5 °C = 11.5 °C All dilute aqueous solutions — and 1 mol L -1 is still considered to be dilute — contain very much more water than they do any other reagent. We therefore assume that the specific heat of the solution is equal to that of water, which is 4.18 J °C -1 g -1. That is, it takes 4.18 J of energy to change the temperature of 1 g of the solution by 1 °C. The mass of 1 mL of water at room temperature is 1 g. We assume that 50 mL of acid has a specific heat equivalent to 50 g of water.

11. Energy = mass of water heated × temperature change ×4.18 J °C -1 g -1

12. Mg(OH) 2 (s) + H 2 SO 4 (aq) → MgSO 4 (aq) + 2H 2 O(l) ∆H = -92.9 kJ mol -1 The temperature rose, so the reaction is exothermic and the ∆H is negative: