1. Determining the enthalpy of a reaction

2. 25.0 mL of 0.500 mol L –1 AgNO 3 solution is mixed with 25.0 mL of 0.500 mol L –1 NaCl solution in a styrofoam cup. The temperature of the mixture rises by 3 °C. Calculate the ΔH for the precipitation of AgCl assuming that the heat capacity of the solution is 4.18 J °C –1 g –1. First calculate the energy absorbed by the liquid. We assume that 1 mL of liquid has a mass of 1 g, as is the case for pure water. Total mass of liquid = 25.0 g + 25.0 g = 50.0 g Energy absorbed = mass of liquid  ΔT  heat capacity = 50.0 g  3 °C  4.18 J °C –1 g –1 = 627 J 50.0 g 3 °C 4.18 J °C –1 g –1 

3. Now calculate the amount, in mol, of reactants present. Remember that we used 25.0 mL of 0.500 mol L –1 solutions of silver nitrate and sodium chloride. n(AgNO 3 ) = n(NaCl) = cV = 0.500 mol L –1  25.0  10 –3 L = 0.0125 mol Next, calculate the amount of energy released per mol of reactants. A total of 627 J was released. 25.0 0.500 mol L –1 =  627 J 0.0125 mol

4. Finally, decide on the sign of the ΔH and write the thermochemical equation. The temperature of the mixture increased, so the reaction is exothermic and the ΔH will be negative. AgNO 3 (aq) + NaCl(aq) → AgCl(s) + NaNO 3 (aq) ΔH = –50.16 kJ mol –1

5. The most common mistake in these calculations is in not selecting the correct value for the mass of liquid absorbing the energy. In this example, the total volume of liquid was 50 mL, so we were heating 50 g of water. In many cases you will be adding a small mass of powder to a liquid (such as 2 g of zinc powder to 100 mL of copper sulfate solution). The mass to use in that energy calculation is 100 g (for the water in the solution), not the 2 g of powder added, nor 102 g total mass of chemicals.