Neutralization reactions occurring when acids and bases are mixed can be very exothermic. Suppose you are investigating how the heat released in various neutralization reactions is related to the structures of certain acids. A constant-volume calorimeter was calibrated by carrying out a reaction known to release 1.78 kJ of heat in 0.100 L of solution in the calorimeter (q = –1.78 kJ), resulting in a temperature rise of 3.65 °C. In a subsequent experiment, 50.0 mL of 0.200 m HCl(aq) and 50.0 mL of 0.200 m NaOH(aq) were mixed in the same calorimeter and the temperature rose by 1.26 °C (that is, ΔT = +1.26 K). What is the change in the internal energy of the neutralization reaction?

ANTICIPATE The rise in temperature in the actual experiment is about one-third the rise in the calibration, so you can estimate that the heat released by the reaction is about one-third of 1.78 kJ, or about 0.6 kJ.

PLAN The calculation has two steps. First, calibrate the calorimeter by calculating its heat capacity with data from the first reaction, C_cal = q_cal/ΔT with q_cal = –q. Second, use that value of C_cal to find the energy change of the neutralization reaction. For the second step, use q = –C_calΔT but with ΔT now the change in temperature observed during the reaction.

Note that the calorimeter contains the same volume of liquid in both cases, so that the temperature change is due only to a difference in the reaction that takes place. Finally, note that ΔU = q.

What should you assume? Because dilute aqueous solutions have approximately the same heat capacities as pure water, assume that the heat capacity for the calorimeter during the reaction is the same as in the calibration. Also, assume that there is no change in volume on mixing, so the final volume is 0.100 L.