Experiment 25 Post Lab: Calorimetry

Part B. The enthalpy of neutralization for all strong acid-strong base reactions should be the same within experimental error. Explain. Will that also be the case for all weak acid-strong base reactions? Explain.

The enthalpy of neutralization for all strong acid-strong base reactions should be the same within experimental error because all those reactions are in simple terms, H+ + OH- = H2O. Although the enthalpy of neutralization for all strong acid-strong base reactions are within experimental error, that is not the case of weak acid-strong base reactions. This is not the case because weak acids do not completely dissociate into ions like strong acids do. Therefore the enthalpy of weak acid-strong base reactions is less than the enthalpy of strong acid-strong base reactions.

Part C.3. If some of the salt remains adhered to the weighing paper (and is therefore not transferred to the calorimeter), will the enthalpy of solution be reported too high or too low? Explain.

The enthalpy of solution would be reported as too low if some of the salt is not transferred to the calorimeter. This happens because the enthalpy of the solution is being measured with a lower mass instead of the original, entire mass.

Part C. The dissolution of ammonium nitrate, NH4NO3, in water is an endothermic process. Since the calorimeter is not a perfect insulator, will the enthalpy of solution, ΔHs, for ammonium nitrate be reported as too high or too low if this heat change is ignored? Explain.

The enthalpy of the solution for ammonium nitrate will be reported as too high if this heat change is ignored. The enthalpy will be too high because the calorimeter does not absorb all the heat. Since it does not absorb all the heat, more heat will have to be put into the system. Therefore, the enthalpy of the solution, ∆Hs, would be reported as too high.

Part B. Heat is lost to the Styrofoam calorimeter. Assuming a 6.22℃ temperature change for the reaction of HCl(aq) with NaOH(aq), calculate the heat loss to the inner 2.35-g Styrofoam cup. The specific heat of Styrofoam is 1.34 J/g•°C.

The heat loss to the inner Styrofoam cup can be calculated as q=mspht∆T. After substituting the values into the equation, we get q= (2.35 g)(1.34 J/g•°C)(6.22°C)= 19.6 J. Therefore, 19.6 J of heat was lost to the inner Styrofoam cup.

Part A.1. The 200-mm test tube also contained some water (besides the metal) that was subsequently added to the calorimeter (in Part A.4). Considering a higher specific heat for water, will the temperature change in the calorimeter be higher, lower, or unaffected by this technique error? Explain.

The temperature change in the calorimeter will be unaffected by this technique error. The original specific heat of water is 4.18 J/g•°C, but if this value increases that means that water will take more energy to change the temperature. Therefore, an increase in the amount of water does not make a difference because the specific heat of water was also increased.