Calorimeter problem

Master coffee-cup calorimeter equation:

q_{p, calorimeter +contents} = ΔH_{calorimeter+contents} = 0

The sum of all the q terms (“heat flows”) is zero (i.e., all heat stays within the calorimeter).

If something in the calorimeter loses heat energy (-q), then something else gains heat energy (+q).

Example:     

In a perfect calorimeter, 0.0125 mol NaOH and 0.0125 mol HCl dissolved in 75.00 mL aqueous solution are allowed to react to completion.   The temperature of the solution increases from 25.00^oC to 27.21^oC.

Find ΔH_rxn for the neutralization reaction in kJ/mol H₂O formed. 

Assume the solution has density = 1.00 g/mL and c = 4.184 J/mol^.oC.  Assume the calorimeter absorbs zero heat energy.

Let’s picture this in terms of ‘system’ and ‘surroundings’.  Since all heat flow stays within the calorimeter, the ‘universe’ in this case is just the NaOH, HCl, and the rest of the solution.

q_system                 +                         q_surroundings         =       0

q_rxn                     +                         q_solution               =       0

HCl + NaOH                                                                    Solution

Gives off heat energy                                                Absorbs heat energy

q_systen is negative                                                         q_surroundings is positive

(exothermic reaction)                                                (temp. inc. of +2.21^oC)

Since ΔH_rxn in kJ/mol H₂O formed is desired, we should watch stoichiometry and units:

NaOH + HCl → H₂O + NaCl  (1 mol NaOH & 1 mole HCl form 1 mol H₂O)

According to the amounts given, 0.0125 mol NaOH & 0.0125 mol HCl react so the temperature change measured is due to the formation of only 0.0125 mol H₂O.  Using density, we see that the solution has a mass of 75.00 g.

           q_rxn                                      +                   q_solution                               =   0

(0.0125mol H₂O)(Δh_rxn)       +  (75.00g)(4.184 J/g^.oC)(2.21^oC)=   0

                                 Δh_rxn             =   -(693.5 J) / (0.0125 mol H₂O)

                                 Δh_rxn             =   -55500 J / mol H₂O

                                 ΔH_rxn            =   -55.5 kJ / mol H₂O

So, the enthalpy change of the NaOH/HCl neutralization reaction is:

ΔH_rxn  =   -55.5 kJ / mol H₂O