Ch. 9 Calorimeter problem #2 (started Monday, 1/14/07)

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, 0.500 g Mg is added to 100.0 g of 1.00 M HCl to give 100.5 g solution.  The temperature of the solution increases from 22.2^oC to 44.8^oC.

Find ΔH_rxn for the reaction Mg_(s)  +  2HCl_(aq)  →  MgCl_2(aq)  + 2 H_2(g) in kJ/mol Mg. 

Assume the solution has c = 4.20 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

Mg + 2 HCl                                                                      Solution

Gives off heat energy                                                Absorbs heat energy

q_systen is negative                                                         q_surroundings is positive

(exothermic reaction)                                                (temp. inc. of +22.6^oC)

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

           Mg_(s)  +  2HCl_(aq)  →  MgCl_2(aq)  + 2 H_2(g) (1 mol Mg, not the 0.500 g in problem)

                                 ΔH_rxn in kJ/mol Mg is an intensive quantity

According to the amounts given, only 0.500 g Mg reacts so the temperature change measured is due to the only 0.500 g Mg reacting (not 1 mol). 

           q_rxn                           +                   q_solution                                          =   0

(0.500 g Mg)(Δh_rxn)       +  (100.5g)(4.20 J/g^.oC)(+22.6^oC)        =   0

                                 Δh_rxn             =   -(9540 J) / (0.500 g Mg)

                                 Δh_rxn             =   -19080 J / g Mg

These units are close, but not quite right.  However, it’s an easy fix to go from g Mg to mol Mg and from J to kJ:

Δh_rxn  =   (-19080 J / g Mg) ´ (24.31 g Mg / 1mol Mg) ´ (1kJ / 1000J)

                                 ΔH_rxn            =   -464 kJ / mol H₂O

So, the enthalpy change of the Mg/HCl reaction is:

ΔH_rxn  =   -464 kJ / mol Mg