Measuring the Enthalpy Change Lecture Demonstrations
To a styrofoam cup calorimeter containing 250 mL of water at 24.0 oC is added a known amount of heat as 250 mL of water from a second styrofoam cup at 32.0 oC. The final temperature, measured by a computer-interfaced thermistor, is 27.7 oC. The computer-generated plot of T vs. time should be projected. What is the calorimeter constant?
ΔT calorimeter = 27.7 oC – 24.0 oC = 3.7 oC
ΔT cold = 27.5 oC – 24.0 oC = 3.7 oC
ΔThot = 27.7 – 32.0 = -4.3 oC
qhot = m x S.H. x ΔT = 250 g x 4.18 J/goC x -4.3oC = -4494 J
qcold = m x S.H. x ΔT = 250 x 4.18 x 3.7 oC = 3867 J
qcalorimeer + qhot + qcold = 0
qcalorimeter -4494 + 3867 = 0
qcalorimeter = 627 J.
C = Qcalorimeter/T = 627 J / 3.7 oC = 169 J/oC
The Ammonium Nitrate "Cold Pack"
A styrofoam cup calorimeter contains 250 mL of water at 25.0oC. Solid NH4NO3 (5 g ) is added, and the temperature falls to 23.8oC. The computer-generated plot of T vs. time should be projected. The calorimeter is found to absorb 169 J to change its temperature 1 oC, so it is said to have a calorimeter constant of 169 J/oC. What is the enthalpy change for the dissolution reaction?
qwater = m x S.H. x ΔT = 250 g x 4.18 J/goC x -1.2oC
= -1254 J
qcalorim = C x ΔT = 169 x -1.2oC = -203 J
q tot = -1254 + -203 = -1457 J
q rxn = +1457 J.
ΔH rxn = q (kJ) / n (mol)
n = m / M = 5 g NH4NO3 / 80 g/mol = 0.057 mol
ΔH rxn = q/n = 1.457 kJ / 0.057 mol = +25.6 kJ/mol
The reaction is spontaneous even though it is endothermic, because of the large positive entropy change resulting from water association with the separate ions in solution.