﻿﻿ Measuring the Enthalpy Change Lecture Demonstrations - ChemPRIME

# Measuring the Enthalpy Change Lecture Demonstrations

## Calorimeter Constant

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[1]. 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"[2]

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[3]. 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.

## References

1. We used Vernier LoggerPro(R) software
2. J. Chem. Educ., 2004, 81 (1), p 64A
3. We used Vernier LoggerPro(R) software