CHM 111 Determine the enthalpy of chemical reaction - hess’s law Report sheet and data analysis

CHM 111 DETERMINE THE ENTHALPY OF CHEMICAL REACTION - HESS’S LAW
REPORT SHEET AND DATA ANALYSIS
PART I, II, AND III

Reaction 1 Reaction 2 Reaction 3

Maximum temperature (°C) 31.4 °C 20.7 °C 30.8 °C

Initial temperature (°C) 19.7 °C 20.1 °C 19.3 °C

Temperature change (∆T) 11.7 0.6 11.5

Heat, q (kJ) 5.04 kJ 0.258 kJ 4.955 kJ
- 50.37 kJ/mol - 2.58 kJ/mol - 49.42 kJ/mol
∆H (kJ/mol)


DATA ANALYSIS
1. Calculate the amount of heat energy, q, produced in each reaction. Use 1.03 g/mL for the
density of all solutions. Use the specific heat of water, 4.18 J/(g•°C), for all solutions.

Reaction 1: q = 4.184 J/g°C x 103 g x 11.7 = 5037.63 J → 5.04 kJ

Reaction 2: q = 4.184 J/g°C x 103 g x 0.6 = 258.57 J → 0.258 kJ

Reaction 3: q = 4.184 J/g°C x 103 g x 11.5 = 4955.95 J → 4.955 kJ

2. Calculate the enthalpy change, ∆H, for each reaction in terms of kJ/mol of
each reactant.

Reaction 1: ∆H : 5..10 = - 50.37 kJ/mol

Reaction 2: ∆H : 0..10 = - 2.58 kJ/mol

Reaction 3: ∆H : 4..10 = - 49.42 kJ/mol

3. Use your answers from 2 above and Hess’s law to determine the experimental molar
enthalpy for Reaction 3.

∆H1 + ∆H2 = -50.37 + (-2.58) = -52.95 kJ/mol (predicted)

Difference between predicted and obtained ∆ = ∣-52.95 - (-49.42)∣ = 3.53 kJ/mol