Hydrogen peroxide slowly decomposes in aqueous solution: H2O2 (aq) → H2O (l) + ½ O2(g) In the presen — Calorimetry Chemistry Question

- Indication that calorimetry will be used to solve the problem
- Excellent plan indicates a strategy to determine the heat capacity for the coffee cup
o Measure temperature change of known quantity of warm/hot water is added to a quantity of room or cold water temperature – plot temperature versus time – extrapolate to determine the initial and final temperatures of each water
o Determine the mass of warm and cold water used
o Calculate the heat capacity for the coffee cup
- Experimental details on calorimetry
o Measure temperature change of known quantity of hydrogen peroxide when iron (III) nitrate is added – plot temperature versus time – extrapolate to determine the initial and final temperatures
o Determine the mass of solutions used
o Calculate the heat of decomposition and enthalpy of decomposition
o Note: the rate of reaction is dependent on the concentration of iron(III) nitrate added. Excellent results are obtained with about 0.08 M iron (III) nitrate added. The enthalpy of decomposition should be independent of the concentration of catalyst used.
- Replicate measurements

Calorimetry Constant – one sample set of data
Room temperature water (50.0 mL): T₁ = 22.4°C and T₂ = 30.5°C
Warm water (50.0 mL): T₁ = 39.6°C and T₂ = 30.5°C

Decomposition of H₂O₂ - sample data
50 mL hydrogen peroxide: Tinitial = 20.0°C
Added 10 mL of 0.5 M iron (III) nitrate
Tfinal = 34.8°C

Student should observe the evolution of bubbles from the mixture and the change of color from pale yellow to darker amber color and back to pale yellow.

Calorimetry Constant – sample set of data
Room temperature water (50.0 mL): T₁ = 22.4°C and T₂ = 30.5°C
Warm water (50.0 mL): T₁ = 39.6°C and T₂ = 30.5°C

q_cold water = m c ΔT = 50.0 g x 4.184 J/g °C x (8.1°C) = 1694.5 J
q_warm water = m c ΔT = 50.0 g x 4.184 J/g °C x (9.1°C) = 1903.7 J
q_calorimeter = 1903.7 − 1694.5 = 209.2 J
C_cal = q_cal / ΔT = 209.2 J / 8.1 °C = 25.8 J/°C
Significant figures are indicated in black.
Acceptable ranges would include 0-60 J/°C

Decomposition of H₂O₂ - sample data
50.0 mL hydrogen peroxide: Tinitial = 20.0°C
Added 10.0 mL of 0.5 M iron (III) nitrate
Tfinal = 34.8°C

q_calorimeter = C_cal(Tf − Ti) = 25.8 J/°C (34.8°C − 20.0°C) = 381.8 J
q_solution = m c (Tf − Ti) = 4.184 J/g °C (60.0 g) (14.8°C) = 3715 J
q_total = q_cal + q_sol = 381.8 + 3715 J = 4097 J

3% Hydrogen peroxide, assume density = 1 g/mL - Students will need to calculate the moles of hydrogen peroxide added:
molarity = (3 g H₂O₂ / 100 g soln) x (1 g soln / 1 mL) x (1000 mL / 1 L) x (1 mol / 34.04 g) = 0.88132 M
moles = 0.88132 mol/L x 0.0500 L = 0.044066 mol

Enthalpy of decomposition
q_rxn = -q_total
ΔH = q_rxn / n = -4097 J / 0.044066 moles = -93,100 J/mol ( -92.975 kJ/mol)
Literature value is -94.6 kJ/mol. Acceptable ranges: -92.7 and 96.5 kJ/mol

Literature value is -94.6 kJ/mol. Acceptable ranges: -92.7 and 96.5 kJ/mol

A careful student would likely not open the coffee cup calorimeter during the experiment. Provided the prompt in the question, a strong student could perform a small scale reaction in a beaker to observe the reaction occurring. The iron (III) nitrate catalyzes the decomposition of the hydrogen peroxide. A good experimental finding that supports this answer is that the rate of bubble formation increases in the presence of the iron(III)nitrate. An excellent answer indicates the color change from pale yellow to darker color back to the pale yellow demonstrating that the catalyst is not consumed in the reaction. Another excellent answer could include providing the mechanism of decomposition (Haber^-Weiss Cycle):
Fe³⁺ + H₂O₂ → [FeIIIOOH]2+ + H⁺
[FeIIIOOH]2+ → [FeVO]3+ + H₂O
[FeVO]3+ + H₂O₂ → Fe³⁺ + H₂O + O₂