Coffee cup calorimetry experiments can be used to obtain ∆Hf˚ for magnesium oxide. — Thermodynamics / Calorimetry Chemistry Question
Problem Context
Coffee cup calorimetry experiments can be used to obtain ∆Hf˚ for magnesium oxide.
Write a balanced equation for the formation of magnesium oxide, whose enthalpy change is ∆Hf˚.
Model Answer
Mg(s) + 1/2 O2(g) → MgO(s)
To determine the heat capacity of the calorimeter, 49.6 mL of 1.01 M HCl are reacted with 50.1 mL of 0.998 M NaOH. The solution's temperature increases by 6.40˚C. Determine the heat capacity of the calorimeter. You may assume the solution's specific heat capacity is 4.025 J·g–1 ·˚C–1 and the enthalpy of neutralization is –55.9 kJ per mole of H2O.
Model Answer
First, determine the limiting reactant:
Mol HCl = 1.01 mol/L × 0.0496 L = 0.00501 mol HCl
Mol NaOH = 0.998 mol/L × 0.0501 L = 0.0500 mol NaOH , so because it is a 1:1 stoichiometry, NaOH is limiting.
Via the enthalpy from the neutralization reaction, HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –55.9 kJ/mol we can calculate,
0.0500 mol NaOH × (–55.9 kJ / 1 mol NaOH) = – 2.795 kJ
Account for heat taken up by the solution, so the rest is taken up by the calorimeter:
Total volume of solution is 49.6 mL + 50.1 mL = 99.7 mL (no information is provided about density, so the simplest assumption is to use 1.00 g) so we have 99.7 g solution. Using the given specific heat capacity the heat absorbed by the solution is,
heat = 99.7 g × 4.025 J·g–1·˚C–1 × 6.40 ˚C = 2568 J (heat absorbed by the solution)
Now we can calculate the heat absorbed by the calorimeter: 2795 J – 2568 J = 227 J absorbed by the calorimeter.
So the heat capacity of the calorimeter is 227 J / 6.40 ˚C = 35.5 J·˚C–1
When 0.221 g of magnesium turnings are added to 49.9 mL of 1.01 M HCl and 49.7 mL of H2O in the same calorimeter, the temperature increases by 9.67˚C. Write a balanced equation for the reaction that occurs and calculate the ∆H per mole of magnesium. (Assume the solution's specific heat capacity is 3.862 J·g–1 ·˚C–1 and the calorimeter constant is the value obtained in b.)
Model Answer
The reaction of magnesium with an acid is: Mg + 2H+ → Mg2+ + H2
Total mass is: 99.6 g solution + 0.221 g Mg = 99.821 g
Total heat is heat absorbed by solution + heat absorbed by calorimeter:
heat solution = 99.821 g × 3.862 J·g–1·˚C–1 × 9.67 ˚C = 3728 J
heat calorimeter = 35.5 J·˚C–1 × 9.67 ˚C = 343 J
Total heat = 3728 J + 343 J = 4071 J
This is heat given off by 0.221 g Mg (using molar mass): 0.221 g Mg × (1 mol / 24.31 g) = 0.00909 mol Mg
Thus, –4071 J / 0.00909 mol = –4.479×10^5 J·mol–1 = –447.9 kJ·mol–1
When 0.576 g of MgO react with 51.0 mL of 1.01 M HCl and 50.1 mL of H2O in the same calorimeter the temperature rises 4.72˚C. Write a balanced equation for this reaction and calculate its ∆H per mole of MgO using the same assumptions as in part c.
Model Answer
The reaction is: MgO + 2H+ → Mg2+ + H2O
First determine moles reacted: 0.576 g MgO × (1 mol / 40.31 g) = 0.0143 mol MgO
Once again, total heat is heat absorbed by solution + heat absorbed by calorimeter: (and solution mass includes MgO)
heat solution = 101.676 g × 3.862 J·g–1·˚C–1 × 4.72 ˚C = 1853 J
heat calorimeter = 35.5 J·˚C–1 × 4.72 ˚C = 168 J
Total heat = 1853 J + 168 J = 2021 J
Thus, –2021 J / 0.0143 mol = –1.413×10^5 J·mol–1 = –141.3 kJ·mol–1
Use the above results and ∆Hf˚ of H2O(l) (–285.8 kJ·mol–1) to calculate ∆Hf˚ of magnesium oxide.
Model Answer
Now construct a series of reactions that when summed are the formation reaction for MgO:
Mg2+ + H2O → MgO + 2H+ ∆H = 141.3 kJ·mol–1
Mg + 2H+ → Mg2+ + H2 ∆H = –447.9 kJ·mol–1
Summed: These reaction yield:
Mg + H2O → MgO + H2 ∆H = –306.6 kJ·mol–1
Now combine this reaction with the heat of formation for water to yield the desired result:
Mg + H2O → MgO + H2 ∆H = –306.6 kJ·mol–1
H2 + 1/2 O2 → H2O ∆H = –285.8 kJ·mol–1
Mg + 1/2 O2 → MgO ∆H = –592.4 kJ·mol–1