D. Energy Changes and Rates of Reaction · Energy Changes and Rates of ReactionInquiry: Testing Hess's Law with the Combustion of Magnesium
D2.6 D2.5 D3.4 A1.4 A1.5 A1.6 A1.8 A1.10 A1.13About 70 minutes
Learning goals- I can explain how two measurable reactions and a known value combine to give the enthalpy of combustion of magnesium.
- I can calculate ΔH values from calorimetry data and combine them using Hess's law.
- I can compare an experimental result with the accepted value and analyse specific sources of error.
- I can describe the safety precautions for working with hydrochloric acid and hydrogen gas.
Words to know
excess reactant
A reactant present in more than the amount needed, so some is left over; the other reactant is limiting.
limiting reactant
The reactant that is used up first and so determines how much reaction happens.
net ionic equation
An equation showing only the ions and molecules that actually change, leaving out spectator ions.
WHMIS
The Workplace Hazardous Materials Information System, which uses pictograms and safety data sheets to communicate hazards.
corrected maximum temperature
The highest temperature found by extrapolating the cooling curve back to the moment of mixing.
Too bright to measure
Magnesium ribbon burns with a dazzling white flame (you may have seen it in SCH3U). The reaction 2Mg(s) + O₂(g) → 2MgO(s) is one of the most exothermic you will meet, but measuring its ΔH directly in a school lab is nearly impossible: the flame is extremely hot, the heat escapes in every direction, and the light is bright enough to harm your eyes. In this lesson you will see how a class tests Hess's law by measuring two gentle reactions in a coffee-cup calorimeter and combining them with a known value to find the enthalpy of combustion of magnesium. (This is expectation D2.6.)
The plan
The target is the molar enthalpy of combustion (and of formation) of magnesium oxide:
Mg(s) + ½O₂(g) → MgO(s) ΔH = ?
We can measure these two reactions safely in a calorimeter with dilute hydrochloric acid:
(1) Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) ΔH₁ (measure)
(2) MgO(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l) ΔH₂ (measure)
and we look up the third:
(3) H₂(g) + ½O₂(g) → H₂O(l) ΔH₃ = −285.8 kJ
Keep (1), reverse (2) and keep (3). Adding them, MgCl₂(aq), 2HCl(aq), H₂(g) and H₂O(l) all cancel, leaving the target. So ΔH = ΔH₁ − ΔH₂ + ΔH₃.
How the three equations combine. Equation (2) is reversed, so its ΔH changes sign; everything except the target equation cancels.The procedure (as done in a school lab)
Safety first: wear goggles and an apron. 1.0 mol/L hydrochloric acid is corrosive (WHMIS: corrosive pictogram); rinse any spill with lots of water. Reaction (1) produces hydrogen gas, which is flammable: no flames in the room, and do the reaction in a ventilated area. Avoid breathing magnesium oxide powder.
- Measure 100.0 mL of 1.0 mol/L HCl(aq) into a coffee-cup calorimeter. Record its temperature every 30 s for 2 min.
- Add about 0.20 g of clean magnesium ribbon (mass measured to 0.001 g), cut into short pieces. Cover, stir, and record the temperature every 30 s until it has peaked and begun to fall.
- Rinse the calorimeter and repeat with fresh acid and about 0.50 g of magnesium oxide powder (mass to 0.001 g).
- The acid is in excess in both trials, so the magnesium or magnesium oxide is the limiting reactant: calculate n from its mass.
| Measurement | Trial 1: Mg | Trial 2: MgO |
|---|
| volume of 1.0 mol/L HCl(aq) | 100.0 mL | 100.0 mL |
| mass of solid | 0.200 g | 0.500 g |
| initial temperature | 21.40 °C | 21.50 °C |
| highest (corrected) temperature | 30.30 °C | 25.80 °C |
Sample class data. Assume the solution has a mass of 100.0 g and c = 4.18 J/(g·°C).Worked exampleExample 1: ΔH₁ for magnesium with acid
Use the Trial 1 data to find ΔH₁ in kJ per mole of Mg (M = 24.31 g/mol).
- ΔT = 30.30 − 21.40 = 8.90 °C.
- q_water = 100.0 g × 4.18 J/(g·°C) × 8.90 °C = 3720 J = 3.720 kJ.
- n(Mg) = 0.200 g ÷ 24.31 g/mol = 0.008227 mol.
- ΔH₁ = −3.720 kJ ÷ 0.008227 mol = −452.2 kJ/mol.
Worked exampleExample 2: ΔH₂ for magnesium oxide with acid
Use the Trial 2 data to find ΔH₂ in kJ per mole of MgO (M = 40.31 g/mol).
- ΔT = 25.80 − 21.50 = 4.30 °C.
- q_water = 100.0 × 4.18 × 4.30 = 1797 J = 1.797 kJ.
- n(MgO) = 0.500 g ÷ 40.31 g/mol = 0.01240 mol.
- ΔH₂ = −1.797 kJ ÷ 0.01240 mol = −144.9 kJ/mol.
Worked exampleExample 3: Combining with Hess's law
Find ΔH for Mg(s) + ½O₂(g) → MgO(s) and the percentage error. Accepted value: −601.6 kJ/mol.
- ΔH = ΔH₁ − ΔH₂ + ΔH₃ = (−452.2) − (−144.9) + (−285.8).
- = −452.2 + 144.9 − 285.8 = −593.1 kJ.
- % error = (−593.1 − (−601.6)) ÷ (−601.6) × 100% = 8.5 ÷ (−601.6) × 100% = −1.4%.
ΔH = −593 kJ/mol of MgO, within about 1.4% of the accepted value. The experiment supports Hess's law.
The accepted values drawn to scale. The left path (ΔH₁ then ΔH₃) and the right path (the target, then ΔH₂) start and end at the same levels, so ΔH₁ + ΔH₃ = target + ΔH₂.Analysing the result
A small percentage error like −1.4% means the experiment supports Hess's law within experimental uncertainty. Specific sources of error to discuss:
- Heat lost from the calorimeter makes both ΔT values too small. Because ΔH₂ is subtracted, the losses partly cancel, which is one reason this lab often works well.
- Oxide coating on the magnesium ribbon (it should be sanded or cleaned first) means less Mg reacted than was weighed, making ΔH₁ too small in size.
- Magnesium oxide absorbing moisture or CO₂ from the air before weighing (making n(MgO) too large).
- Mass of solution: using 100.0 g ignores the dissolved solid and the density of 1.0 mol/L HCl (slightly above 1.00 g/mL).
- Hydrogen bubbles carrying away a little heat and a little spray.
When writing your conclusion, answer the question you asked: Do the measured values combine to give the accepted ΔH, within experimental error?
Hands-on activityModel the lab with kitchen chemistry
This home version uses the same logic with harmless reactions. You need: your foam-cup calorimeter from the earlier lesson, a kitchen thermometer, baking soda (sodium hydrogen carbonate), vinegar, water, a measuring cup and measuring spoons.
- Reaction A: 50 mL of vinegar + 1 level teaspoon of baking soda in the calorimeter. Record the starting temperature and the lowest temperature.
- Reaction B: 50 mL of water + 1 level teaspoon of baking soda (it just dissolves). Record the temperature change.
- Think of the reaction in vinegar as two steps: first the baking soda dissolves (like B), then the dissolved hydrogen carbonate reacts with the acid. By Hess's law, ΔT(A) ≈ ΔT(B) + ΔT(second step), when the same masses and volumes are used.
- Calculate q for A and B (assume 50 g, c = 4.18 J/(g·°C)) and use Hess's law to estimate q for the second step.
What to look for: both A and B are endothermic (the temperature falls); A usually falls more. The second step's q comes from the difference. Your answer will be rough because CO₂ foam escapes and the volumes are small, which is a good example to write about in an error analysis. Do the reaction over a sink.
Just for funMagnesium burns so brightly that early photographers used magnesium "flash powder" to light their pictures. The chemistry teacher's version of "say cheese" is "say MgO".
Your turn
1Why is the combustion of magnesium not measured directly in a school calorimeter?
The flame is extremely hot and bright: the heat escapes in all directions (cannot be captured accurately), and the intense light can harm your eyes.
2Show how equations (1), reversed (2) and (3) cancel to give Mg(s) + ½O₂(g) → MgO(s).
Sum: Mg + 2HCl + MgCl₂ + H₂O + H₂ + ½O₂ → MgCl₂ + H₂ + MgO + 2HCl + H₂O. MgCl₂, 2HCl, H₂ and H₂O appear on both sides and cancel, leaving Mg(s) + ½O₂(g) → MgO(s).
30.150 g of Mg reacts with excess HCl in 100.0 g of solution; ΔT = +6.60 °C. Find ΔH₁.
q = 100.0 × 4.18 × 6.60 = 2759 J. n = 0.150 ÷ 24.31 = 0.006170 mol. ΔH₁ = −2.759 ÷ 0.006170 = −447 kJ/mol.
40.400 g of MgO reacts with excess HCl in 100.0 g of solution; ΔT = +3.50 °C. Find ΔH₂.
q = 100.0 × 4.18 × 3.50 = 1463 J. n = 0.400 ÷ 40.31 = 0.009923 mol. ΔH₂ = −1.463 ÷ 0.009923 = −147 kJ/mol.
5Use the answers to the last two questions and ΔH₃ = −285.8 kJ to find ΔH for Mg + ½O₂ → MgO and the % error (accepted −601.6 kJ/mol).
ΔH = −447 − (−147) + (−285.8) = −586 kJ/mol. % error = (−586 + 601.6) ÷ (−601.6) × 100% = −2.6%.
6Why is the acid used in excess in both trials?
So that all the Mg (or MgO) reacts; then the solid is the limiting reactant and n can be calculated from its mass.
More practice: warm-up
1Which gas is produced when magnesium reacts with hydrochloric acid?
2Which equation is reversed when the three are combined?
Equation (2), MgO(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l).
3What is the WHMIS hazard of hydrochloric acid?
4Write the target equation of this lab.
5Which ΔH value is taken from a table rather than measured?
ΔH₃, the formation of liquid water (−285.8 kJ).
More practice: on your own
1A student forgot to sand the oxide coating off the magnesium ribbon. How does this affect ΔH₁ and the final answer?
Part of the weighed "magnesium" was really MgO, which releases less heat per gram. So q is too small for the n calculated: ΔH₁ is too small in size, and the final ΔH (= ΔH₁ − ΔH₂ + ΔH₃) is also too small in size.
2Explain why heat loss affects the final answer less than you might expect in this lab.
Heat loss makes both ΔH₁ and ΔH₂ too small in size. Since ΔH₂ is subtracted, part of the error cancels out.
3Using the accepted values ΔH₁ = −466.9 kJ/mol and ΔH₂ = −151.1 kJ/mol, predict the temperature rise when 0.250 g of Mg reacts in 100.0 g of solution.
n = 0.250 ÷ 24.31 = 0.01028 mol; heat = 0.01028 × 466.9 = 4.801 kJ = 4801 J. ΔT = 4801 ÷ (100.0 × 4.18) = 11.5 °C.
4Write the net ionic equations for reactions (1) and (2).
(1) Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g); (2) MgO(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂O(l). (Cl⁻ is a spectator ion.)
5A group gets ΔH = −640 kJ/mol. Calculate the % error and suggest one error that would make the result too large in size.
% error = (−640 + 601.6) ÷ (−601.6) × 100% = +6.4%. A cause: ΔH₂ measured too small in size (for example, MgO absorbed water and CO₂ from the air before weighing, so less MgO reacted than calculated). Subtracting a too-small ΔH₂ makes the final answer too large in size.
Challenge
1Design a similar Hess's law experiment to find ΔH for Zn(s) + ½O₂(g) → ZnO(s). Write the three equations and the combination rule.
(1) Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g), ΔH₁ (measure); (2) ZnO(s) + 2HCl(aq) → ZnCl₂(aq) + H₂O(l), ΔH₂ (measure); (3) H₂(g) + ½O₂(g) → H₂O(l), −285.8 kJ. ΔH = ΔH₁ − ΔH₂ + ΔH₃.
2If the 1.0 mol/L HCl has a density of 1.02 g/mL, recalculate ΔH₁ from Example 1 using a solution mass of 102.0 g + 0.200 g of Mg. Does this improve the final answer?
m = 102.2 g; q = 102.2 × 4.18 × 8.90 = 3802 J; ΔH₁ = −3.802 ÷ 0.008227 = −462.1 kJ/mol. For MgO (m = 102.5 g): q = 102.5 × 4.18 × 4.30 = 1842 J; ΔH₂ = −1.842 ÷ 0.01240 = −148.5 kJ/mol. ΔH = −462.1 + 148.5 − 285.8 = −599.4 kJ/mol, error about −0.4%. Yes, it improves the answer (the specific heat capacity of the solution is also slightly different from water's, which limits how far this correction can be trusted).
3Why is it acceptable to use a ΔH₃ value from a table in an experiment designed to test Hess's law?
Hess's law is tested by checking whether measured steps add up to the accepted overall value. ΔH₃ (forming water from hydrogen and oxygen) is itself a well-established, independently measured value. The experiment still tests whether the enthalpy changes of different paths agree.
Check your understandingQuiz
1. In the magnesium Hess's law lab, the target ΔH equals
2. Why is no flame allowed in the room during reaction (1)?
3. 0.0100 mol of Mg reacts and the water absorbs 4.60 kJ. What is ΔH₁?
4. A class finds ΔH = −590 kJ/mol; the accepted value is −601.6 kJ/mol. The best conclusion is
5. Which error would make ΔH₁ too small in size?
For parents and teachersBig ideas (D2.6, D2.5, D3.4): Hess's law can be tested by measuring the enthalpy changes of two reactions that are easy to measure (magnesium and magnesium oxide each reacting with acid), combining them with a known value, and comparing the result with the accepted enthalpy of combustion of magnesium.
How to help: Ask your teen to explain why equation (2) is reversed and how that changes the sign. Ask them to list two specific sources of error and say whether each makes the result too large or too small.
Common difficulties:
- Sign errors when subtracting a negative ΔH₂ (−452 − (−145) = −307).
- Using the mass of the solid instead of the mass of the solution in q = mcΔT.
- Forgetting that the acid is in excess, so moles come from the solid.
- Writing a conclusion that does not answer the question asked.
Activity safety: the home version uses only vinegar, baking soda and water; do it over a sink. The real lab uses corrosive acid and makes flammable hydrogen, so it is done only at school with goggles and supervision.