SME
Grade 10 Science · Problem type
Hess’s law: adding steps
Use this for ΔH of this reaction, C + O₂ → CO₂, from C + ½O₂ → CO and CO₂ → CO + ½O₂.
Diagram
ΔH₁: C + O₂ → CO + ½O₂, −110.5 kJ
ΔH₂: CO + ½O₂ → CO₂, −283 kJ (the equation reversed, so +283 became −283)
ΔH = ΔH₁ + ΔH₂ = (−110.5) + (−283)
ΔH = −393.5 kJ
Down the ladder: heat is given off (exothermic).
This is an example. Type your own values to start a new problem.
ΔH (kJ)
=
ΔH₁ (kJ)
−
(
ΔHgiven (kJ)
)
ΔH₂
Step 2 reversed: CO + ½O₂ → CO₂
worked out (not typed) · kJ
Formulas
ΔH₂ = −ΔHgiven
−283 = −283
ΔH = ΔH₁ + ΔH₂
−393.5 = −110.5 + (−283)
Assumptions
•
ΔH depends only on where a reaction starts and ends, not on the path.
•
Reversing an equation flips the sign of its ΔH.
•
The steps add up to the overall equation: the CO made in step 1 is used in step 2.
Step-by-step
WE KNOW
ΔH₁ = −110.5 kJ, ΔHgiven = 283 kJ
FIND
step 2 reversed: CO + ½O₂ → CO₂ (ΔH₂), overall: C + O₂ → CO₂ (ΔH)
Step 1 · Find step 2 reversed: CO + ½O₂ → CO₂ (ΔH₂)
Use ΔH₂ = −ΔHgiven
The step is used backwards, so its ΔH changes sign.
ΔH₂ = −283 kJ
Step 2 · Find overall: C + O₂ → CO₂ (ΔH)
Use ΔH = ΔH₁ + ΔH₂
Hess’s law: add the steps’ ΔH.
ΔH = −110.5 − 283
ΔH = −393.5 kJ
Check
Put the numbers back in. Both sides should match.
−283 = −283 ✓
−393.5 = −110.5 − 283 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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Related lessons
Thermochemistry: heat, enthalpy and calorimetry
Main lesson
Calorimetry: q = mcΔT
Use this for “A reaction warms 100 g of water from 21 °C to 27.5 °C. How much heat did it give off?”, or a metal’s specific heat from its heat, mass and warming.
A cold pack: ΔH from a calorimeter
Use this for the ΔH per mole of a salt that cools the water as it dissolves.
Heat to turn ice into steam
Use this for the heat to take ice all the way to steam: warm it to 0 °C, melt it, warm the water to 100 °C and boil it away.
ΔH from heats of formation
Use this for ΔH of this reaction, CH₄ + 2O₂ → CO₂ + 2H₂O, from the heats of formation.