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6.9Thermochemistry

Hess’s Law — AP CHEMISTRY Revision Notes & Topical Questions

5 min read 9 parts Lesson 14 of 14 1 quiz

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Lesson summary

6.9 Hess's Law — AP CHEMISTRY notes: Use Hess's law to combine reactions and calculate an unknown enthalpy change.Hess's law — the total enthalpy change of a reaction is the same whether it occurs in one step or severalIt follows from conservation of energy (first law of thermody…

  • Propane combustion
  • Reverse the second
  • State function
  • Reversed equation
  • Target equation
  • It lets ΔH be found for reactions that cannot be measured directly

Section 1 of 9Statement of the law

  • Hess's law
    • the total enthalpy change of a reaction is the same whether it occurs in one step or several
    • It follows from conservation of energy (first law of thermodynamics).
    • It lets ΔH be found for reactions that cannot be measured directly.

Section 2 of 9Rules for combining equations

Operation on equationEffect on ΔH
Reverse the equationChange the sign of ΔH
Multiply the coefficients by a factorMultiply ΔH by the same factor
Add equationsAdd the ΔH values

Section 3 of 9Reversing and scaling

Propane combustion
  • ΔH = −2044 kJ per mole, so the reverse reaction is +2044 kJ
Doubling
  • 2C3H8 + 10O2 → 6CO2 + 8H2O has ΔH = −4088 kJ
Halving
  • ½C3H8 amounts give −1022 kJ

Section 4 of 9Method

1
Find the given equation containing the required product or reactant.
2
Reverse it if the substance is on the wrong side and change the sign of ΔH.
3
Multiply it so the number of moles matches the target equation.
4
Repeat for each substance until all given equations are used.
5
Add the equations, cancel species appearing on both sides, and add the ΔH values.

Section 5 of 9Worked pattern: nitrogen dioxide

Given
  • N2 + O2 → 2NO, ΔH = +180 kJ; 2NO2 → 2NO + O2, ΔH = +112 kJ
Reverse the second
  • 2NO + O2 → 2NO2, ΔH = −112 kJ
Add
  • N2 + 2O2 → 2NO2, ΔH = +180 − 112 = +68 kJ

Section 6 of 9Worked pattern: ethane from ethene

1Target
  • C2H4 + H2 → C2H6
2Equations
  • Equations combustion of ethene (−1411), reversed combustion of ethane (+1560), and formation of water (−286)
3Result
  • ΔH = −1411 + 1560 − 286 = −137 kJ

Section 7 of 9More from the specification

1
Bonds formed
  • Product bonds, energy released
2
Standard conditions
  • 298 K and 101 kPa.
3
Bonds broken
  • Reactant bonds, energy absorbed
4
Haber process
  • Industrial manufacture of ammonia from nitrogen and hydrogen.
5
Exothermic
  • A reaction that transfers thermal energy to the surroundings.
6
Oxygen
  • A gas that relights a glowing splint.
7
Element
  • A substance made of only one type of atom.
8
Endothermic
  • A reaction that takes in thermal energy from the surroundings.

Section 8 of 9Key terms

TermMeaning
Hess's lawTotal enthalpy change is independent of the route.
State functionProperty that depends only on initial and final states.
Reversed equationEquation with products and reactants swapped and ΔH sign changed.
Target equationThe overall reaction whose ΔH is to be found.

Section 9 of 9Exam technique

1
Not changing the sign when reversing an equation.
2
Not multiplying ΔH when the equation is multiplied.
3
Failing to cancel species that appear on both sides before checking the target equation.
4
Saying route matters — ΔH is independent of the pathway.

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