4.2.3Additional lessons
Hess’s Law Calculations — IB Chemistry SL Revision Notes & Topical Questions
5 min read 6 parts Lesson 53 of 98
Part 1 of 6Using Formation Data
Tabulated data turn Hess's law into a short calculation.
Formula
- ΔH° (reaction) = Σ ΔHf°(products) − Σ ΔHf°(reactants), each multiplied by its coefficient in the balanced equation.
Cycle idea
- The reactants are broken down to elements, and these elements are used to build the products;
- the two steps give the formula above.
Elements
- Elements in their standard states have ΔHf° = 0 and do not contribute, which simplifies many calculations.
Coefficients
- The enthalpy of formation is per mole, so it must be multiplied by the number of moles of each substance in the equation.
Part 2 of 6Using Combustion Data
Using Combustion Data
Formula
- ΔH° (reaction) = Σ ΔHc°(reactants) − Σ ΔHc°(products), the reverse of the formation formula.
Cycle idea
- Reactants and products are both burned to the same combustion products, so the cycle links them through CO₂ and H₂O.
When to use it
- Combustion data are common for organic compounds, whose formation from elements cannot be carried out directly.
Check
- Combustion enthalpies of CO₂ and H₂O are zero, since they cannot burn, so they do not contribute.
| Data given | Formula |
|---|---|
| ΔHf° | Σ products − Σ reactants |
| ΔHc° | Σ reactants − Σ products |
Part 3 of 6Step-by-Step Method
1
Write a fully balanced equation with state symbols for the reaction whose ΔH is required.
2
Decide whether the data given are enthalpies of formation or of combustion and choose the matching formula.
3
Multiply each tabulated value by its coefficient and keep track of the sign of every number.
4
Insert the totals into the formula, being careful with the subtraction of a negative value.
5
State the final ΔH in kJ mol⁻¹ and check that its sign matches whether the reaction should be exothermic or endothermic.
Part 4 of 6Common Applications
Common Applications
Combustion of fuels
- The enthalpy of combustion can be checked from the formation data of the fuel, CO₂ and H₂O.
Formation of hydrocarbons
- Carbon and hydrogen do not combine directly to give ethane, so ΔHf° is calculated from combustion data.
Decomposition reactions
- Formation data give the energy needed to decompose a compound into simpler products.
Comparing methods
- Results from Hess cycles agree better with experiment than bond enthalpy estimates.
Part 5 of 6Worked Example
Question Use ΔHf°: CH₄(g) = −75, CO₂(g) = −394, H₂O(l) = −286 kJ mol⁻¹ to find ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).
- Σ ΔHf°(products) = (−394) + 2(−286) = −394 − 572 = −966 kJ mol⁻¹.
- Σ ΔHf°(reactants) = (−75) + 2(0) = −75 kJ mol⁻¹, since O₂ is an element.
- ΔH = −966 − (−75) = −891 kJ mol⁻¹.
Answer ΔH = −891 kJ mol⁻¹
Part 6 of 6Exam Checklist & Tips
4.2.3 – Key Points for the Exam
- Standard enthalpy of formation The enthalpy change when one mole of a compound forms from its elements in their standard states.
- Standard enthalpy of combustion The enthalpy change when one mole of a substance burns completely in oxygen under standard conditions.
- Σ (sigma) The sum of the values, each multiplied by its coefficient in the balanced equation.
- Reaction enthalpy The enthalpy change for the amounts of substances shown in the balanced equation.
Common Mistakes
- Using the wrong formula, subtracting reactants from products for combustion data.
- Forgetting to multiply by the stoichiometric coefficient before summing.
- Losing a negative sign when subtracting a negative number.
- Including a non-zero value for an element in its standard state.
Last updated:
\n