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Hess's law and bond energies: TCE Chemistry (Tasmania)

Syllabus dot point

“Apply Hess's law and bond enthalpy data to calculate enthalpy changes for reactions.”

9 min read

Quick answer

Hess's law states that the total enthalpy change for a reaction is the same regardless of the route taken. This lets you add the enthalpies of known steps (reversing a step flips its sign, multiplying a step multiplies its ΔH\Delta H), or use bond enthalpies (bonds broken minus bonds made), to find an enthalpy change you cannot measure directly.

What this dot point is asking

TASC expects you to use Hess's law with energy cycles and standard enthalpies of formation, and to estimate enthalpy changes from average bond enthalpies.

Hess's law

Because enthalpy is a state function, the energy change depends only on the start and end points, not the path between them.

Hess's law

The total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final conditions are the same.

You can therefore construct an energy cycle: if a direct reaction is hard to measure, go via intermediate steps whose enthalpies are known and add them up. Reversing a step changes the sign of its ΔH\Delta H; multiplying a step multiplies its ΔH\Delta H.

Using enthalpies of formation

The standard enthalpy of formation ΔHf∘\Delta H_f^\circ is the enthalpy change when one mole of a compound forms from its elements in their standard states. Elements in their standard state have ΔHf∘=0\Delta H_f^\circ = 0.

Enthalpy from formation data

ΔHrxn∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_{rxn}^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})

Multiply each value by its coefficient in the balanced equation.

Using bond enthalpies

A reaction breaks bonds in the reactants (requiring energy) and makes bonds in the products (releasing energy):

ΔH≈∑(bonds broken)−∑(bonds made).\Delta H \approx \sum(\text{bonds broken}) - \sum(\text{bonds made}).

Bond enthalpies are averages over many molecules, so this method gives an estimate, not an exact value, and it strictly applies only to gaseous species.

Enthalpy from formation data

Use standard enthalpies of formation to calculate ΔH\Delta H for the combustion of methane: CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l). Data: ΔHf∘(CH4)=−75\Delta H_f^\circ(\text{CH}_4) = -75, ΔHf∘(CO2)=−394\Delta H_f^\circ(\text{CO}_2) = -394, ΔHf∘(H2O(l))=−286\Delta H_f^\circ(\text{H}_2\text{O}(l)) = -286, ΔHf∘(O2)=0\Delta H_f^\circ(\text{O}_2) = 0 (all in kJ mol−1\text{kJ mol}^{-1}).

Step 1: Identify the formation enthalpies of each species

Standard formation enthalpies are tabulated for one mole of each compound formed from its elements in their standard states. Elements in their standard state (such as O2(g)\text{O}_2(g)) always have ΔHf∘=0\Delta H_f^\circ = 0 and contribute nothing to the calculation.

Step 2: Sum the formation enthalpies of the products

Multiply each product's ΔHf∘\Delta H_f^\circ by the coefficient from the balanced equation, then add. Here there is one mole of CO2\text{CO}_2 and two moles of H2O\text{H}_2\text{O}:

∑ΔHf∘(products)=1(−394)+2(−286)=−394−572=−966 kJ.\sum \Delta H_f^\circ(\text{products}) = 1(-394) + 2(-286) = -394 - 572 = -966\ \text{kJ}.

Step 3: Sum the formation enthalpies of the reactants

Apply the same process to the reactants. Oxygen has ΔHf∘=0\Delta H_f^\circ = 0, so only methane contributes:

∑ΔHf∘(reactants)=1(−75)+2(0)=−75 kJ.\sum \Delta H_f^\circ(\text{reactants}) = 1(-75) + 2(0) = -75\ \text{kJ}.

Step 4: Subtract reactants from products

The formula ΔHrxn∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_{rxn}^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}) reflects the fact that the reaction breaks apart reactants (reversing their formation) and then forms products. Subtracting gives:

ΔH=−966−(−75)=−891 kJ mol−1.\Delta H = -966 - (-75) = -891\ \text{kJ mol}^{-1}.

The large negative value confirms that methane combustion is strongly exothermic.

Final answer: ΔH=−891 kJ mol−1\Delta H = -891\ \text{kJ mol}^{-1}

Bonds broken minus bonds made

Breaking bonds absorbs energy (positive) and forming bonds releases energy (negative). If more energy is released making bonds than is absorbed breaking them, the reaction is exothermic. Note this order is the reverse of the formation-data order.

Why bond enthalpies give only an estimate

Bond enthalpy data are average values measured across many different molecules, so the energy of, say, a C−H\text{C}-\text{H} bond in methane differs slightly from one in ethanol. When you use average bond enthalpies you are therefore approximating, and the result usually differs by a few percent from a value calculated using exact enthalpies of formation. Bond enthalpies also apply strictly to species in the gas phase, because they describe breaking bonds in isolated gaseous molecules; if a reactant or product is a liquid or solid, the additional enthalpy of vaporisation or sublimation is ignored, adding further error. For an exact answer, formation data or Hess cycles built from measured combustion enthalpies are preferred.

Building a Hess cycle

When formation data are not directly available, you can construct an energy cycle from any set of reactions that share common intermediates, typically combustion reactions or formation reactions. Write the target reaction, then arrange the known equations (reversing or scaling as needed) so that adding them reproduces the target exactly. The sum of the adjusted enthalpies is the target ΔH\Delta H. This is the method used in the benzene and propan-1-ol questions above, and it works because enthalpy is a state function: the route does not matter, only the start and end states.

Reversing the subtraction order

A frequent error is computing reactants minus products. The correct order is products minus reactants for formation data, but bonds broken minus bonds made for bond enthalpies (because those quantities already carry the opposite physical meaning). Keep the two formulae straight, and remember to flip the sign of any step you reverse in a Hess cycle.

In the exam, write the balanced equation first, label which method you are using, multiply every term by its coefficient, and check the sign against whether you expect the reaction to be exothermic or endothermic.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice3 marks
The enthalpy of formation of benzene, 3H2(g)+6C(s)→C6H6(l)3\text{H}_2(g) + 6\text{C}(s) \rightarrow \text{C}_6\text{H}_6(l) (Equation 1), cannot be measured directly. Use Hess's law with: Equation 2, C6H6(l)+152O2(g)→6CO2(g)+3H2O(l)\text{C}_6\text{H}_6(l) + \tfrac{15}{2}\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 3\text{H}_2\text{O}(l), ΔH=−3267 kJ mol−1\Delta H = -3267\ \text{kJ mol}^{-1}; Equation 3, H2(g)+12O2(g)→H2O(l)\text{H}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l), ΔH=−286 kJ mol−1\Delta H = -286\ \text{kJ mol}^{-1}; Equation 4, C(s)+O2(g)→CO2(g)\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g), ΔH=−394 kJ mol−1\Delta H = -394\ \text{kJ mol}^{-1}. Calculate ΔH\Delta H for Equation 1.
Show worked answer →

Combine: target =(6×Eq 4)+(3×Eq 3)−Eq 2= (6 \times \text{Eq 4}) + (3 \times \text{Eq 3}) - \text{Eq 2}. Reversing Equation 2 flips the sign of its ΔH\Delta H. (1 mark)

ΔH=6(−394)+3(−286)−(−3267)=−2364−858+3267=+45 kJ mol−1.\Delta H = 6(-394) + 3(-286) - (-3267) = -2364 - 858 + 3267 = +45\ \text{kJ mol}^{-1}.

(2 marks)

So the enthalpy of formation of benzene is about +45 kJ mol−1+45\ \text{kJ mol}^{-1} (endothermic, consistent with benzene being less stable than its elements).

Practice3 marks
For N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g), use the average bond energies to determine the theoretical ΔH\Delta H. H−H=436\text{H}-\text{H} = 436; N−H=391\text{N}-\text{H} = 391; N≡N=941 kJ mol−1\text{N}\equiv\text{N} = 941\ \text{kJ mol}^{-1}.
Show worked answer →

ΔH=∑(bonds broken)−∑(bonds made)\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made}).

Bonds broken: one N≡N=941\text{N}\equiv\text{N} = 941, plus three H−H=3×436=1308\text{H}-\text{H} = 3 \times 436 = 1308. Total =2249 kJ= 2249\ \text{kJ}. (1 mark)

Bonds made: each NH3\text{NH}_3 has three N−H\text{N}-\text{H} bonds and there are two molecules, so 6×391=2346 kJ6 \times 391 = 2346\ \text{kJ}. (1 mark)

ΔH=2249−2346=−97 kJ\Delta H = 2249 - 2346 = -97\ \text{kJ} per mole of reaction. The reaction is exothermic, close to the accepted −92 kJ mol−1-92\ \text{kJ mol}^{-1}. (1 mark)

Practice3 marks
Propan-1-ol combusts: C3H8O(l)+92O2(g)→3CO2(g)+4H2O(g)\text{C}_3\text{H}_8\text{O}(l) + \tfrac{9}{2}\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g). Using C(s)+O2(g)→CO2(g)\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g), ΔH=−393.5\Delta H = -393.5; 3C(s)+4H2(g)+12O2(g)→C3H8O(l)3\text{C}(s) + 4\text{H}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{C}_3\text{H}_8\text{O}(l), ΔH=−303.0\Delta H = -303.0; H2(g)+12O2(g)→H2O(g)\text{H}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(g), ΔH=−241.8 kJ mol−1\Delta H = -241.8\ \text{kJ mol}^{-1}, show that the heat of combustion is about −1850 kJ mol−1-1850\ \text{kJ mol}^{-1}.
Show worked answer →

Combine: target =(3×CO2 eq)+(4×H2O eq)−(formation of propan-1-ol)= (3 \times \text{CO}_2 \text{ eq}) + (4 \times \text{H}_2\text{O eq}) - (\text{formation of propan-1-ol}). Reverse the formation so its sign flips. (1 mark)

ΔH=3(−393.5)+4(−241.8)−(−303.0)=−1180.5−967.2+303.0=−1844.7 kJ mol−1.\Delta H = 3(-393.5) + 4(-241.8) - (-303.0) = -1180.5 - 967.2 + 303.0 = -1844.7\ \text{kJ mol}^{-1}.

(1 mark)

This rounds to approximately −1850 kJ mol−1-1850\ \text{kJ mol}^{-1}, as required. (1 mark)

Practise this

Sources & how we know this

Based on the official TASC course document. Not yet individually reviewed by a person; always check TASC and your school's materials for current rules.

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