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What is the enthalpy of reaction, and how is it used in stoichiometric (thermochemical) calculations?

Topic 6.6 Introduction to Enthalpy of Reaction: interpret the enthalpy of reaction as a state function and use thermochemical equations to relate the heat of a reaction to the amount of substance reacted.

A focused answer to AP Chemistry Topic 6.6, covering the enthalpy of reaction as a state function, thermochemical equations, the meaning of the sign of delta H, and how to scale the heat of a reaction with the amount reacted, with full worked examples.

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  1. What this topic is asking
  2. Enthalpy of reaction as a state function
  3. Thermochemical equations
  4. Scaling, reversing and the sign
  5. Try this

What this topic is asking

The College Board (Topic 6.6) wants you to interpret the enthalpy of reaction ΔH\Delta H as a state function and use thermochemical equations to relate the heat of a reaction to the amount of substance reacted. This is the point where the heat measured in calorimetry becomes a property you can scale, reverse and combine.

Enthalpy of reaction as a state function

Because it is a state function, ΔH\Delta H behaves predictably under manipulation: reverse the reaction and the sign flips; scale the reaction and the value scales; add reactions and the enthalpies add (Hess's law). This is what makes thermochemistry quantitative and additive.

Thermochemical equations

So N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3, ΔH=−92 kJ\Delta H = -92\ \text{kJ} means 92 kJ92\ \text{kJ} is released per mole of N2\text{N}_2 (or per two moles of NH3\text{NH}_3). Reacting half as much N2\text{N}_2 releases half as much heat. The enthalpy is tied to the amounts in the equation, which is why you must track moles carefully.

Scaling, reversing and the sign

The sign convention is the same as before: exothermic reactions have ΔH<0\Delta H < 0 (heat out), endothermic reactions have ΔH>0\Delta H > 0 (heat in). Reversing the reaction reverses the sign, because the products and reactants swap roles. Scaling multiplies ΔH\Delta H by the same factor as the coefficients. These three operations (scale, reverse, add) are the toolkit for Hess's law and for relating the heat of a reaction to any given mass or amount.

Try this

Q1. A reaction has ΔH=−50. kJ\Delta H = -50.\ \text{kJ} as written. Calculate the heat released when one third of the molar amount reacts. [2 points]

  • Cue. q=13(−50.)=−16.7 kJq = \tfrac{1}{3}(-50.) = -16.7\ \text{kJ} (16.7 kJ released).

Q2. State the ΔH\Delta H for the reverse of a reaction whose forward ΔH=+120 kJ\Delta H = +120\ \text{kJ}. [1 point]

  • Cue. ΔHreverse=−120 kJ\Delta H_\text{reverse} = -120\ \text{kJ} (same magnitude, opposite sign).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). For 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), ΔH=−890. kJ mol−1\Delta H = -890.\ \text{kJ mol}^{-1} of methane. (a) State whether the reaction is endothermic or exothermic and justify. (b) Calculate the heat released when 0.2500.250 mol of CH4\text{CH}_4 burns. (c) Calculate the heat released when 8.008.00 g of CH4\text{CH}_4 burns (M=16.0M = 16.0 g/mol). (d) State the ΔH\Delta H for the reverse reaction and justify.
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A 4-point quantitative FRQ on enthalpy of reaction.

(a) Classification (1 point): ΔH<0\Delta H < 0, so the combustion is exothermic (it releases heat to the surroundings).
(b) Heat from 0.250 mol (1 point): q=nΔH=(0.250)(−890.)=−223 kJq = n\Delta H = (0.250)(-890.) = -223\ \text{kJ} (223 kJ released).
(c) Heat from 8.00 g (1 point): n=8.0016.0=0.500n = \dfrac{8.00}{16.0} = 0.500 mol; q=(0.500)(−890.)=−445 kJq = (0.500)(-890.) = -445\ \text{kJ} (445 kJ released).
(d) Reverse reaction (1 point): ΔHreverse=+890. kJ mol−1\Delta H_\text{reverse} = +890.\ \text{kJ mol}^{-1}, because enthalpy is a state function and reversing the reaction reverses the sign.

Markers reward the exothermic classification, the heat for 0.250 mol, the heat for 8.00 g via moles, and the reversed sign.

AP 2021 (style)1 marksSection I (multiple choice). For a reaction with ΔH=−150 kJ\Delta H = -150\ \text{kJ} as written, doubling all the coefficients changes ΔH\Delta H to (A) −75 kJ-75\ \text{kJ} (B) −150 kJ-150\ \text{kJ} (C) −300 kJ-300\ \text{kJ} (D) +150 kJ+150\ \text{kJ}. Justify your choice.
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A 1-point conceptual MCQ. The answer is (C).

Enthalpy of reaction is an extensive quantity: it scales with the amount of substance. Doubling the coefficients doubles the amount reacting, so ΔH\Delta H doubles to −300 kJ-300\ \text{kJ}. The trap is treating ΔH\Delta H as independent of the amount; it scales directly with it.

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