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What is a standard enthalpy of formation, and how is it used to calculate the enthalpy of a reaction?

Topic 6.8 Enthalpy of Formation: use standard enthalpies of formation to calculate the enthalpy of a reaction as the sum for products minus the sum for reactants.

A focused answer to AP Chemistry Topic 6.8, covering the standard enthalpy of formation, the zero value for elements in their standard states, and calculating the enthalpy of a reaction as products minus reactants, with full worked examples.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. Standard enthalpy of formation
  3. The products-minus-reactants formula
  4. Applying the method
  5. Try this

What this topic is asking

The College Board (Topic 6.8) wants you to use standard enthalpies of formation to calculate the enthalpy of a reaction as the sum for the products minus the sum for the reactants, each weighted by its coefficient. This is the most direct route to an accurate ΔH\Delta H when formation data are tabulated.

Standard enthalpy of formation

So ΔHf∘(O2,g)=0\Delta H_f^\circ(\text{O}_2, g) = 0, ΔHf∘(C, graphite)=0\Delta H_f^\circ(\text{C, graphite}) = 0, but ΔHf∘(CO2,g)=−394 kJ mol−1\Delta H_f^\circ(\text{CO}_2, g) = -394\ \text{kJ mol}^{-1} because forming carbon dioxide from graphite and oxygen releases energy. The zero for elements is the reference point that makes the whole scheme self-consistent.

The products-minus-reactants formula

This works because enthalpy is a state function. Imagine breaking all the reactants down to their elements (the reverse of their formation, so minus their formation enthalpies) and then building the products from those elements (plus their formation enthalpies). The net is products minus reactants, regardless of the real mechanism. Elements in their standard states contribute zero, simplifying the arithmetic.

Applying the method

The procedure is mechanical once the data are in hand: list each species with its coefficient and ΔHf∘\Delta H_f^\circ, sum the products, sum the reactants, and subtract. Watch the signs (many formation enthalpies are negative) and remember to apply the coefficients. The sign of the final answer classifies the reaction as exothermic (ΔH∘<0\Delta H^\circ < 0) or endothermic (ΔH∘>0\Delta H^\circ > 0). This is usually the cleanest of the three enthalpy methods, alongside calorimetry and bond enthalpies.

Try this

Q1. For a reaction, the products' formation enthalpies sum to −500 kJ-500\ \text{kJ} and the reactants' sum to −300 kJ-300\ \text{kJ}. Calculate ΔH∘\Delta H^\circ. [2 points]

  • Cue. ΔH∘=−500−(−300)=−200 kJ\Delta H^\circ = -500 - (-300) = -200\ \text{kJ} (exothermic).

Q2. State the standard enthalpy of formation of solid copper metal and explain. [1 point]

  • Cue. Zero, because solid copper is the element in its standard state.

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 C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)\text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(l), the standard enthalpies of formation (in kJ mol−1\text{kJ mol}^{-1}) are: C2H4(g) +52\text{C}_2\text{H}_4(g)\ +52, CO2(g) −394\text{CO}_2(g)\ -394, H2O(l) −286\text{H}_2\text{O}(l)\ -286, O2(g) 0\text{O}_2(g)\ 0. (a) State the value of ΔHf∘\Delta H_f^\circ for O2(g)\text{O}_2(g) and justify. (b) Calculate the sum for the products. (c) Calculate the standard enthalpy of the reaction. (d) State whether the reaction is endothermic or exothermic.
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A 4-point quantitative FRQ on enthalpies of formation.

(a) Oxygen (1 point): ΔHf∘(O2)=0\Delta H_f^\circ(\text{O}_2) = 0, because O2(g)\text{O}_2(g) is the standard state of the element oxygen, and an element in its standard state has a formation enthalpy of zero by definition.
(b) Products sum (1 point): 2(−394)+2(−286)=−788−572=−1360 kJ2(-394) + 2(-286) = -788 - 572 = -1360\ \text{kJ}.
(c) Reaction enthalpy (1 point): reactants sum =(+52)+3(0)=+52 kJ= (+52) + 3(0) = +52\ \text{kJ}; ΔH∘=(products)−(reactants)=−1360−52=−1412 kJ mol−1\Delta H^\circ = (\text{products}) - (\text{reactants}) = -1360 - 52 = -1412\ \text{kJ mol}^{-1}.
(d) Classification (1 point): ΔH∘<0\Delta H^\circ < 0, so the reaction is exothermic.

Markers reward the zero for oxygen with reasoning, the products sum, the reaction enthalpy as products minus reactants, and the exothermic classification.

AP 2021 (style)1 marksSection I (multiple choice). The standard enthalpy of formation of N2(g)\text{N}_2(g) is (A) a large negative number (B) a large positive number (C) zero (D) equal to its bond enthalpy. Justify your choice.
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A 1-point conceptual MCQ. The answer is (C).

N2(g)\text{N}_2(g) is the standard state of the element nitrogen, and the standard enthalpy of formation of an element in its standard state is defined to be zero. The trap is (D): the bond enthalpy is the energy to break the N-N bond, not the formation enthalpy.

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