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AP Chemistry Notes

6.6.4 Calculating Reaction Heat from Moles and Molar Enthalpy

AP Syllabus focus: ‘Calculate heat absorbed or released in a reaction using the amount of reacting substance in moles and the molar enthalpy of reaction.’

These notes explain how to convert between the amount of reacting substance (moles) and the heat absorbed or released by a reaction using a molar enthalpy value and correct stoichiometry.

Core idea: scale heat by chemical amount

In AP Chemistry, reaction heat is commonly reported “per mole” of reaction as written. To find the actual heat change for a specific quantity, you scale the molar value by the number of moles that actually react.

Molar enthalpy as a proportionality constant

Molar enthalpy of reaction: The enthalpy change associated with the reaction as written, per mole of the specified substance or per “mole of reaction,” typically in kJ mol1\text{kJ mol}^{-1}.

A key skill is identifying what the given ΔH\Delta H is “per mole of”:

  • Sometimes it is explicitly per mole of a named reactant/product (e.g., “kJ per mol of O2\text{kJ per mol of } \text{O}_2 consumed”).

  • Sometimes it is tied to the balanced thermochemical equation (meaning the stated ΔH\Delta H corresponds to the stoichiometric amounts shown).

Relationship between heat and moles

qrxn=n,ΔHrxn q_{\text{rxn}} = n ,\Delta H_{\text{rxn}}

qrxn q_{\text{rxn}} = heat absorbed or released by the reaction for the amount that reacts, in kJ\text{kJ} (or J\text{J})

n n = moles of the specified substance (or moles of “reaction as written”), in mol\text{mol}

ΔHrxn \Delta H_{\text{rxn}} = molar enthalpy of reaction, in kJ mol1\text{kJ mol}^{-1} (or J mol1\text{J mol}^{-1})

Always keep units consistent: if ΔHrxn\Delta H_{\text{rxn}} is in kJ mol1\text{kJ mol}^{-1}, then qrxnq_{\text{rxn}} comes out in kJ.

How to choose the correct “n”

Correctly determining nn is the most important step. The reaction may involve multiple reactants, and the enthalpy value is only directly proportional to the reaction progress defined by stoichiometry.

Use the balanced equation as the reference

Treat the balanced chemical equation like a “recipe.” The coefficients tell you mole-to-mole relationships:

  • If the enthalpy is given for the reaction as written, then “1 mole of reaction” corresponds to the full set of coefficients.

  • If you know moles of some species, convert that amount to the equivalent “moles of reaction” using the coefficient ratio.

Practical approach (no arithmetic shown here):

  • Start from the amount you’re given (mass, particles, volume and molarity, etc.) and convert to moles.

  • Use the mole ratio from the balanced equation to convert to the moles that match the enthalpy basis.

Limiting reactant determines heat actually produced/absorbed

If multiple reactants are provided, the heat is determined by the reactant that runs out first.

Use this logic:

  • Convert each reactant amount to a possible extent of reaction (how many “moles of reaction” each could support).

  • The smallest extent corresponds to the limiting reactant.

  • Use that extent as nn in qrxn=nΔHrxnq_{\text{rxn}} = n\Delta H_{\text{rxn}}.

Sign conventions and interpreting the result

The sign on ΔHrxn\Delta H_{\text{rxn}} carries meaning and should be preserved when computing qrxnq_{\text{rxn}}:

Pasted image

Energy-level style thermochemical diagrams for an exothermic and an endothermic reaction, with the corresponding ΔH\Delta H values labeled. The figure makes the sign convention concrete: exothermic processes have ΔH<0\Delta H<0 because enthalpy decreases, while endothermic processes have ΔH>0\Delta H>0 because enthalpy increases. Source

  • Negative ΔHrxn\Delta H_{\text{rxn}}: heat is released by the reaction; qrxnq_{\text{rxn}} is negative for the amount reacted.

  • Positive ΔHrxn\Delta H_{\text{rxn}}: heat is absorbed; qrxnq_{\text{rxn}} is positive.

Common interpretation checkpoints:

  • If the problem asks “how much heat is released,” report the magnitude but keep your internal sign logic consistent.

  • If the reaction amount is doubled (in moles), the heat change doubles; if halved, the heat change halves.

Common AP Chemistry pitfalls to avoid

  • Forgetting stoichiometric scaling: Using moles of the wrong species directly with ΔHrxn\Delta H_{\text{rxn}} when the enthalpy is based on a different reactant/product or on the whole equation.

  • Ignoring limiting reactant: Calculating heat from a reactant that is actually in excess.

  • Unit mismatch: Mixing J\text{J} and kJ\text{kJ} or reporting kJ mol1\text{kJ mol}^{-1} when the question asks for total heat in kJ\text{kJ}.

  • Sign errors: Changing the sign because the question uses words like “released” or “absorbed,” instead of letting the given ΔHrxn\Delta H_{\text{rxn}} determine the sign.

What you should be able to do on sight

  • Identify whether ΔHrxn\Delta H_{\text{rxn}} is tied to a thermochemical equation or to 1 mol of a specific species.

  • Convert given quantities to moles and relate them to the enthalpy basis using the balanced equation.

  • Use qrxn=nΔHrxnq_{\text{rxn}} = n\Delta H_{\text{rxn}} with correct sign and units to state heat absorbed or released for the amount that reacts.

FAQ

Look for wording like “for the reaction as written” or a thermochemical equation with a single $\Delta H$. If it says “per mol of $\text{O}_2$” (etc.), the basis is that species.

Yes. Reversing the chemical change reverses the direction of energy transfer, so $\Delta H_{\text{rxn}}$ changes sign, and $q_{\text{rxn}}$ follows.

That usually means it applies to the exact stoichiometric amounts in the written equation. Treat it as “per 1 mole of reaction as written,” then scale by the extent of reaction.

“Released” typically wants a positive magnitude in words, but the thermochemical sign remains negative. For example: $q = -50\ \text{kJ}$ means “$50\ \text{kJ}$ released.”

Yes, as long as you use the balanced equation to convert product moles to the correct enthalpy basis (moles of reaction or the specified species).

Practice Questions

(2 marks) For a reaction with ΔHrxn=125 kJ mol1\Delta H_{\text{rxn}} = -125\ \text{kJ mol}^{-1} (per mole of reaction as written), determine the heat change when 0.400 mol0.400\ \text{mol} of reaction occurs.

  • Uses q=nΔHq = n\Delta H: 1 mark

  • q=0.400×(125)=50.0 kJq = 0.400 \times (-125) = -50.0\ \text{kJ} with correct sign and units: 1 mark

(5 marks) For N2(g)+3H2(g)2NH<em>3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}<em>3(g), ΔH</em>rxn=92.4 kJ\Delta H</em>{\text{rxn}} = -92.4\ \text{kJ} (for the equation as written). If 1.50 mol N21.50\ \text{mol}\ \text{N}_2 reacts with 3.00 mol H23.00\ \text{mol}\ \text{H}_2, calculate the heat change.

  • Determines limiting reagent by comparing required ratio (needs 3 mol H23\ \text{mol}\ \text{H}_2 per 1 mol N21\ \text{mol}\ \text{N}_2): 1 mark

  • Identifies H2\text{H}_2 as limiting (only supports 1.001.00 “mole of reaction”): 1 mark

  • Converts to extent of reaction n=1.00n = 1.00 (or equivalent reasoning): 1 mark

  • Uses q=nΔHq = n\Delta H: 1 mark

  • q=1.00×(92.4)=92.4 kJq = 1.00 \times (-92.4) = -92.4\ \text{kJ} with sign and units: 1 mark

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