SCIENCE · INVESTIGATE AND EXPLAIN

Distinguish successive bond breaking from a mean bond value

Build a gas-phase enthalpy ledger from reference values, cancel shared intermediates, and test a bond-energy claim.

Learning outcome: Calculate specific dissociation, atomization and mean enthalpies while preserving species, phases, reaction direction and shared uncertainty.

Starting recap

A formation enthalpy refers to making a substance from reference elements. Reaction enthalpy is the sum for products minus the sum for reactants, using balanced amounts and stated phases.

All lessons are open directly. Your saved experiments remain when you leave and return.

Try

Two equivalent bonds in CO₂ do not require equal energies in successive dissociation steps: after the first step, the remaining molecule is CO.

Open the experiment

Predict

Predict the sign and approximate magnitude of the first gas dissociation. Predict whether its value equals half the total atomization or the next step.

Your prediction and reason

Experiment

Enter a signed prediction rounded to ten kJ/mol and capture your explanation. Compare the first, second, total, mean and reversed reactions. Combine the coefficient ledgers before computing a bound, then transfer to two CO molecules with atomic carbon as a product.

Audit an empirical dissociation-energy claim

Frozen formation-enthalpy references · calculated reaction ledger · Selected NIST CODATA 1984 standard gas formation enthalpies at 298.15 K. Derived reaction enthalpies are not zero-kelvin dissociation energies, potential-well depths or activation barriers.

Reaction or mean-value comparison: First dissociation: CO₂(g) → CO(g) + O(g) case Your signed enthalpy prediction, nearest 10 kJ/mol: 0 kJ/mol

Reset keeps your prediction, captures and practice history.

Use products minus reactants, with phases attached

First dissociation: CO₂(g) → CO(g) + O(g). Positive coefficients below denote products; negative coefficients denote reactants. Each source row is a selected standard gas-phase formation enthalpy at 298.15 K, in kJ/mol. C(g) is an isolated carbon atom, not solid graphite; O(g) is atomic oxygen, not O₂.

Species	Source ΔfH° ± uncertainty / kJ mol⁻¹	Signed coefficient ν	ν × ΔfH°	\|ν\| × uncertainty
CO₂(g)	-393.51 ± 0.13	-1	393.51	0.13
CO(g)	-110.53 ± 0.17	1	-110.53	0.17
C(g)	716.68 ± 0.45	0	0	0
O(g)	249.18 ± 0.1	1	249.18	0.1

ΔrH° = Σ νᵢ ΔfH°ᵢ = 532.16 kJ/mol of reaction as written. Carbon and oxygen balance independently. Your prediction is 0; prediction minus reference calculation is -532.16 kJ/mol. The nearest-ten arithmetic check gives 530; rounding is not measurement uncertainty or a grade for your explanation.

The product CO has a different electronic structure from the starting CO₂. This is a specific gas-phase dissociation enthalpy. The endpoint envelope is [531.76, 532.56] kJ/mol, treating the tabulated ± values as input bounds for this exercise. No coverage probability or covariance is supplied here; the envelope is not a claimed 95% confidence interval.

One state function, two unequal steps

First step 532.16 plus second step 1076.39 equals 1608.55 kJ/mol. The mean is 804.275 kJ per mole of bonds. Equivalent bonds in the initial CO₂ do not imply equal successive enthalpies: after the first step, the molecule is CO. These levels show thermodynamic states, not a reaction trajectory, bond length or activation barrier.

Adding the two separate bound widths would give 1.12. Combining the coefficient ledgers first cancels CO and gives 0.78. The repeated CO reference is the same uncertain input in both expressions, not two independent measurements. The same rule cancels O for the transfer reaction.

Reference definitions and limits

Selected NIST Chemistry WebBook CODATA review rows (Cox, Wagman et al., 1984), accessed 28 October 2026; species links are in the ledger. CODATA's key-value definition identifies the 298.15 K formation-enthalpy reference. Standard gas-state quantities are used, not a newly measured gas mixture. They do not supply zero-kelvin D₀, a potential minimum Dₑ, rate constants or barriers. The IUPAC mean bond-enthalpy definition distinguishes a species mean from a particular dissociation. Capture your prediction and explain which quantity the question actually requests.

Derived enthalpy 532.160 kJ/mol; endpoint envelope ±0.400 kJ/mol.

The product CO has a different electronic structure from the starting CO₂. This is a specific gas-phase dissociation enthalpy.

Observation to attach to a capture

Captures retain the selected reaction and your signed prediction. Replay reconstructs the formation-enthalpy ledger; it does not measure a new dissociation, activation barrier or independent reasoning score.

Understand

Ask which energy the claim means

A student divides the energy needed to separate CO₂ into gaseous atoms by two and labels the answer “the energy needed to remove the first oxygen.” Predict whether that is justified. Two bonds may be equivalent in the initial molecule, but after removing one oxygen the remaining molecule is CO, with different electronic structure.

This investigation derives gas-phase reaction enthalpies from selected historical NIST CODATA formation-enthalpy references. It adds empirical thermochemical evidence to the original potential curves and isotope stiffness comparison. A local curvature, a well depth and an enthalpy change answer different questions.

Build a ledger before subtracting

The selected standard formation enthalpies at 298.15 K, in kJ/mol, are CO₂(g), −393.51 ± 0.13, CO(g), −110.53 ± 0.17, C(g), 716.68 ± 0.45, and O(g), 249.18 ± 0.10. All four selected rows cite the same CODATA review; they are not newly measured by the browser or silently combined with different JANAF rows.

For CO₂(g) → CO(g) + O(g), use products minus reactants:

ΔrH° = −110.53 + 249.18 − (−393.51) = +532.16 kJ/mol.

The subscript r means reaction; f in the source table means formation. The degree sign denotes standard-state quantities. Positive means the products have higher enthalpy for the balanced reaction as written. It is not the height of a transition-state barrier. Enter your own prediction rounded to the nearest ten, capture the explanation, then compare its sign and magnitude with the ledger.

C(g) means a gaseous carbon atom. It is not graphite, whose formation enthalpy is the elemental-reference zero. O(g) is not O₂(g). Ignoring either phase or species changes the reaction, not just its notation.

Compare successive breaking with the mean

The next reaction, CO(g) → C(g) + O(g), requires 716.68 + 249.18 − (−110.53) = 1076.39 kJ/mol. This differs from 532.16 because the starting species is now CO. The original CO₂ symmetry is not contradicted.

Add the two reactions. CO appears as a product and a reactant, so it cancels. Total atomization is CO₂(g) → C(g) + 2 O(g), with enthalpy 1608.55 kJ/mol of CO₂. Dividing by two gives 804.275 kJ per mole of bonds as the mean within this species. Neither actual successive step equals that mean.

Use the fixed-scale state diagram to explain this with three levels. The horizontal spacing only separates labels; it is not bond length, time or a proposed reaction coordinate. No barrier is drawn. A mean for CO₂ also is not a universal value for every C–O bond in every molecule.

Cancel shared inputs before evaluating bounds

The same CO reference occurs with opposite signs in the two steps. Treating its uncertainty as two unrelated errors would lose that cancellation. The first-step bound width is 0.40 and the second is 0.72 kJ/mol. Adding those widths gives 1.12, but the combined atomization ledger contains only C, two O and CO₂: 0.45 + 2(0.10) + 0.13 = 0.78 kJ/mol.

This endpoint exercise treats each tabulated ± value as an input interval. It is not a 95% confidence statement and assumes no independent Gaussian samples. Other covariance is not supplied. Cancel identical inputs algebraically first; then apply the declared bound rule to the net coefficients. The doubled oxygen coefficient uses the same reference value twice, so its bound contribution doubles.

Reverse and transfer

Reverse the first reaction. Your prediction must change sign: CO(g) + O(g) → CO₂(g) has −532.16 kJ/mol. That thermodynamic release does not guarantee a rapid process.

For the unfamiliar transfer, subtract the first dissociation from the second. Write the species yourself before selecting the case. The result is 2 CO(g) → CO₂(g) + C(g). Oxygen atoms cancel, and the derived enthalpy is 544.23 kJ/mol of reaction as written, not a value for producing solid carbon. Explain why the gas-phase product label matters and why CO’s coefficient is now −2.

Finally, apply the public changed-species practice case: if successive steps require 400 and 700 kJ/mol, derive the total and mean without assuming equal successive breaking. Your captures should contain a numerical prediction, a balanced species ledger and an explanation of one misleading energy claim.

Keep the thermodynamic boundary visible

These standard gas enthalpies refer to 298.15 K. They are not spectroscopic D₀ values at zero kelvin or Dₑ values measured from a potential minimum. Thermal and zero-point contributions matter when converting between those definitions. The spectroscopic Dₑ column discussed in the earlier isotope lesson is also a centrifugal-distortion constant, so even the symbol must be read in context.

The HS reaction-energy lesson introduces the break-minus-form bookkeeping with assigned teaching numbers. Here the new action is selecting empirical references, preserving species and conditions, and distinguishing a specific reaction from a mean. Completing the arithmetic or submitting prose does not establish independent reasoning quality, full Unit 2 completion or practical laboratory evidence.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Use an energy-level diagram and balanced species ledger to explain the unequal successive steps. Distinguish this 298.15 K enthalpy from a potential well depth or activation barrier, and explain why the repeated intermediate uncertainty cancels.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Construct a material representation without overclaiming evidence Next: Test when adding substance changes vapor pressure

Save a science backup or import one

Science work is stored locally in this browser, separately from mathematics and AI. It does not sync automatically. Use this science-only backup; this checkout's older all-subject export does not include science.

A legacy file is available only when the complete snapshot fits the 36 MB limit. For larger snapshots, prepare a complete file set below; never discard notebooks to make a smaller file.

Choose a science backup (maximum 36 MB)

Matching imported notebook drafts replace their current drafts. Omitted lessons and known first attempts remain. Export your current work first if you want to retain both drafts.

Complete file sets and course backups

A file set can hold more work than one science file. Keep its manifest and every part together. Each file is limited to 36 MB. This format needs this version of LearnAI or a compatible newer version; older versions can use only the legacy single-file backup above.

Export scope

The set includes every saved science course and global continuation at this moment. Later edits need a new export. A conflicting tab exports its own visible work, which may differ from saved work in another tab.

Choose the manifest and all parts together

0 file(s) selected. File order and filenames do not determine identity. Missing, changed or conflicting parts block the entire restore; no good part is applied by itself.

Science storage: loading. No account or remote computation is required.