Compare thermal and mechanical changes in equilibrium

Heating changes more than the pressure

A gas mixture starts at equilibrium at 300K. Heat it to 350K while keeping volume fixed. Even before composition changes, every partial pressure rises by 350/300. For N₂O₄⇌2NO₂, the squared product-pressure term means Q also rises by 350/300. But the equilibrium constant is no longer the old one.

Predict the sign of the chemical response and enter a signed extent in the lab. Positive extent consumes N₂O₄ and creates twice as much NO₂; negative extent associates NO₂ into N₂O₄. Inspect the inventory before deciding whether your proposal is physically possible.

Reference constants and an original batch

The NIST-JANAF tables give base 10 logarithms of formation constants for each species at stated temperatures. Combining twice the NO₂ formation equation with the reversed N₂O₄ formation equation cancels the elemental reference substances. Therefore log₁₀K(dissociation)=2log₁₀Kf(NO₂)−log₁₀Kf(N₂O₄).

At 300K this is 2(−8.944)−(−17.122)=−0.766, so K≈0.17140. At 350K the same calculation gives log₁₀K=0.652 and K≈4.48745. These are calculations from rounded evaluated reference inputs, not newly measured equilibrium constants. No interpolation is offered.

The teaching batch starts with 0.100mol N₂O₄ and 0.100mol NO₂. Its volume is chosen so both partial pressures equal K(300K)×1bar, making Q=K at 300K. Total conserved inventory is 0.300mol of NO₂ units. This batch and all its predicted changes are original ideal-gas calculations; they are not taken from an experimental run.

Compare the actual held constraint

At fixed volume, the composition-frozen Q increases with temperature for this reaction. At fixed total pressure, a piston expands during heating and the unchanged mole fractions keep partial pressures—and immediate Q—unchanged. In both cases the new K is larger than immediate Q, so dissociation is favored within this model. The final volumes and compositions differ because the mechanical constraints differ.

The final state must satisfy conservation and the new-temperature equation Q=K simultaneously. The solver searches only physically admissible extents between complete consumption of the available NO₂ in reverse and complete consumption of the available N₂O₄ forward. Your proposal may lie outside that interval; the model keeps and explains it instead of silently clipping it.

The composition-frozen intermediate state is a reasoning device that separates thermal/mechanical and chemical changes. It is not a claim that an actual apparatus resolves these events in that order or on an observed time scale. Neither reference K nor a computed extent supplies a reaction rate.

Test a reverse path and transfer

Compare heating 300→400K with cooling the corresponding400K equilibrium back to 300K under the same constraint. The original state is recovered by this reversible equilibrium calculation. That result does not establish a real finite-rate experiment, heat requirement or absence of hysteresis outside the model.

For another reaction X⇌Y, equal gas coefficients make the common pressure factor cancel in Q. Use that structural difference in the practice case instead of memorizing that heating always multiplies a quotient by temperature.

Source and limits

NIST-JANAF Fourth Edition nitrogen tables, printed 1605 and 1630, selected 300/350/400K rows, p°=0.1MPa=1bar; database provenance. These are evaluated historical ideal-gas functions. Printed digits are not independent uncertainty bounds. The model excludes other species, condensation and nonideal interactions. Endothermic dissociation is consistent with the increasing reference K, but no unlisted temperature or constant-enthalpy extrapolation is claimed.

Return to gas constraints for disturbances at fixed temperature, or construct the quotient to explain why different stoichiometries respond differently.

Predict an observable, not only a shift arrow

For a separate fictional solution equilibrium A + B ⇌ C, stipulate that only C absorbs the selected visible light, with absorbance proportional to its concentration at fixed path length. At fixed temperature, adding B without appreciable volume change lowers Q immediately and favors net C formation, so the eventual absorbance increases. This prediction needs the stated optical selectivity; a darker solution alone does not prove a particular reaction.

Changed-case action: Instead of adding B, double the solution volume with an ideal inert solvent while holding temperature fixed. All three concentrations immediately halve. Show that Q = [C]c°/([A][B]) doubles, so net reverse reaction follows. Separate the immediate absorbance change from the chemical response. Does a forward shift after some different intervention necessarily restore the original color?

Compare the optical and chemical changes

Immediate absorbance halves by dilution, before any composition response. Since Q becomes 2K, the reverse response then consumes C, reducing absorbance further under the stated model. A shift direction alone does not specify the final absolute concentration or color: the volume change and an actual optical calibration matter.

For a weak-acid equilibrium HA ⇌ H⁺ + A⁻, a supplied common-A⁻ addition at nearly fixed volume raises Q and favors association; under the ideal single-acid, negligible-water-ionization premises, [H⁺] falls and pH rises. For a separately stipulated exothermic forward reaction, a net forward response releases heat; an insulated sample may warm and thereby change K further, whereas a thermostatted sample exports heat and holds temperature. A fixed-temperature equilibrium solver cannot predict the temperature change without a heat-capacity and energy balance. These are conceptual supplied cases, not reagent additions or heating instructions.

OpenStax shifting equilibria, checked 31 October 2026, supports the conditional concentration/temperature reasoning. The optical comparison is an original fictional task, not an observation of the nitrogen-oxide reference system above.