SCIENCE · INVESTIGATE AND EXPLAIN

Transform the equation and its equilibrium constant together

Construct a target from scaled reversible equations, inspect cancellation and verify multiplicative constants against independent activity expressions.

Learning outcome: Justify reciprocal, power and product transformations while refusing to assign a constant to a mismatched species equation.

Starting recap

A quotient is the product of dimensionless activities raised to signed stoichiometric coefficients. Negative exponents put reactant activities in the denominator.

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Try

A ⇌ B has K = 4 and B ⇌ C has K = 9 at the same temperature. Is the constant for A ⇌ C 13, 36 or a value that depends on how quickly B reacts?

Open the experiment

Predict

Predict what reversing, doubling or halving the entire equation does to K and to the physical equilibrium mixture.

Your prediction and reason

Experiment

Construct A ⇌ C with both multipliers +1, then reverse, double and halve the target. Inspect B cancellation and the two independently calculated quotients. Deliberately use +1 and −1 to make a unit-conserving but wrong species equation. Capture the failed construction and a repair.

Construct the equation before assigning its equilibrium constant

Worked teaching model · synthetic values · Original hypothetical isomers A, B and C have identical elemental composition. Assigned dimensionless K values 4 and 9 share one fixed temperature and compatible standard states. All quotient examples use explicitly positive ideal activities; no experimental constants, elementary mechanism, rates or new equilibrium composition are inferred. Fractional equation coefficients rescale a molar reaction, not individual particles.

Target equilibrium equation: A ⇌ C target

Reset keeps your prediction, captures and practice history.

Build the target: A ⇌ C

A, B and C represent hypothetical isomers with the same elemental composition. All constants refer to the same temperature and standard-state convention. Change each source multiplier; every species coefficient and every quotient exponent must change together.

Reaction multipliersSource 1: A ⇌ B, K = 4Source 2: B ⇌ C, K = 9

Species	Target exponent	Residual
A	-1	1
B	0	0
C	1	-1

Conserved isomer-unit residual: 0. Every source combination conserves these units, but the full species equation does not match. Cancellation of B is essential for these targets.

Source	Multiplier	Transformed factor
K1 = 4	0	4⁰ = 1.00000
K2 = 9	0	9⁰ = 1.00000

Product of factors = 1.00000. This belongs only to the proposed net equation; the target value is withheld. The zero combination is the identity, with quotient 1. It does not mean the target reaction has K = 1.

Verify quotient algebra at two stated compositions

Activities are dimensionless ratios to a standard state, not raw amounts. Q for any equation is the product of each activity raised to that species’ signed coefficient. These two hypothetical compositions are independent examples, not successive times or conserved-batch trajectories.

Off-equilibrium illustrative state

Activities A/B/C: 0.1 / 0.2 / 0.3.

Q1 = aB/aA	Q2 = aC/aB	Transformed product Q1ᵐ¹Q2ᵐ²	Direct product from net exponents	Q of target
2.00000	1.50000	1.00000	1.00000	3.00000

Compatible equilibrium reference

Activities A/B/C: 0.01 / 0.04 / 0.36.

Q1 = aB/aA	Q2 = aC/aB	Transformed product Q1ᵐ¹Q2ᵐ²	Direct product from net exponents	Q of target
4.00000	9.00000	1.00000	1.00000	36.0000

The two routes for the proposed quotient agree even away from equilibrium. Agreement does not prove that the proposal matches the target. In the equilibrium reference, Q1 = 4 and Q2 = 9, so every transformed combination satisfies its own Q = K.

Why powers, and what the construction does not show

Reversing swaps numerator and denominator: K becomes 1/K. Doubling every coefficient doubles every exponent: K becomes K², not 2K. Combining equations adds exponents, which multiplies their factors. A half-sized equation uses the positive square root because equilibrium constants are positive. Neither doubling nor halving the written equation changes the physical equilibrium mixture.

These algebraic sources need not be elementary steps or a mechanism. K supplies no equilibration time. No rate is inferred from Q, and actual species concentrations still require conserved amounts and a physical model. All displayed activities are positive, so zero-activity and missing-phase quotient limits are outside this packet.

The net species equation does not match the target: do not assign this K to the target.

Proposed K = product of transformed source factors = 1.000000. Both sources are omitted: the identity equation has quotient 1, not the requested reaction.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Try: cancel the intermediate species

Two hypothetical isomers are related by A ⇌ B, with K1 = 4. A second equation B ⇌ C has K2 = 9. The constants refer to the same fixed temperature and standard states. Construct A ⇌ C by selecting +1 for each source. The B coefficients cancel. The product K1 × K2 = 36 belongs to the resulting equation.

The letters stand for hypothetical species of identical elemental composition. The assigned constants are original teaching premises, not measurements or named chemical systems.

Predict: which operation follows an equation change?

If the equation is reversed, numerator and denominator of its quotient exchange places. Predict 1/36 for C ⇌ A. If the equation is doubled, every exponent doubles, so predict 36² = 1296 for 2A ⇌ 2C. Halving the equation gives the positive square root, 6, for ½A ⇌ ½C. Rescaling a written molar equation does not physically create a different equilibrium mixture or half an individual particle.

Experiment: an apparently convincing wrong answer

Use multipliers +1 and −1 while the target remains A ⇌ C. The net equation is A + C ⇌ 2B. It still conserves isomer units, and its calculated constant 4/9 is mathematically valid for that equation. It is not the target constant. Check every species residual before accepting the result.

Try both multipliers zero. The identity has quotient 1, but it supplies no reaction and cannot satisfy any displayed target. Capture a wrong construction and then a repaired half-sized or reversed target.

Understand: the same algebra works for Q

Let aA, aB and aC denote dimensionless activities relative to compatible standard states. Q1 = aB/aA and Q2 = aC/aB. Their product simplifies to aC/aA because aB cancels. More generally, scaling an equation by m raises its quotient to the power m. Adding scaled equations adds the signed species exponents, so the combined quotient is the product of their powered quotients.

At the illustrative activities 0.1, 0.2 and 0.3, Q1 = 2 and Q2 = 1.5; the combined Q is 3, not 36. The algebra still holds away from equilibrium. At the separate compatible equilibrium reference 0.01, 0.04 and 0.36, Q1 = 4 and Q2 = 9, so the combined Q equals 36. These states have different inventories and are not a simulated time sequence.

The table calculates the proposed quotient two ways: multiply transformed source quotients, and independently use the net species exponents. Both must agree. Their agreement verifies the algebra; matching the requested target is a separate check.

Apply: unfamiliar constants

Complete the practice with constants 0.20 and 5.0, a reversed doubled target, and an intermediate species whose equilibrium activity is not equal to the endpoints. Explain why K = 1 for the endpoint equation does not establish that every species has equal concentration. Keep the same-temperature and compatible-standard-state conditions explicit.

Continue: representation is not a mechanism

The previous composition construction combines equilibrium with conserved material. This lesson does not calculate a new equilibrium composition, equilibration time or elementary reaction mechanism. All quotient evaluations here use positive activities; missing-phase and zero-activity limits require separate treatment.

AP Chemistry CED topic 7.6, printed page 134 specifies these equation transformations and their application to Q. OpenStax Chemistry 2e, equilibrium constants provides the quotient framework and transformed-equation rules. This original activity addresses a bounded part of Unit 7, not full course proficiency.

Apply

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

Explain why equation coefficients add but constants multiply, why doubling is a power rather than 2K, and why a quotient identity can hold for an equation that does not match the target.

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 composition that satisfies both equilibrium and conservation Next: Respect finite solid inventory when applying a solubility product

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