Chemistry · Lesson
Equilibrium calculations and conservation constraints
Combine an equilibrium expression with the conserved material inventory. Reject impossible roots and check the original equation before rounding a final answer.
Back to the Chemistry mapBy the end, you can
- Build initial/change/equilibrium bookkeeping from stoichiometry.
- Keep normalized concentrations and algebraic numerical values distinct.
- Check nonnegative concentrations, constituent balance and quotient residuals.
The constant needs an inventory to determine composition
An equilibrium constant supplies one relation between activities. Initial amounts, volume and conservation supply additional constraints. For a hypothetical A ⇌ B isomerization at fixed volume, initially [A] = 0.100 mol L⁻¹ and [B] = 0, let x be the concentration converted. Then equilibrium concentrations are 0.100 − x and x.
With ideal normalized activities and c° = 1 mol L⁻¹, K° = 4.00 gives x/(0.100 − x) = 4.00 when numerical concentrations are expressed in mol L⁻¹. The solution x = 0.0800 mol L⁻¹ leaves 0.0200 mol L⁻¹ A and 0.0800 mol L⁻¹ B. Their sum remains 0.100 mol L⁻¹.
Follow every stoichiometric change
Our separate original association model is A + B ⇌ AB. All three species are neutral; A and B carry distinct conserved building blocks, and AB contains both. At fixed volume, the initial concentrations are [A] = [B] = 0.100 mol L⁻¹ and [AB] = 0. An amount x of AB formed consumes the same concentration x of each reactant.
Thus [A]eq = [B]eq = 0.100 mol L⁻¹ − x and [AB]eq = x. The A-building-block balance is [A] + [AB] = 0.100 mol L⁻¹, and the B balance is [B] + [AB] = 0.100 mol L⁻¹. Do not use the sum of all molecule concentrations as a conserved count when association changes the number of molecules.
| Species | Initial / mol L⁻¹ | Change / mol L⁻¹ | Equilibrium / mol L⁻¹ |
|---|---|---|---|
| A | 0.100 | −x | 0.0618 |
| B | 0.100 | −x | 0.0618 |
| AB | 0 | +x | 0.0382 |
| A-building-block total | 0.100 | 0 | 0.100 |
| B-building-block total | 0.100 | 0 | 0.100 |
Original neutral A + B ⇌ AB ideal model at fixed volume, initial A=B=0.100 mol L⁻¹, AB=0, c°=1 mol L⁻¹, K°=10.0. Here x is the admissible unrounded concentration change. Displayed equilibrium values are rounded; A+AB and B+AB are separately conserved.
Keep the reference concentration in the equation
For the stipulated ideal model, c° = 1 mol L⁻¹ and K° = 10.0. The dimensionless equation is (x/c°)/[((0.100 mol L⁻¹ − x)/c°)²] = 10.0. Define y = x/c°; then y/(0.100 − y)² = 10.0 and 10y² − 3y + 0.100 = 0.
The roots are approximately y = 0.0381966 and y = 0.2618034. The second exceeds the initial dimensionless reactant inventory 0.100 and would make both reactant concentrations negative. Reject it. The admissible result is x ≈ 0.0382 mol L⁻¹, with [A]eq = [B]eq ≈ 0.0618 mol L⁻¹.
Check the solution in the original problem
Before rounding, substitute the admissible root into the normalized quotient: it returns approximately 10.0, and each building-block total remains 0.100 mol L⁻¹. Nonnegative species values, conservation and the quotient all agree. A merely algebraic root is not sufficient evidence of a physical state.
If a small-change approximation is used, check its effect on the omitted terms and the final quotient. A classroom five-percent criterion is a rough heuristic, not a guarantee of every claimed digit. Coupled ionic calculations may also require charge balance and additional equilibrium relations.
Worked examples
Select the physically admissible quadratic root
For the stated ideal association model with K° = 10.0 and c° = 1 mol L⁻¹, calculate the equilibrium composition.
- Use dimensionless y = x/c° and y/(0.100 − y)² = 10.0.
- Expand to 10y² − 3y + 0.100 = 0.
- The roots are (3 ± √5)/20, approximately 0.0381966 and 0.2618034.
- Reject the larger root: y > 0.100 would leave negative A and B.
- Use the smaller unrounded root to check Q ≈ 10.0 and both conserved building-block totals.
[AB]eq ≈ 0.0382 mol L⁻¹; [A]eq = [B]eq ≈ 0.0618 mol L⁻¹. Rounding follows the conservation and quotient checks.
Try it yourself
Are initial stock concentrations generally the final equilibrium concentrations?
Show the answer and reasoning
No.
The reaction changes composition; its equilibrium relation must be solved with the inventory constraints.
Why reject the association root y ≈ 0.2618034?
Show the answer and reasoning
It exceeds the initial reactant inventory and makes A and B negative.
An algebraic root must also be physically admissible.
Which association inventory is conserved: [A]+[AB] or necessarily the sum of all molecule concentrations?
Show the answer and reasoning
[A]+[AB], for the stipulated A building block.
Association changes molecule count while conserving each constituent.
What should be checked before rounding an equilibrium result?
Show the answer and reasoning
The original quotient, conserved inventories and physical nonnegativity.
Together they test both the equilibrium relation and the material constraints.
What to carry forward
- Combine equilibrium relations with conserved material inventories.
- Define normalized numerical variables before simplifying units away.
- Reject roots that violate inventory or nonnegativity.
- Back-substitute and assess approximations before claiming answer precision.
References
Teaching resources and reference material supporting this lesson’s explanations and examples.