Reaction rates, stoichiometry and kinetic models

Start with amounts and a rate basis

Use the imaginary reaction A → 2B. Each mole of A consumed forms two moles of B if that is the only reaction. A could stand for a molecule that splits into two fragments; its mass equals the combined product mass. The change in total molecule count does not create mass. Real components require their actual balanced equation and appropriate atomic or elemental checks.

Define r as the positive rate of progress per unit liquid volume. Then the species formation rates are r_A=−r and r_B=2r. Their units might be mol L⁻¹ min⁻¹. A negative formation rate denotes consumption. A rate per catalyst mass or surface area would have a different basis and units; it cannot be inserted unchanged into a volume balance.

The rate law is additional information

For an original teaching model, suppose r=k C_A, with supplied k=0.20 min⁻¹ at a fixed temperature. This is first order in A: doubling its concentration doubles this model rate. The coefficient 2 on B in the equation does not establish a second-order law. Reaction orders generally come from a justified mechanism or experimental evidence, rather than from an overall balanced equation alone.

Assume a closed, well-mixed, constant-volume liquid with no B initially, no side reaction and no relevant reverse reaction over the exercise. The material balance gives dC_A/dt=−k C_A. Its solution is C_A=C_A0 exp(−kt). With C_A0=1.00 mol/L, the corresponding product concentration is C_B=2(C_A0−C_A). The two curves therefore share a stoichiometric account; they are not two independently fitted predictions.

A fit must survive a relevant comparison

At 5 minutes the model predicts C_A≈0.3679 mol/L. An actual investigation would retain measured values, timebase, temperature, sampling method and uncertainty to compare with such a prediction. A residual is an observation minus the predicted value under specified conditions. A systematic residual pattern can expose an omitted effect even when an overall fit looks close.

If volume changes, concentration can change through expansion or dilution as well as reaction. If temperature changes, the supplied constant k may no longer apply. Mixing and transport can also make a measured overall rate differ from intrinsic chemistry. Record the fitted range and assumptions before using a convenient law to predict a different process.

Link rates and accumulated amounts

Initially r=0.20×1.00=0.20 mol L⁻¹ min⁻¹, so A’s formation rate is−0.20 and B’s is+0.40 in those units. At 5 minutes exp(−1)=0.367879, giving C_B=2(1−0.367879)=1.26424 mol/L and an instantaneous A consumption rate of 0.073576 mol L⁻¹ min⁻¹.

The average A consumption rate over the first five minutes is (1−0.367879)/5≈0.12642 mol L⁻¹ min⁻¹. It differs from the final instantaneous rate because concentration falls throughout the interval. The sum C_A+C_B rises, while the fragment account C_A+C_B/2 stays at 1.00 mol/L; neither sum is a mass calculation without molar masses.

Try the reasoning

1. Does A → 2B establish a second-order rate law?

   Show the answer and reasoning

   No. Stoichiometry relates species changes; kinetic order requires separate evidence or an applicable elementary mechanism.

2. Why is the five-minute average consumption rate larger than the final rate?

   Show the answer and reasoning

   The supplied first-order rate was higher earlier, when more A remained.

3. Would dilution alone prove that A reacted?

   Show the answer and reasoning

   No. Concentration can change through volume or flow changes, so an amount balance is needed.

What to carry forward

Define the transformation and rate basis separately, then connect the rate law to a material balance and a checkable history.