Pseudo-First-Order Conditions

Introduction

A reaction can be second order overall yet look first order in one monitored reactant. If B is present in a large excess and changes negligibly while A is consumed, rate = k[A][B] becomes rate ≈ kobs[A], where kobs=k[B]approximately. This pseudo-first-order approximation simplifies measurement but must be tied to the actual concentration conditions.

Core explanation

Consider A+B → products with measured law −d[A]/dt=k[A][B]. If [B]0 is much larger than [A]0 and one B is consumed per A, then the maximum decrease of B is about the initial amount of A. For [B]0=2.0 M and [A]0=0.010 M, complete consumption of A changes B by at most 0.010 M, or 0.5% of its start. Treating B as constant at about 2.0 M may be reasonable within experimental uncertainty.

Define kobs=k[B]0. Because k has units M⁻¹ s⁻¹ and [B]0 has M, kobs has s⁻¹. The approximate law is −d[A]/dt=kobs[A], giving ln([A]t/[A]0)=−kobs t and apparent half-life 0.693/kobs. The true underlying law remains second order in the pair of reactants; only the selected experiment behaves approximately first order in A.

To test the model, repeat the experiment with different excess B concentrations. If kobs doubles when [B]0 doubles while other conditions stay fixed, this supports first-order dependence on B in the underlying law. A plot of kobs versus [B]0 should be approximately linear through the origin for this simple model. If it bends or saturates, another mechanism or concentration effect may be present.

“Large excess” is relative to the amount consumed, not merely a visually large number. If [B]0=0.10 M and [A]0=0.08 M, B could change by up to 80%, so treating it as constant is poor. Even [B]0 ten times [A]0 may lead to a 10% change, which might or might not be acceptable depending on precision. State the expected fractional depletion.

Water often serves as an excess reactant or solvent in hydrolysis. Its effective concentration is nearly constant in a dilute aqueous reaction, so a rate law involving water can appear first order in the dissolved substrate. The observed constant then incorporates water's effective concentration and depends on solvent conditions. Do not use an actual 55.5 M water concentration blindly if activity or mechanism requires a different treatment.

Pseudo-order approximations can apply to other underlying orders too. If rate=k[A][B]² with B constant, the observed law is still first order in A but kobs=k[B]². Varying B across experiments reveals the square dependence. The word pseudo refers to the observed order of a particular experiment, not a claim that the chemistry is fake.

Step-by-step reasoning

1. Write the full measured rate law before approximation. 2. Estimate the maximum fractional change in the excess reactant. 3. Replace its concentration with an approximately constant starting value if justified. 4. Define kobs and verify its units. 5. Vary the excess concentration to recover the hidden order.

Visual explanation

Draw two concentration curves: A falls steeply from 0.010 M toward zero, while B remains nearly flat near 2.0 M. Put rate=k[A][B] above them and an arrow replacing [B] with [B]0, leading to rate≈kobs[A].

Real-world analogy

One teaspoon of dye added to a swimming pool barely changes the pool's water amount. A process depending on both dye and water can appear to depend only on dye because the water amount stays nearly constant during the observation.

Real-world example

Hydrolysis of a dilute substrate in water may be followed by substrate absorbance. Water participates chemically but its concentration changes negligibly, allowing a first-order-looking decay curve for the substrate.

Why?

Why does an overall second-order reaction show a constant apparent half-life in excess B? With B nearly unchanged, its concentration is absorbed into kobs, leaving a first-order differential equation for A over that experimental range.

Common misconception

“A pseudo-first-order plot proves the elementary mechanism is unimolecular.” The hidden excess reactant can still participate in the rate law; changing its starting concentration tests that dependence.

Worked example

For rate=k[A][B] with k=0.40 M⁻¹ min⁻¹, [A]0=0.010 M and [B]0=1.0 M, B changes by at most about 1% if all A reacts. Set kobs≈0.40×1.0=0.40 min⁻¹. The apparent half-life is 0.693/0.40=1.73 min. If B were doubled to 2.0 M under otherwise comparable conditions, kobs would become about 0.80 min⁻¹ and half-life about 0.866 min.

Quick check

1. What is kobs for rate=k[A][B] when B is effectively constant? Answer: kobs=k[B]approximately, commonly evaluated near [B]0.

Exam focus

Show the full law, justify small excess-reactant depletion numerically, derive kobs with units and explain how varying excess concentration tests the hidden order.

Advanced insight

The approximation's error grows as B is depleted. One can fit the full two-reactant integrated law when precision demands it, rather than forcing a straight pseudo-first-order line across the entire reaction.

Summary

An excess reactant can make a higher-order law appear first order in the monitored species. kobs combines the true k with the nearly constant excess concentration. The approximation is valid only while depletion is small.

Practice questions

1. If [B]0 doubles in rate=k[A][B], what happens to kobs under the approximation? Answer: It doubles. 2. Why is [B]0=0.10 M poor excess for [A]0=0.08 M in a one-to-one reaction? Answer: B could fall by 80%, so it is far from constant. 3. Does pseudo-first-order behavior change the underlying overall order? Answer: No. It is an experimental approximation for one concentration regime.