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Pseudo First Order Reactions

General Chemistry • Chemical Kinetics

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Pseudo–First-Order Reaction: Effective First-Order Rate Law

Many reactions are second order overall, such as \(\mathrm{A + B \rightarrow products}\) with rate law \(\text{rate} = k[A][B]\). If one reactant (usually the solvent) is present in large excess, its concentration remains essentially constant and we can write \(\,k' = k[B]_0\). The reaction then behaves as a pseudo-first-order process in \(A\): \[ \text{rate} = k'[A], \qquad \ln\!\big([A]_t\big) = -k' t + \ln\!\big([A]_0\big). \] This tool uses that effective first-order law to compute concentrations, pseudo-rate constants, and half-life.

1. Reaction and excess reactant

Consider \(\mathrm{A + B \rightarrow products}\) with true rate law \(\text{rate} = k[A][B]\). Reactant \(B\) is present in large excess so that \([B] \approx [B]_0\) is effectively constant. The effective first-order rate constant is \[ k' = k[B]_0. \] You may enter either \(k'\) directly, or enter \(k\) and \([B]_0\) and let the calculator compute \(k'\).

For aqueous reactions with water as the solvent, \([B]_0 \approx 55.5~\text{M}\).

Units: \(\text{L}\,\text{mol}^{-1}\,\text{(time)}^{-1}\) (for example, \(\text{M}^{-1}\text{s}^{-1}\)).

Units: \(\text{(time)}^{-1}\), such as \(\text{s}^{-1}\) or \(\text{min}^{-1}\). If left empty, it will be computed as \(k' = k[B]_0\).

2. Initial concentration of A and time

If you supply an experimental value of \([A]_t\), the calculator will estimate \(k'\) from \(\ln([A]_t) = -k' t + \ln([A]_0)\) and compare it with the \(k'\) based on \([B]_0\) and \(k\).

Example (hydrolysis of ethyl acetate in excess water)

Load a pseudo–first-order example for \(\mathrm{CH_3COOC_2H_5 + H_2O \rightarrow CH_3COOH + CH_3CH_2OH}\). In aqueous solution, \([H_2O] \approx 55.5~\text{M}\) and is effectively constant, so the reaction behaves as first order in \(\mathrm{CH_3COOC_2H_5}\) with \(k' = k[H_2O]\).

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Frequently Asked Questions

What is a pseudo-first-order reaction?

It is a reaction that is truly second order (rate = k[A][B]) but behaves like first order in A because B is in large excess and stays approximately constant. The rate becomes rate = k'[A] where k' = k[B]0.

How do I calculate the pseudo-first-order rate constant k'?

When B is effectively constant, use k' = k[B]0. Use consistent concentration units so k[B]0 has units of 1/time.

How is the concentration [A]t found at a given time?

The integrated pseudo-first-order law is ln([A]t/[A]0) = -k' t, so [A]t = [A]0 e^(-k' t). This lets you compute the amount remaining after any time t.

What is the half-life for a pseudo-first-order process?

The half-life has the first-order form t1/2 = ln(2)/k' (approximately 0.693/k'). Under pseudo-first-order conditions it does not depend on [A]0.

When is the pseudo-first-order approximation valid?

It is valid when B is in large excess so its concentration changes negligibly during the time interval of interest. If B is not truly constant, ln([A]) versus time may not stay linear and k' = k[B]0 will not accurately describe the kinetics.