Pseudo-first-order kinetics

What is Pseudo-first-order Kinetics?

WHY does this matter?

Simplifies complex kinetics into easily measurable first-order form
Common in biochemistry (water as solvent is in huge excess)
Allows us to isolate and study the effect of one reactant at a time
Used in analytical chemistry to determine rate constants

HOW does it work? Consider a general second-order reaction: $A + B \rightarrow \text{Products}$

True rate law: $\text{Rate} = k[A][B]$

If $[B] \gg [A]$ (B is in large excess, say100× or1000× more), then:

As is consumed, [A] decreases significantly
But [B] barely changes: $[B] \approx [B]_0$ (initial concentration)

We can treat $[B]$ as a constant and absorb it into the rate constant:

$\text{Rate} = k[A][B]_0 = k'[A]$

where k′ (pseudo-first-order rate constant) $= k[B]_0$

Now the reaction looks first-order in A!

Derivation from First Principles

Starting point: True second-order reaction $A + B \rightarrow \text{Products}$

Step 1: Write the true rate law $-\frac{d[A]}{dt} = k[A][B]$

Why this step? The rate of disappearance of A depends on both concentrations.

Step 2: Apply the excess condition If $[B]_0 \gg [A]_0$ (typically50-100 times more), then: $[B] \approx [B]_0 \text{ (constant)}$

Why can we do this? If B starts at 1.0 M and A at 0.01 M, when all A is consumed, B only drops to 0.99 M (1% change).

Step 3: Substitute into rate law $-\frac{d[A]}{dt} = k[B]_0[A]$

Why this step? We replace the variable [B] with the constant $[B]_0$ .

Step 4: Define pseudo-first-order rate constant $k' = k[B]_0$ $-\frac{d[A]}{dt} = k'[A]$

Why introduce k′? It bundles the constant parts into one parameter we can measure directly.

Step 5: Integrate to get the pseudo-first-order equation $\int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = -k' \int_0^t dt$ $\ln[A] - \ln[A]_0 = -k't$

When is the Approximation Valid?

The excess condition requires: $\frac{[B]_0}{[A]_0} \geq 50 \text{ to } 100$

WHY this threshold?

If $[B]_0 =100[A]_0$ , consuming all A changes B by only 1%
Error in treating [B] as constant is ~1%, acceptable for most purposes
Below 10× excess, significant error appears

Worked Examples

Worked example Example 1: Ester Hydrolysis

The hydrolysis of ethyl acetate is pseudo-first-order: $\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$ Given:

Initial ester concentration: $[A]_0 = 0.01$ M
Water (solvent): $[H_2O] \approx 55.5$ M (constant)
Pseudo-first-order rate constant: $k' = 2.5 \times 10^{-4}$ s⁻¹

Find: (a) True second-order rate constant, (b) Time for 75% completion

Solution:

(a) Find true rate constant $k$ : $k = \frac{k'}{[B]_0} = \frac{2.5 \times 10^{-4}}{55.5} = 4.5 \times 10^{-6} \text{ M}^{-1}\text{s}^{-1}$

Why this step? We divide out the constant water concentration to get the true bimolecular rate constant.

(b) For 75% completion,25% remains: $[A] = 0.25[A]_0$ Using $\ln[A] = \ln[A]_0 - k't$ : $\ln(0.25[A]_0) = \ln[A]_0 - k't$ $\ln(0.25) = -k't$ $t = -\frac{\ln(0.25)}{k'} = -\frac{-1.386}{2.5 \times 10^{-4}} = 5544 \text{ s} \approx 92.4 \text{ min}$

Why use ln(0.25)? After 75% reaction, [A]/[A]₀ = 0.25, so we solve for when this ratio is reached.

Worked example Example 2: Inversion of Sucrose

Acid-catalyzed inversion of sucrose in aqueous solution: $\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_{12}\text{O}_6 + \text{C}_6\text{H}_{12}\text{O}_6$ Given: At25°C, $k' = 6.0 \times 10^{-5}$ s⁻¹

Initial sucrose concentration: 0.5 M Find: Concentration after 2 hours

Solution:

Time $t = 2 \times 3600 = 7200$ s

$[A] = [A]_0 e^{-k't}$ $[A] = 0.5 \times e^{-(6.0 \times 10^{-5})(7200)}$ $[A] = 0.5 \times e^{-0.432}$ $[A] = 0.5 \times 0.649 = 0.325 \text{ M}$

Why use the exponential form? It directly gives concentration at time t without needing logarithms.

Check: Fraction remaining = 0.325/0.5 = 0.65 = 65% This makes sense: in ~2 hours with this rate constant, about 35% has reacted.

Worked example Example 3: Determining Rate Order

A reaction between A and B is studied under two conditions: Experiment 1: $[A]_0 = 0.01$ M, $[B]_0 = 0.01$ M Plot of $\ln[A]$ vs $t$ is curved

Experiment 2: $[A]_0 = 0.01$ M, $[B]_0 = 1.0$ M Plot of $\ln[A]$ vs $t$ is linear with slope = $-0.05$ s⁻¹ Conclusion:

Why is Experiment 1 curved? Both reactants change significantly, so the reaction is truly second-order. The integrated second-order equation doesn't give a straight line in the ln[A] vs t plot. Why is Experiment 2 linear? B is in 100× excess, making it pseudo-first-order. The straight line confirms first-order behavior.

From Experiment 2:

$k' = 0.05$ s⁻¹
$k = k'/[B]_0 = 0.05/1.0 = 0.05$ M⁻¹s⁻¹ Why this matters? This method of isolation lets us determine the true order and rate constant by systematically varying excess conditions.

Common Mistakes and Misconceptions

Practical Applications

1. Enzyme Kinetics When substrate concentration $[S] \ll K_M$ (Michaelis constant), enzyme reactions become pseudo-first-order in substrate.

2. Analytical Chemistry Used in spectrophotometric analysis where one reagent is in huge excess, simplifying data analysis.

3. Environmental Chemistry Pollutant degradation in large bodies of water (excess water and oxygen makes kinetics pseudo-first-order).

4. Industrial Processes Allows engineers to design reactors treating complex reactions as simpler first-order for flow calculations.

Recall Explain to a 12-year-old

Imagine you're making chocolate milk. You have a tiny spoon of chocolate powder and a huge galon of milk.

Normally, to make chocolate milk, both the powder AND the milk need to mix together. The speed of mixing should depend on how much powder AND how much milk you have. But here's the trick: you have SO MUCH milk (like a swimming pool) that even after you mix in all the powder, the amount of milk barely changes. It's like the milk amount is "frozen" because it's so huge. So instead of worying about both things changing, you only need to watch the powder disappear. The mixing speed now only depends on the powder amount. The milk amount is so big, it might as well be a constant number.

Scientists call this "pseudo-first-order" because it LOOKS LIKE it only depends on one thing (the powder), even though technically both are needed. "Pseudo" means "fake" – it's fake-simple because we made one ingredient so huge that it doesn't matter if we use a bit of it.

Why do scientists like this trick? Because it makes measuring reaction speeds WAY easier. Instead of tracking two things changing, you only track one!

Connections

Integrated Rate Laws - pseudo-first-order follows first-order math
Second-Order Reactions - the true underlying kinetics
Method of Isolation - experimental technique using pseudo-order
Enzyme Kinetics - Michaelis-Menten can show pseudo-first-order limits
Half-life - pseudo-first-order has constant $t_{1/2} = \frac{\ln 2}{k'}$
Activation Energy - k′ still follows Arrhenius equation with temperature
Buffer Solutions - maintain constant [H⁺] for pseudo-order acid catalysis
Collision Theory - excess doesn't change collision mechanism, only the math

#flashcards/chemistry

What is a pseudo-first-order reaction? :: A higher-order reaction (typically second-order) where one or more reactants are in such large excess that their concentrations remain effectively constant, making the reaction appear first-order in the limiting reactant.

What is the mathematical requirement for pseudo-first-order conditions?

The excess reactant must be at least 50-100 times more concentrated than the limiting reactant:

[B]_0/[A]_0 \geq 50

to 100.

How do you convert pseudo-first-order rate constant k′ to true second-order rate constant k?

Divide by the concentration of the excess reactant:

k = k'/[B]_0

. Note the units change from s⁻¹ to M⁻¹s⁻¹.

What is the integrated rate law for pseudo-first-order kinetics?

\ln[A] = \ln[A]_0 - k't

or equivalently

[A] = [A]_0 e^{-k't}

, where

k' = k[B]_0

Why does hydrolysis in aqueous solution always show pseudo-first-order kinetics?

Water is the solvent at ~55.5 M concentration, which is in vast excess over any solute. Its concentration remains effectively constant throughout the reaction.

What is the slope of a ln[A] vs time plot for a pseudo-first-order reaction? :: The slope is $-k'$ (negative pseudo-first-order rate constant).

If a pseudo-first-order reaction has k′ = 0.02 s⁻¹, what is its half-life?

t_{1/2} = \ln(2)/k' = 0.693/0.02 = 34.65

seconds. Pseudo-first-order reactions have constant half-life like true first-order.

What are the units of the pseudo-first-order rate constant k′?

s⁻¹ (or min⁻¹, hr⁻¹), same as true first-order, different from the true second-order constant which has units M⁻¹s⁻¹.

What experimental technique uses pseudo-first-order conditions to determine reaction order?

The method of isolation – running experiments with different reactants in large excess to isolate the effect of each reactant one at a time.

Why can't you use 5× excess and still get accurate pseudo-first-order behavior? :: If the limiting reactant is 50% consumed, the "excess" reactant changes by 10% (from 5× to 4.5×), introducing significant error in treating it as constant.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Pseudo-first-order kinetics ek bahut clever trick hai chemistry mein. Dekho, normally agar do reactants A aur B reaction kar rahe hain, toh rate dono pe depend karti hai. Lekin agar

What is Pseudo-first-order Kinetics?

Derivation from First Principles

When is the Approximation Valid?

Worked Examples

Common Mistakes and Misconceptions

Practical Applications

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Chemical Kinetics

Connections