Integrated rate laws — half-life t₁ - ₂ for each order

Why Half-Life Matters

Half-life isn't just an arbitrary metric—it's the most intuitive way to compare reaction speeds and predict concentration decay:

In medicine: drug clearance times
In nuclear chemistry: radioactive decay
In environmental science: pollutant persistence
In manufacturing: shelf-life predictions

Deriving Half-Life for Each Order

Zero-Order Reactions (Rate = k)

Derivation of t₁/₂:

Starting from the integrated rate law, at half-life t = t₁/₂, we have [A]_t = [A]₀/2:

$\frac{[A]_0}{2} = [A]_0 - kt_{1/2}$

Why this step? We're setting the final concentration to exactly half the initial value—that's the definition of half-life.

Rearranging: $kt_{1/2} = [A]_0 - \frac{[A]_0}{2} = \frac{[A]_0}{2}$

Physical meaning: Zero-order kinetics occur when the reaction rate is saturated (e.g., enzyme fully saturated, surface catalysis with full coverage). The rate is constant regardless of how much reactant remains, so halving a larger amount takes proportionally longer.

First-Order Reactions (Rate = k[A])

Derivation of t₁/₂:

At t = t₁/₂, [A]_t = [A]₀/2:

$\ln\left(\frac{[A]_0}{2}\right) = \ln[A]_0 - kt_{1/2}$

Why this step? Substituting the half-life condition into the integrated rate law.

Expanding the logarithm: $\ln[A]_0 - \ln 2 = \ln[A]_0 - kt_{1/2}$

Why this step? Using the logarithm property ln(a/b) = ln(a) - ln(b).

The ln[A]₀ terms cancel: $-\ln 2 = -kt_{1/2}$

Physical meaning: In first-order reactions, the rate is always proportional to the current concentration. Whether you start with 100 M or 0.01 M, it takes the same time to halve because the rate scales proportionally with the amount present.

Second-Order Reactions (Rate = k[A]²)

Derivation of t₁/₂:

At t = t₁/₂, [A]_t = [A]₀/2:

$\frac{1}{[A]_0/2} = \frac{1}{[A]_0} + kt_{1/2}$

Why this step? Applying the half-life condition to the integrated rate law.

Simplifying the left side: $\frac{2}{[A]_0} = \frac{1}{[A]_0} + kt_{1/2}$

Why this step? 1/(a/2) = 2/a by algebra.

Solving for t₁/₂: $kt_{1/2} = \frac{2}{[A]_0} - \frac{1}{[A]_0} = \frac{1}{[A]_0}$

Physical meaning: Second-order reactions require two molecules to collide. At higher concentrations, collisions are more frequent, so the reaction proceeds faster. As concentration drops, collisions become rare, and it takes increasingly longer to halve the remaining amount.

Worked example Second-Order Half-Life Calculation

A second-order reaction has k = 0.54 M⁻¹s and [A]₀ = 0.20 M.

(a)* Calculate the first half-life. (b)* What is the concentration after the first half-life? (c)* Calculate the second half-life.

Solution: (a)* First half-life: $t_{1/2,1} = \frac{1}{k[A]_0} = \frac{1}{(0.54 \text{ M}^{-1}\text{s}^{-1})(0.20 \text{ M})} = \frac{1}{0.108 \text{ s}^{-1}} = 9.26 \text{ s}$

Why this step? Direct substitution into the second-order half-life formula.

(b)* After first half-life: $[A] = \frac{[A]_0}{2} = \frac{0.20 \text{ M}}{2} = 0.10 \text{ M}$

Why this step? By definition of half-life.

(c) Second half-life (starting from 0.10 M): $t_{1/2,2} = \frac{1}{k[A]} = \frac{1}{(0.54 \text{ M}^{-1}\text{s}^{-1})(0.10 \text{ M})} = \frac{1}{0.054 \text{ s}^{-1}} = 18.5 \text{ s}$

Why this step? The "new" initial concentration for the second half-life is 0.10 M, not 0.20 M.

Key insight: The second half-life (18.5 s) is exactly twice the first (9.26 s) because concentration halved. Each successive half-life doubles for second-order reactions.

Figure — Integrated rate laws — half-life t₁ - ₂ for each order

Comparison Table

| Order | Integrated Rate Law | Half-Life Formula | Dependence on [A]₀ | Half-Life Trend | |-------|---------------------|----------------|---------------| | 0 | [A]ₜ = [A]₀ - kt | $\frac{[A]_0}{2k}$ | Proportional | Decreases | | 1 | ln[A]ₜ = ln[A]₀ - kt | $\frac{0.693}{k}$ | Independent | Constant | | 2 | 1/[A]ₜ = 1/[A]₀ + kt | $\frac{1}{k[A]_0}$ | Inversely proportional | Increases |

Worked example Experimental Determination of Reaction Order via Half-Life

You measure the time for a reaction to go from [A]₀ to [A]₀/2, then from [A]₀/2 to [A]₀/4:

Experiment 1: [A]₀ = 0.80 M, first t₁/₂ = 50 s, second t₁/₂ = 50 s
Experiment 2: [A]₀ = 0.40 M, first t₁/₂ = 50 s What is the reaction order? Solution: The half-life is constant regardless of concentration. This is diagnostic of first-order kinetics.

Why this step? Constant t₁/₂ is the unique signature of first-order reactions.

We can calculate k: $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{50 \text{ s}} = 0.0139 \text{ s}^{-1}$

Check: If it were zero-order, doubling [A]₀ from 0.40 to 0.80 M would double t₁/₂ (from 25 s to 50 s). If it were second-order, doubling [A]₀ would halve t₁/₂ (from 100 s to 50 s). Neither matches our observation.

The "Number of Half-Lives" Concept

For first-order reactions only, we can predict the fraction remaining after n half-lives:

After1 half-life: 1/2 remains After 2 half-lives: 1/4 remains After 3 half-lives: 1/8 remains After n half-lives: (1/2)ⁿ remains

Total time: t = n × t₁/₂

Worked example Drug Clearance (First-Order)

A drug has a first-order elimination half-life of 4 hours. If a patient receives a 200 mg dose, how much remains after 12 hours?

Solution: Number of half-lives: n = 12 h / 4 h = 3

Why this step? Counting how many half-life periods fit into the elapsed time.

Fraction remaining: (1/2)³ = 1/8

Amount remaining: 200 mg × 1/8 = 25 mg

Why this step? Each half-life halves the amount, so three half-lives multiply by 1/2 three times.

Alternative approach using the integrated rate law: $[A]_t = [A]_0 e^{-kt}$

First find k: $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{4 \text{ h}} = 0.173 \text{ h}^{-1}$ Then: $[A]_{12} = 200 \text{ mg} \times e^{-(0.173)(12)} = 200 \times e^{-2.08} = 200 \times 0.125 = 25 \text{ mg}$

Both methods give the same answer, validating our approach.

Recall Feynman Explanation (Explain to a 12-year-old)

Imagine you're eating a big bowl of candy, and we want to know how long it takes to eat half of it.

Zero-order (constant rate): You eat exactly10 candies every minute, no matter how many are left. If you start with 100 candies, it takes 5 minutes to eat 50. If you start with 200 candies, it takes 10 minutes to eat 100. More candies = longer to eat half. The "half-time" depends on how much you started with.

First-order (proportional rate): You always eat 10% of what's left each minute. If you have 100, you eat 10. Next minute you have 90, you eat 9. Next minute 81, you eat 8.1. The cool thing? It ALWAYS takes the same time to eat half of whatever you have—about 7 minutes to go from 100 to 50, but also 7 minutes to go from 50 to 25, and another 7 minutes to go from 25 to 12.5. The "half-time" is constant.

Second-order (squared rate): The depends on candies meeting each other (like you need two candies to bump into each other for you to notice them and eat them). When you have lots of candies, they bump into each other often, so you eat fast. As you run out, bumps are rare, so eating slows down. The "half-time" gets longer and longer as you run out.

Connections

Integrated rate laws — derivation and graphical analysis
Rate constant k — units and temperature dependence
Reaction order determination from experimental data
Radioactive decay — first-order kinetics
Pharmacokinetics — drug elimination
Method of initial rates vs. integrated rate laws

Flashcards

#flashcards/chemistry

What is the definition of half-life (t₁/₂) in chemical kinetics? :: The time required for the concentration of a reactant to decrease to exactly half of its initial value.

What is the half-life formula for a zero-order reaction?

t₁/₂ = [A]₀/(2k), which shows half-life is directly proportional to initial concentration.

What is the half-life formula for a first-order reaction?

t₁/₂ = 0.693/k = ln(2)/k, which shows half-life is independent of initial concentration.

What is the half-life formula for a second-order reaction?

t₁/₂ = 1/(k[A]₀), which shows half-life is inversely proportional to initial concentration.

How does half-life change for a zero-order reaction as the reaction proceeds?

Half-life decreases as the reaction proceeds (as [A] decreases).

How does half-life change for a first-order reaction as the reaction proceeds?

Half-life remains constant throughout the reaction.

How does half-life change for a second-order reaction as the reaction proceeds?

Half-life increases as the reaction proceeds (as [A] decreases).

What are the units of the rate constant k for a second-order reaction?

M⁻¹s⁻¹ or L·mol⁻¹·s⁻¹

If a first-order reaction has k = 0.0462 min⁻¹, what is its half-life?

t₁/₂ = 0.693/0.0462 = 15.0 min

A second-order reaction starts at [A]₀ = 0.50 M with k = 0.20 M⁻¹s⁻¹. What is the first half-life?

t₁/₂ = 1/(0.20 × 0.50) = 10s

For the same second-order reaction in the previous card, what is the second half-life?

20 s (double the first half-life, since [A] is now 0.25 M)

After 3 half-lives of a first-order reaction, what fraction of the original reactant remains?

(1/2)³ = 1/8 or 12.5%

If you experimentally measure that t₁/₂ is constant regardless of initial concentration, what is the reaction order?

First-order

If you experimentally measure that t₁/₂ doubles when initial concentration is halved, what is the reaction order?

Second-order

What is the value of ln(2) used in first-order half-life calculations? :: 0.693

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Half-life ek bahut important concept hai chemical kinetics mein. Iska matlab hai ki kitna time lagega reactant ki concentration ko exactly half karne mein. Lekin sabse interesting baat yeh hai ki har order ke reaction ka half-life ka behavior bilkul alag hota hai.

Zero-order reaction mein, jaise ki surface catalysis ya enzyme-saturated reactions, rate constant rehta hai chahe kitna bhi reactant bacha ho. Toh agar apke pas zyada reactant hai initially, toh half karne mein zyada time lagega. Formula hai t₁/₂ = [A]₀/(2k), matlab directly proportional hai initial concentration ke sath. First-order reactions sabse unique hain—inki half-life bilkul constant rehti hai! Chahe ap 100 M se start karo ya 0.01 M se, halfone mein same time lagega. Yeh radioactive decay ka principle hai, aur formula hai t₁/₂ = 0.693/k. Yeh 0.693 ata hai ln(2) se, jo natural logarithm hai.

Second-order reactions mein do molecules ko collide karna padta hai reaction ke liye. Jab concentration high hai, collisions zyada frequent hote hain, toh reaction fast hota hai aur half-life chhoti hoti hai. Lekin jaise-jaise concentration kam hoti jati hai, collisions rare ho jate hain aur half-life badhti jati hai. Formula hai t₁/₂ = 1/(k[A]₀), matlab inversely proportional. Yeh behavior experimentally bahut useful hai—agar ap measure karo ki successive half-lives constant hain, toh pata chal gaya ki first-order hai. Agar half-lives increase ho rahi hain, toh second-order hai. Yeh diagnostic tool hai reaction order determine karne ke liye!

Integrated rate laws — half-life t₁ - ₂ for each order

Why Half-Life Matters

Deriving Half-Life for Each Order

Zero-Order Reactions (Rate = k)

First-Order Reactions (Rate = k[A])

Second-Order Reactions (Rate = k[A]²)

Comparison Table

The "Number of Half-Lives" Concept

Connections

Flashcards

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Chemical Kinetics

Connections