Integrated Rate Laws and Half-Life

Master integrated rate laws for zero, first, and second order reactions, calculate concentrations over time, and understand half-life concepts.

Integrated Rate Laws and Half-Life

Overview

Integrated rate laws relate concentration to time directly, allowing us to:

  • Predict reactant concentration at any time
  • Determine reaction order from experimental data
  • Calculate rate constants from concentration-time data

These equations are derived from differential rate laws using calculus.

Zero Order Reactions

Rate law: Rate = k

Integrated form:

[A]t=[A]0โˆ’kt[A]_t = [A]_0 - kt

Key characteristics:

  • Concentration decreases linearly with time
  • Plot [A] vs t gives straight line with slope = -k
  • Half-life depends on initial concentration

Half-life formula:

t1/2=[A]02kt_{1/2} = \frac{[A]_0}{2k}

Example: Surface-catalyzed reactions where catalyst surface is saturated


First Order Reactions

Rate law: Rate = k[A]

Integrated form:

lnโก[A]t=lnโก[A]0โˆ’kt\ln[A]_t = \ln[A]_0 - kt

Alternative form:

[A]t=[A]0eโˆ’kt[A]_t = [A]_0 e^{-kt}

Key characteristics:

  • Concentration decreases exponentially
  • Plot ln[A] vs t gives straight line with slope = -k
  • Half-life is constant (independent of concentration)

Half-life formula:

t1/2=0.693k=lnโก2kt_{1/2} = \frac{0.693}{k} = \frac{\ln 2}{k}

Common examples: Radioactive decay, many decomposition reactions


Second Order Reactions

Rate law: Rate = k[A]ยฒ

Integrated form:

1[A]t=1[A]0+kt\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt

Key characteristics:

  • Plot 1/[A] vs t gives straight line with slope = +k
  • Half-life increases as reaction proceeds

Half-life formula:

t1/2=1k[A]0t_{1/2} = \frac{1}{k[A]_0}

Example: Gas-phase dimerization reactions (2A โ†’ Aโ‚‚)


Comparison at a Glance (Mobileโ€‘Friendly)

Rather than a cramped table, hereโ€™s a stacked, easyโ€‘scan summary for each key property:

Rate law

  • Zero order: Rate = k
  • First order: Rate = k[A]
  • Second order: Rate = k[A]ยฒ

Integrated law

  • Zero order: [A]t=[A]0โˆ’kt[A]_t = [A]_0 - kt
  • First order: lnโก[A]t=lnโก[A]0โˆ’kt\ln[A]_t = \ln[A]_0 - kt (equivalently, [A]t=[A]0eโˆ’kt[A]_t = [A]_0 e^{-kt})
  • Second order: 1[A]t=1[A]0+kt\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt

Linear plot for straight line

  • Zero order: [A] vs t (slope = โˆ’k)
  • First order: ln[A] vs t (slope = โˆ’k)
  • Second order: 1/[A] vs t (slope = +k)

Halfโ€‘life

  • Zero order: t1/2=[A]02kt_{1/2} = \dfrac{[A]_0}{2k} (depends on [A]โ‚€)
  • First order: t1/2=0.693kt_{1/2} = \dfrac{0.693}{k} (constant)
  • Second order: t1/2=1k[A]0t_{1/2} = \dfrac{1}{k[A]_0} (depends on [A]โ‚€)

Units of k

  • Zero order: Mยทsโปยน
  • First order: sโปยน
  • Second order: Mโปยนยทsโปยน

Determining Reaction Order Graphically

Method:

  1. Collect data: Measure [A] at various times
  2. Make three plots:
    • [A] vs t
    • ln[A] vs t
    • 1/[A] vs t
  3. Identify the linear plot - this reveals the order:
    • [A] vs t is linear โ†’ Zero order
    • ln[A] vs t is linear โ†’ First order
    • 1/[A] vs t is linear โ†’ Second order

Extracting the rate constant:

  • Zero/First order: k = -slope (note negative!)
  • Second order: k = +slope (positive)

Half-Life Concepts

Definition: Time required for concentration to decrease to half its initial value

After n half-lives:

[A]=[A]02n[A] = \frac{[A]_0}{2^n}

Examples:

  • After 1 half-life: [A] = [A]โ‚€/2 = 50% remaining
  • After 2 half-lives: [A] = [A]โ‚€/4 = 25% remaining
  • After 3 half-lives: [A] = [A]โ‚€/8 = 12.5% remaining

Key insight: Only first-order reactions have a constant half-life!


Practice Example

Problem: A reaction has the following data:

| Time (s) | [A] (M) | |----------|---------| | 0 | 1.00 | | 50 | 0.61 | | 100 | 0.37 | | 150 | 0.22 |

Determine the order and rate constant.

Solution:

Calculate ln[A] and 1/[A]:

| Time (s) | [A] (M) | ln[A] | 1/[A] (Mโปยน) | |----------|---------|-------|-------------| | 0 | 1.00 | 0.00 | 1.00 | | 50 | 0.61 | -0.49 | 1.64 | | 100 | 0.37 | -0.99 | 2.70 | | 150 | 0.22 | -1.51 | 4.55 |

Plot ln[A] vs t โ†’ Linear! (slope โ‰ˆ -0.01)

Conclusion: First order reaction with k = 0.01 sโปยน

Half-life: tโ‚/โ‚‚ = 0.693/0.01 = 69.3 seconds


Real-World Applications

Radioactive Dating:

  • Carbon-14 dating: tโ‚/โ‚‚ = 5,730 years (first order)
  • Used to date organic materials up to ~50,000 years old

Pharmacology:

  • Drug metabolism follows first-order kinetics
  • Constant half-life allows predictable dosing schedules
  • Example: Aspirin tโ‚/โ‚‚ โ‰ˆ 2-3 hours

Environmental Chemistry:

  • Pollutant degradation rates
  • Predicting cleanup timelines

Chemical Engineering:

  • Reactor design and optimization
  • Determining optimal reaction times

๐Ÿ“š Practice Problems

1Problem 1easy

โ“ Question:

The decomposition of Nโ‚‚Oโ‚… is first order with k = 5.0 ร— 10โปโด sโปยน. If [Nโ‚‚Oโ‚…]โ‚€ = 0.200 M, find: (a) [Nโ‚‚Oโ‚…] after 100 seconds, (b) time to reach 0.050 M, (c) half-life.

๐Ÿ’ก Show Solution

Given:

  • First order: k = 5.0 ร— 10โปโด sโปยน
  • [Nโ‚‚Oโ‚…]โ‚€ = 0.200 M

First order integrated law: ln[A]_t = ln[A]โ‚€ - kt


(a) Find [Nโ‚‚Oโ‚…] after t = 100 s

ln[Nโ‚‚Oโ‚…]โ‚โ‚€โ‚€ = ln(0.200) - (5.0 ร— 10โปโด)(100) ln[Nโ‚‚Oโ‚…]โ‚โ‚€โ‚€ = -1.609 - 0.050 = -1.659

[Nโ‚‚Oโ‚…]โ‚โ‚€โ‚€ = eโปยนยทโถโตโน = 0.190 M

Answer: 0.190 M


(b) Time to reach 0.050 M

Rearrange: t = (1/k)ln([A]โ‚€/[A]_t)

t = (1/(5.0 ร— 10โปโด))ln(0.200/0.050) t = 2000 ร— ln(4) = 2000 ร— 1.386 t = 2772 s = 46.2 min

Answer: 2772 s or 46 min


(c) Half-life

For first order: tโ‚/โ‚‚ = 0.693/k

tโ‚/โ‚‚ = 0.693/(5.0 ร— 10โปโด) = 1386 s = 23.1 min

Answer: 1386 s or 23 min

Check: In part (b), 0.200 M โ†’ 0.050 M is a factor of 4 = 2ยฒ So time = 2 half-lives = 2(1386) = 2772 s โœ“

2Problem 2medium

โ“ Question:

Given concentration vs time data, determine the order and rate constant: t(s): 0, 10, 20, 30, 40 | A: 1.00, 0.63, 0.46, 0.36, 0.29

๐Ÿ’ก Show Solution

Data: | t (s) | [A] (M) | ln[A] | 1/[A] (Mโปยน) | |-------|---------|-------|-------------| | 0 | 1.00 | 0.000 | 1.00 | | 10 | 0.63 | -0.462| 1.59 | | 20 | 0.46 | -0.777| 2.17 | | 30 | 0.36 | -1.022| 2.78 | | 40 | 0.29 | -1.238| 3.45 |


Test for zero order: Plot [A] vs t

  • Not linear (curved)

Test for first order: Plot ln[A] vs t

  • Points: (0, 0.000), (10, -0.462), (20, -0.777), (30, -1.022), (40, -1.238)
  • Check linearity: ฮ”(ln[A])/ฮ”t โ‰ˆ -0.031 per second (constant!)
  • Linear! โ†’ First order

Test for second order: Plot 1/[A] vs t

  • Not as linear as ln[A] plot

Conclusion: First order

Calculate k from slope:

slope = ฮ”(ln[A])/ฮ”t = (-1.238 - 0.000)/(40 - 0) = -0.031 sโปยน

For first order: slope = -k k = 0.031 sโปยน

Verification: Calculate tโ‚/โ‚‚ = 0.693/k = 0.693/0.031 = 22.4 s Check data: At t โ‰ˆ 22 s, [A] should be โ‰ˆ 0.50 M (half of 1.00) Interpolating between t=20 (0.46 M) and t=30 (0.36 M) gives โ‰ˆ 0.48 M โœ“

Answer: First order, k = 0.031 sโปยน

3Problem 3hard

โ“ Question:

A second-order reaction has k = 0.54 Mโปยนยทsโปยน and [A]โ‚€ = 0.10 M. (a) What is [A] after 2.0 s? (b) What is the first half-life? (c) What is the second half-life?

๐Ÿ’ก Show Solution

Given:

  • Second order: k = 0.54 Mโปยนยทsโปยน
  • [A]โ‚€ = 0.10 M

Second order integrated law: 1/[A]_t = 1/[A]โ‚€ + kt


(a) [A] after 2.0 s

1/[A]โ‚‚ = 1/0.10 + (0.54)(2.0) 1/[A]โ‚‚ = 10 + 1.08 = 11.08 Mโปยน

[A]โ‚‚ = 1/11.08 = 0.0903 M

Answer: 0.090 M


(b) First half-life

For second order: tโ‚/โ‚‚ = 1/(k[A]โ‚€)

tโ‚/โ‚‚ = 1/((0.54)(0.10)) = 1/0.054 = 18.5 s

Answer: 18.5 s

Verification: After 18.5 s, [A] should be 0.050 M

1/[A]โ‚โ‚ˆ.โ‚… = 10 + (0.54)(18.5) = 10 + 10 = 20 Mโปยน [A]โ‚โ‚ˆ.โ‚… = 1/20 = 0.050 M โœ“


(c) Second half-life

Key point: For second order, half-life increases!

After first tโ‚/โ‚‚: [A] = 0.050 M (new starting point)

Second tโ‚/โ‚‚ = 1/(k ร— 0.050) = 1/((0.54)(0.050)) Second tโ‚/โ‚‚ = 1/0.027 = 37.0 s

Answer: 37.0 s

Important: Second half-life (37 s) is twice the first (18.5 s)

  • For second order: each successive half-life doubles
  • Total time for 2 half-lives: 18.5 + 37.0 = 55.5 s
  • After 55.5 s: [A] = 0.025 M (1/4 of original)

Contrast with first order:

  • First order: all half-lives equal
  • Second order: each half-life = 2 ร— previous
  • Zero order: each half-life = 1/2 ร— previous

4Problem 4medium

โ“ Question:

Zero-order decay: A โ†’ products with k = 2.5ร—10โปยณ Mยทsโปยน. If [A]_0 = 0.300 M, (a) write [A]_t, (b) time to 0.120 M, (c) half-life, (d) how much remains after 4.0 minutes?

๐Ÿ’ก Show Solution

Zero order: [A]_t = [A]_0 โˆ’ kt. (a) [A]_t = 0.300 โˆ’ (2.5ร—10โปยณ)t (M). (b) t = (0.300 โˆ’ 0.120)/(2.5ร—10โปยณ) = 0.180/0.0025 = 72 s. (c) tโ‚/โ‚‚ = [A]_0/(2k) = 0.300/(2ร—2.5ร—10โปยณ) = 60 s. (d) t = 4.0 min = 240 s โ‡’ [A] = 0.300 โˆ’ (2.5ร—10โปยณ)(240) = 0.300 โˆ’ 0.600 = negative โ†’ zero-order model predicts exhaustion at t = [A]_0/k = 120 s, so after 240 s, [A] โ‰ˆ 0 (reaction complete).

5Problem 5easy

โ“ Question:

First-order: A first-order reaction has k = 0.035 sโปยน. (a) Time to reach 25% of [A]_0? (b) Fraction remaining after 1.0 minute?

๐Ÿ’ก Show Solution

First order: [A]_t = [A]_0 e^{โˆ’kt}. (a) Set [A]_t/[A]_0 = 0.25 = e^{โˆ’kt} โ‡’ t = (1/k)ln(1/0.25) = (1/0.035)ln(4) โ‰ˆ 28.6ร—1.386 โ‰ˆ 39.7 s. (b) t = 60 s โ‡’ fraction = e^{โˆ’0.035ร—60} = e^{โˆ’2.10} โ‰ˆ 0.122 = 12.2% remains.

6Problem 6medium

โ“ Question:

Order identification from data: t (s) = 0, 50, 100, 150; [A] (M) = 0.80, 0.57, 0.41, 0.30. Determine order and k.

๐Ÿ’ก Show Solution

Compute ln[A]: 0.80โ†’โˆ’0.223; 0.57โ†’โˆ’0.562; 0.41โ†’โˆ’0.894; 0.30โ†’โˆ’1.204. ฮ”(ln[A]) per 50 s โ‰ˆ โˆ’0.339, โˆ’0.332, โˆ’0.310 (nearly constant) โ‡’ ln[A] vs t is ~linear โ‡’ first-order. Slope โ‰ˆ (โˆ’1.204 โˆ’ (โˆ’0.223))/(150 โˆ’ 0) = (โˆ’0.981)/150 = โˆ’0.00654 sโปยน โ‡’ k โ‰ˆ 6.5ร—10โปยณ sโปยน. Check 1/[A] vs t curvature to rule out second order.

7Problem 7hard

โ“ Question:

Second-order: For 2A โ†’ products with k = 0.22 Mโปยนยทsโปยน and [A]_0 = 0.50 M, how long to reach 0.20 M? What are the first and second half-lives?

๐Ÿ’ก Show Solution

Second order: 1/[A]_t = 1/[A]_0 + kt. Solve for t to [A]_t = 0.20: t = (1/[A]_t โˆ’ 1/[A]_0)/k = (1/0.20 โˆ’ 1/0.50)/0.22 = (5.00 โˆ’ 2.00)/0.22 = 3.00/0.22 โ‰ˆ 13.6 s. Half-lives: tโ‚/โ‚‚ = 1/(k[A]_0) = 1/(0.22ร—0.50) = 1/0.11 โ‰ˆ 9.09 s; second half-life = 1/(kร—0.25) = 1/0.055 โ‰ˆ 18.2 s (double).