⚗️ Reaction Order Calculator
Solve zero order, first order, and second order reaction kinetics — calculate concentration at any time, find rate constant k, compute half-life for all reaction orders, and determine reaction order from experimental data with full step-by-step integrated rate law working.
Key Equations:
[A] = [A]₀ − kt
ln[A] = ln[A]₀ − kt
1/[A] = 1/[A]₀ + kt
t½ = [A]₀/(2k)
t½ = 0.693/k
t½ = 1/(k[A]₀)
🟦 Zero Order Reactions
rate = k
Independent of concentration
[A] = [A]₀ − kt
[A] = [A]₀ − kt
🟨 First Order Reactions
rate = k[A]
Exponential decay
ln[A] = ln[A]₀ − kt
ln[A] = ln[A]₀ − kt
🟪 Second Order Reactions
rate = k[A]²
Hyperbolic decay
1/[A] = 1/[A]₀ + kt
1/[A] = 1/[A]₀ + kt
📐 Zero Order Integrated Rate Law — [A] = [A]₀ − kt
Zero order rate law: rate = k |
Integrated rate law: [A] = [A]₀ − kt |
Half-life: t½ = [A]₀/(2k)
📐 First Order Integrated Rate Law — ln[A] = ln[A]₀ − kt
First order rate law: rate = k[A] |
Integrated rate law: ln[A] = ln[A]₀ − kt |
Exponential form: [A] = [A]₀ × e^(−kt) |
Half-life: t½ = 0.693/k
📐 Second Order Integrated Rate Law — 1/[A] = 1/[A]₀ + kt
Second order rate law: rate = k[A]² |
Integrated rate law: 1/[A] = 1/[A]₀ + kt |
Half-life: t½ = 1/(k[A]₀) |
Units of k: M⁻¹s⁻¹ or L/(mol·s)
🔍 Determine Reaction Order from Experimental Data
Enter concentration vs time data — the calculator tests zero order ([A] vs t) and first order (ln[A] vs t) linearity, computes R² values, and identifies the reaction order.
| Time (s) | Concentration [A] (M) |
|---|
Zero Order vs First Order Reactions — Key Differences
The most important distinction: zero order half-life depends on initial concentration while first order half-life is constant and independent of concentration. This single fact separates the two reaction orders in every kinetics course.
| Property | 🟦 Zero Order | 🟨 First Order |
|---|---|---|
| Rate Law | rate = k |
rate = k[A] |
| Integrated Rate Law | [A] = [A]₀ − kt |
ln[A] = ln[A]₀ − kt |
| Concentration Form | Linear: [A] vs t | Exponential: [A] = [A]₀e^(−kt) |
| Graph [A] vs t | Straight line (negative slope −k) | Exponential decay curve |
| Graph ln[A] vs t | Curved (non-linear) | Straight line (slope = −k) |
| Units of k | M·s⁻¹ or mol/(L·s) | s⁻¹ or 1/s |
| Half-life Formula | t½ = [A]₀/(2k) |
t½ = 0.693/k |
| Half-life depends on [A]₀? | ✅ YES — Depends on [A]₀ | ❌ NO — Always constant |
| Successive half-lives | Each one is SHORTER (less reactant) | Each one is IDENTICAL (constant) |
| Examples | Photochemistry, surface reactions, saturated enzymes | Radioactive decay, drug metabolism, organic decomposition |
Zero Order Reactions — Rate = k
In a zero order reaction (also written zeroth order), the reaction rate is completely independent of the reactant concentration. No matter how much or how little [A] is present, the reaction proceeds at the same constant rate k. This typically occurs when a catalyst, enzyme, or light-source is saturated — the rate is limited by something other than the amount of reactant.
rate = k
[A] = [A]₀ − kt
t½ = [A]₀/(2k)
Derivation of [A] = [A]₀ − kt:
rate = −d[A]/dt = k
Integrate: −∫d[A] = ∫k dt
−([A] − [A]₀) = kt
[A] = [A]₀ − kt ✓
rate = −d[A]/dt = k
Integrate: −∫d[A] = ∫k dt
−([A] − [A]₀) = kt
[A] = [A]₀ − kt ✓
Graphical Signature:
Plot [A] vs time → straight line
Slope = −k (units: M/s)
y-intercept = [A]₀
x-intercept = t_complete = [A]₀/k
Plot [A] vs time → straight line
Slope = −k (units: M/s)
y-intercept = [A]₀
x-intercept = t_complete = [A]₀/k
Zero Order Half-Life intuition:
Because concentration drops at a constant rate (−k M/s), halving a larger starting amount requires
consuming more reactant at the same constant speed — so it takes longer.
If [A]₀ = 1 M and k = 0.05 M/min: t½ = 1.0/(2×0.05) = 10 minutes.
If [A]₀ = 2 M with same k: t½ = 2.0/(2×0.05) = 20 minutes.
Double the starting amount → double the half-life.
| Example | Values | Calculation | Result |
|---|---|---|---|
| Find [A] at t=10 s | [A]₀=0.5M, k=0.02 M/s | [A]=0.5−(0.02×10)=0.5−0.2 | 0.3 M |
| Find half-life | [A]₀=1.0M, k=0.05 M/min | t½=1.0/(2×0.05)=1.0/0.1 | 10 min |
| Find k | [A]₀=0.8M, [A]=0.5M, t=15s | k=(0.8−0.5)/15=0.3/15 | 0.02 M/s |
First Order Reactions — Rate = k[A]
A first order reaction has a rate directly proportional to the reactant concentration [A]. As [A] decreases, the rate decreases proportionally — producing the characteristic exponential decay curve. First order kinetics is the most common reaction order in chemistry and biology.
rate = k[A]
ln[A] = ln[A]₀ − kt
[A] = [A]₀ × e^(−kt)
t½ = 0.693/k
Derivation of ln[A] = ln[A]₀ − kt:
rate = −d[A]/dt = k[A]
Separate: −d[A]/[A] = k dt
Integrate: −ln[A] = kt + C
At t=0: C = −ln[A]₀
ln[A] = ln[A]₀ − kt ✓
rate = −d[A]/dt = k[A]
Separate: −d[A]/[A] = k dt
Integrate: −ln[A] = kt + C
At t=0: C = −ln[A]₀
ln[A] = ln[A]₀ − kt ✓
Graphical Signatures:
[A] vs t → exponential curve
ln[A] vs t → straight line
Slope of ln[A] plot = −k (s⁻¹)
y-intercept of ln[A] plot = ln[A]₀
[A] vs t → exponential curve
ln[A] vs t → straight line
Slope of ln[A] plot = −k (s⁻¹)
y-intercept of ln[A] plot = ln[A]₀
Constant Half-Life — The Signature of First Order:
Because rate is proportional to concentration, the fractional change per unit time is constant.
Halving from 1.0 M to 0.5 M takes the same time as halving from 0.5 M to 0.25 M.
For k = 0.05 s⁻¹: t½ = 0.693/0.05 = 13.86 s always.
After 13.86 s: 50% remains. After 27.72 s: 25%. After 41.58 s: 12.5%.
| Example | Values | Calculation | Result |
|---|---|---|---|
| Find [A] at t=10 s | [A]₀=1.0M, k=0.1 s⁻¹ | [A]=1.0×e^(−0.1×10)=e^(−1) | 0.368 M (36.8%) |
| Find time t | [A]₀=0.5M, [A]=0.125M, k=0.2 min⁻¹ | t=ln(0.5/0.125)/0.2=ln(4)/0.2 | 6.93 min |
| Find half-life | k=0.05 hour⁻¹ | t½=0.693/0.05 | 13.86 hours |
| Find k | [A]₀=1.0M, [A]=0.25M, t=20 s | k=ln(1.0/0.25)/20=ln(4)/20 | 0.0693 s⁻¹ |
Half-Life Formula: Zero Order vs First Order
🟦 Zero Order Half-Life
t½ = [A]₀ / (2k)
- DEPENDS on initial concentration [A]₀
- More starting material → longer half-life
- 1 M takes 10 min; 2 M takes 20 min (same k)
- Each successive half-life is shorter
- Reaction reaches zero in finite time
🟨 First Order Half-Life
t½ = 0.693 / k
- INDEPENDENT of initial concentration [A]₀
- Same half-life for any starting amount
- 1 M or 0.001 M: same t½
- Every half-life is identical
- Concentration approaches zero asymptotically
The formulas t½ = [A]₀/(2k) and t½ = 0.693/k capture the fundamental difference: zero order half-life depends on [A]₀; first order half-life does not.
| [A]₀ | k (M/s zero; s⁻¹ first) | Zero t½ (s) | First t½ (s) |
|---|---|---|---|
| 0.5 M | 0.05 | 5.0 s | 13.86 s |
| 1.0 M | 0.05 | 10.0 s | 13.86 s |
| 2.0 M | 0.05 | 20.0 s | 13.86 s |
| 5.0 M | 0.05 | 50.0 s | 13.86 s |
Same k = 0.05 for both orders. Zero order t½ changes with [A]₀; first order t½ stays constant at 13.86 s.
Integrated Rate Law Equations — Complete Formula Reference
The integrated rate law equations are derived by integrating the differential rate law. The key relationships: [A] = [A]₀ − kt (zero order), ln[A] = ln[A]₀ − kt (first order), and 1/[A] = 1/[A]₀ + kt (second order).
| Reaction Order | Rate Law | Integrated Form | Alternative Form | Linear Plot | Slope |
|---|---|---|---|---|---|
| Zero Order | rate = k |
[A] = [A]₀ − kt |
— | [A] vs t | −k |
| First Order | rate = k[A] |
ln[A] = ln[A]₀ − kt |
[A] = [A]₀e^(−kt) |
ln[A] vs t | −k |
| Second Order | rate = k[A]² |
1/[A] = 1/[A]₀ + kt |
[A] = 1/(1/[A]₀ + kt) |
1/[A] vs t | +k |
How to Find Rate Constant k — All Three Reaction Orders
| Reaction Order | Formula for k | Units of k | How to find from graph |
|---|---|---|---|
| Zero Order | k = ([A]₀ − [A])/t |
M·s⁻¹ | Slope of [A] vs t (negative) |
| First Order | k = ln([A]₀/[A])/t |
s⁻¹ | −Slope of ln[A] vs t |
| Second Order | k = (1/[A] − 1/[A]₀)/t |
M⁻¹·s⁻¹ | Slope of 1/[A] vs t (positive) |
Zero Order k example:
[A]₀=1.0M, [A]=0.6M at t=20s
k = (1.0−0.6)/20 = 0.4/20 = 0.020 M/s
[A]₀=1.0M, [A]=0.6M at t=20s
k = (1.0−0.6)/20 = 0.4/20 = 0.020 M/s
First Order k example:
[A]₀=1.0M, [A]=0.368M at t=10s
k = ln(1.0/0.368)/10 = 1.000/10 = 0.100 s⁻¹
[A]₀=1.0M, [A]=0.368M at t=10s
k = ln(1.0/0.368)/10 = 1.000/10 = 0.100 s⁻¹
Second Order k example:
[A]₀=0.5M, [A]=0.357M at t=8s
k = (1/0.357−1/0.5)/8 = (2.80−2.0)/8 = 0.100 M⁻¹s⁻¹
[A]₀=0.5M, [A]=0.357M at t=8s
k = (1/0.357−1/0.5)/8 = (2.80−2.0)/8 = 0.100 M⁻¹s⁻¹
Zero, First and Second Order Reactions — Complete Comparison
Master reference table for all three reaction orders — the most important table in chemical kinetics.
| Property | 🟦 Zero Order | 🟨 First Order | 🟪 Second Order |
|---|---|---|---|
| Rate law | rate = k | rate = k[A] | rate = k[A]² |
| Integrated rate law | [A] = [A]₀ − kt | ln[A] = ln[A]₀ − kt | 1/[A] = 1/[A]₀ + kt |
| Linear plot | [A] vs t | ln[A] vs t | 1/[A] vs t |
| Slope of linear plot | −k | −k | +k |
| Units of k | M·s⁻¹ | s⁻¹ | M⁻¹·s⁻¹ |
| Half-life formula | t½ = [A]₀/2k | t½ = 0.693/k | t½ = 1/k[A]₀ |
| Half-life depends on [A]₀? | ✅ YES | ❌ NO (constant) | ✅ YES |
| Successive half-lives | Decrease (shorter each time) | Constant (same every time) | Increase (double each time) |
| Real examples | Photochemistry, surface reactions | Radioactive decay, drug metabolism | NO₂ decomposition, ozone |
How to Determine Reaction Order — Method Explained
The graphical method for determining reaction order uses the integrated rate laws as tests. The plot that produces a straight line identifies the reaction order.
Step 1: Collect [A] vs time data at several time points.
Step 2: Plot [A] vs t. If straight → zero order. Curved → not zero order.
Step 3: Plot ln[A] vs t. If straight → first order. Curved → not first order.
Step 4: Plot 1/[A] vs t. If straight → second order.
📊 R² Method: The R² value measures how well data fits a straight line (R²=1.00 is perfect).
If R² for [A] vs t = 0.85 but R² for ln[A] vs t = 0.99, the reaction is definitely first order.
The calculator above uses linear regression to compute R² automatically.
| Time (s) | [A] (M) | ln[A] | 1/[A] (M⁻¹) |
|---|---|---|---|
| 0 | 1.000 | 0.000 | 1.000 |
| 10 | 0.905 | −0.100 | 1.105 |
| 20 | 0.819 | −0.200 | 1.221 |
| 30 | 0.741 | −0.300 | 1.350 |
| 50 | 0.607 | −0.500 | 1.647 |
| → ln[A] vs t is perfectly linear: this is a FIRST ORDER reaction, k = 0.010 s⁻¹ | |||
Real-World Applications — Where Zero and First Order Reactions Happen
🟦 Zero Order Examples
☀️ Photochemistry
When light intensity limits the reaction, rate equals the photon flux regardless of reactant concentration. Photosynthesis approaches zero order when CO₂ is abundant and light is limiting.
When light intensity limits the reaction, rate equals the photon flux regardless of reactant concentration. Photosynthesis approaches zero order when CO₂ is abundant and light is limiting.
🧬 Enzyme-Catalyzed Reactions
When enzyme active sites are fully saturated with substrate, adding more substrate has no effect. The rate is k (determined by enzyme turnover) — zero order with respect to substrate. [A] = [A]₀ − kt applies perfectly.
When enzyme active sites are fully saturated with substrate, adding more substrate has no effect. The rate is k (determined by enzyme turnover) — zero order with respect to substrate. [A] = [A]₀ − kt applies perfectly.
🏭 Surface Catalysis
Heterogeneous catalysis on a solid surface proceeds at zero order when the surface is fully covered. The rate depends on catalyst surface area, not reactant concentration. The zero order formula rate = k describes this precisely.
Heterogeneous catalysis on a solid surface proceeds at zero order when the surface is fully covered. The rate depends on catalyst surface area, not reactant concentration. The zero order formula rate = k describes this precisely.
🟨 First Order Examples
☢️ Radioactive Decay (classic example)
Carbon-14 has a constant half-life of 5,730 years. No matter how much C-14 you start with, half decays in 5,730 years. After 11,460 years: 25% remains. After 17,190 years: 12.5%. This constant half-life — from t½ = 0.693/k — is why C-14 dating works precisely.
Carbon-14 has a constant half-life of 5,730 years. No matter how much C-14 you start with, half decays in 5,730 years. After 11,460 years: 25% remains. After 17,190 years: 12.5%. This constant half-life — from t½ = 0.693/k — is why C-14 dating works precisely.
💊 Drug Metabolism
Most drugs are eliminated from the bloodstream at a rate proportional to their concentration — first order kinetics. Ibuprofen half-life: ~2 hours. After 10 hours (5 half-lives): only 3.1% remains. Dosing schedules use t½ = 0.693/k to maintain therapeutic levels.
Most drugs are eliminated from the bloodstream at a rate proportional to their concentration — first order kinetics. Ibuprofen half-life: ~2 hours. After 10 hours (5 half-lives): only 3.1% remains. Dosing schedules use t½ = 0.693/k to maintain therapeutic levels.
🌎 Ozone Decomposition
Ozone (O₃) decomposition in the stratosphere follows first order kinetics. The rate = k[O₃] means ozone concentration decays exponentially as [A] = [A]₀e^(−kt). Catalytic destruction by CFCs follows the same first order mathematics.
Ozone (O₃) decomposition in the stratosphere follows first order kinetics. The rate = k[O₃] means ozone concentration decays exponentially as [A] = [A]₀e^(−kt). Catalytic destruction by CFCs follows the same first order mathematics.
Worked Examples
1. Zero Order — Find [A] at time t
[A]₀=0.5M, k=0.02 M/s, t=10s. Formula: [A] = [A]₀ − kt. Substitute: [A] = 0.5 − (0.02×10) = 0.5 − 0.2 = 0.30 M. Concentration reduced by 40% in 10 seconds at constant rate of 0.02 M per second.
[A]₀=0.5M, k=0.02 M/s, t=10s. Formula: [A] = [A]₀ − kt. Substitute: [A] = 0.5 − (0.02×10) = 0.5 − 0.2 = 0.30 M. Concentration reduced by 40% in 10 seconds at constant rate of 0.02 M per second.
2. Zero Order — Find k from data
[A]₀=0.8M, [A]=0.6M at t=10s. Rearrange [A]=[A]₀−kt: k=([A]₀−[A])/t = (0.8−0.6)/10 = 0.20/10 = 0.020 M/s. Units of k for zero order are always M/s (or mol·L⁻¹·s⁻¹).
[A]₀=0.8M, [A]=0.6M at t=10s. Rearrange [A]=[A]₀−kt: k=([A]₀−[A])/t = (0.8−0.6)/10 = 0.20/10 = 0.020 M/s. Units of k for zero order are always M/s (or mol·L⁻¹·s⁻¹).
3. Zero Order Half-Life
[A]₀=1.0M, k=0.05 M/min. Formula: t½=[A]₀/(2k) = 1.0/(2×0.05) = 1.0/0.10 = 10.0 minutes. At t=10 min: [A]=0.5M. At t=20 min: [A]=0M. Total reaction time = [A]₀/k = 1.0/0.05 = 20 minutes.
[A]₀=1.0M, k=0.05 M/min. Formula: t½=[A]₀/(2k) = 1.0/(2×0.05) = 1.0/0.10 = 10.0 minutes. At t=10 min: [A]=0.5M. At t=20 min: [A]=0M. Total reaction time = [A]₀/k = 1.0/0.05 = 20 minutes.
4. First Order — Find [A] using [A] = [A]₀e^(−kt)
[A]₀=1.0M, k=0.1 s⁻¹, t=10s. [A]=1.0×e^(−0.1×10)=e^(−1.0)=0.368M. Using ln form: ln[A]=ln(1.0)−(0.1)(10)=0−1=−1, so [A]=e^(−1)=0.368M. 36.8% of initial concentration remains.
[A]₀=1.0M, k=0.1 s⁻¹, t=10s. [A]=1.0×e^(−0.1×10)=e^(−1.0)=0.368M. Using ln form: ln[A]=ln(1.0)−(0.1)(10)=0−1=−1, so [A]=e^(−1)=0.368M. 36.8% of initial concentration remains.
5. First Order — Solve ln[A] = ln[A]₀ − kt for time
[A]₀=0.5M, [A]=0.125M, k=0.2 min⁻¹. Rearrange: kt=ln([A]₀/[A])=ln(0.5/0.125)=ln(4)=1.386. t=1.386/0.2=6.93 minutes. Equivalently: 0.125/0.5=0.25=(1/2)², so 2 half-lives → t=2×(0.693/0.2)=6.93 min ✓
[A]₀=0.5M, [A]=0.125M, k=0.2 min⁻¹. Rearrange: kt=ln([A]₀/[A])=ln(0.5/0.125)=ln(4)=1.386. t=1.386/0.2=6.93 minutes. Equivalently: 0.125/0.5=0.25=(1/2)², so 2 half-lives → t=2×(0.693/0.2)=6.93 min ✓
6. Determine reaction order from a graph
Plot [A] vs t → curved exponential. Plot ln[A] vs t → straight line with R²=0.9998. Conclusion: first order reaction. From slope of ln[A] plot: slope=−k=−0.050, so k=0.050 s⁻¹. Half-life=0.693/0.050=13.86 s.
Plot [A] vs t → curved exponential. Plot ln[A] vs t → straight line with R²=0.9998. Conclusion: first order reaction. From slope of ln[A] plot: slope=−k=−0.050, so k=0.050 s⁻¹. Half-life=0.693/0.050=13.86 s.
7. First Order Half-Life from t½=0.693/k
k=0.05 hour⁻¹. t½=0.693/0.05=13.86 hours. This is identical regardless of whether [A]₀ is 0.001M or 1000M. After 10 half-lives: (1/2)^10=0.00098=0.098% remains. Time=10×13.86=138.6 hours=5.77 days.
k=0.05 hour⁻¹. t½=0.693/0.05=13.86 hours. This is identical regardless of whether [A]₀ is 0.001M or 1000M. After 10 half-lives: (1/2)^10=0.00098=0.098% remains. Time=10×13.86=138.6 hours=5.77 days.
8. Exponential form — [A] = [A]₀ × e^(−kt)
k=0.693 hour⁻¹, [A]₀=2.0M, t=2 hours. [A]=2.0×e^(−0.693×2)=2.0×e^(−1.386)=2.0×0.250=0.500M. Check: t½=0.693/0.693=1.0 hour. After 2 hours=2 half-lives: 2.0×(0.5)²=2.0×0.25=0.500M ✓
k=0.693 hour⁻¹, [A]₀=2.0M, t=2 hours. [A]=2.0×e^(−0.693×2)=2.0×e^(−1.386)=2.0×0.250=0.500M. Check: t½=0.693/0.693=1.0 hour. After 2 hours=2 half-lives: 2.0×(0.5)²=2.0×0.25=0.500M ✓
9. How many half-lives to reach X%?
Target: reach 5% of initial. n=ln(0.05)/ln(0.5)=−2.996/−0.693=4.32 half-lives. For k=0.1 s⁻¹: t½=6.93s, time=4.32×6.93=29.9 seconds. Formula: n=ln(fraction remaining)/ln(0.5)=log₂(1/fraction).
Target: reach 5% of initial. n=ln(0.05)/ln(0.5)=−2.996/−0.693=4.32 half-lives. For k=0.1 s⁻¹: t½=6.93s, time=4.32×6.93=29.9 seconds. Formula: n=ln(fraction remaining)/ln(0.5)=log₂(1/fraction).
10. Find k from half-life
First order: t½=5,730 years (Carbon-14). k=0.693/t½=0.693/5730=1.209×10⁻⁴ year⁻¹. In seconds: k=1.209×10⁻⁴/(3.156×10⁷)=3.83×10⁻¹² s⁻¹. This allows calculation of C-14 remaining in any archaeological sample.
First order: t½=5,730 years (Carbon-14). k=0.693/t½=0.693/5730=1.209×10⁻⁴ year⁻¹. In seconds: k=1.209×10⁻⁴/(3.156×10⁷)=3.83×10⁻¹² s⁻¹. This allows calculation of C-14 remaining in any archaeological sample.
Frequently Asked Questions
What is the difference between zero order and first order reactions?
In a zero order reaction, rate = k — the rate is constant and independent of concentration. The integrated rate law is [A] = [A]₀ − kt (linear). In a first order reaction, rate = k[A] — the rate is proportional to concentration. The integrated rate law is ln[A] = ln[A]₀ − kt or [A] = [A]₀e^(−kt) (exponential decay). The half-life for zero order depends on [A]₀; for first order it is constant.
What is the half-life formula for a zero order reaction?
The zero order half-life formula is t½ = [A]₀/(2k). Unlike first order, this DEPENDS on the initial concentration. If [A]₀ = 1.0 M and k = 0.05 M/s, then t½ = 1.0/(2×0.05) = 10 seconds. If [A]₀ doubles to 2.0 M, t½ also doubles to 20 seconds. The half-life formula t½ = [A]₀/(2k) shows this directly.
What is the half-life formula for a first order reaction?
The first order half-life formula is t½ = 0.693/k (or exactly ln(2)/k = 0.6931/k). For k = 0.05 s⁻¹, t½ = 0.693/0.05 = 13.86 seconds — regardless of whether [A]₀ is 0.001 M or 100 M. This constant half-life is the defining characteristic of first order kinetics. The formula contains no [A]₀ term, proving independence from initial concentration.
Why is the half-life constant for first order reactions?
Because the rate is proportional to concentration: rate = k[A]. As concentration decreases, rate decreases proportionally, so the time to halve any amount is always the same fraction of the current amount. Mathematically, integrating rate = k[A] gives ln[A] = ln[A]₀ − kt. Setting [A] = [A]₀/2 and solving for t gives t½ = ln(2)/k = 0.693/k — no [A]₀ in the result.
How do you determine reaction order from experimental data?
The graphical method: (1) Plot [A] vs time — if straight line, zero order. (2) Plot ln[A] vs time — if straight line, first order. (3) Plot 1/[A] vs time — if straight line, second order. The plot giving R² closest to 1.00 identifies the reaction order. The slope of the straight line gives k (negative slope for zero and first order; positive for second order). Use the determination tool above to run this analysis automatically.
What is the integrated rate law equation?
The integrated rate law depends on reaction order. Zero order: [A] = [A]₀ − kt. First order: ln[A] = ln[A]₀ − kt (equivalently [A] = [A]₀e^(−kt)). Second order: 1/[A] = 1/[A]₀ + kt. These equations are derived by integrating the differential rate law and relate concentration directly to time, allowing calculation of [A] at any point during the reaction.
Why does zero order reaction half-life depend on initial concentration?
Because zero order reactions consume reactant at a constant rate k (M/s). Halving a larger starting amount requires consuming more moles of reactant — but the consumption rate is the same. From [A] = [A]₀ − kt: setting [A] = [A]₀/2 gives [A]₀/2 = [A]₀ − kt½, so t½ = [A]₀/(2k). The [A]₀ in the numerator confirms the direct dependence on initial concentration.
What are real examples of zero order reactions?
Zero order (zeroth order) reactions include: photochemical reactions where light intensity (not reactant concentration) limits rate; enzyme-catalyzed reactions when the enzyme is saturated with substrate; surface-catalyzed reactions where the solid catalyst surface is fully covered; and certain drug metabolism processes (e.g., alcohol metabolism by alcohol dehydrogenase at high concentrations).
What are real examples of first order reactions?
First order reactions are extremely common: radioactive decay (carbon-14 t½ = 5,730 years; iodine-131 t½ = 8 days), drug elimination from blood (most pharmaceuticals follow first order pharmacokinetics), decomposition of hydrogen peroxide, ozone decomposition in the atmosphere, N₂O₅ decomposition, and many organic elimination reactions. The constant half-life is the experimental signature.
How do you solve the first order integrated rate law for time?
Start with ln[A] = ln[A]₀ − kt. Rearrange: kt = ln[A]₀ − ln[A] = ln([A]₀/[A]). Therefore t = ln([A]₀/[A])/k. Example: [A]₀=1.0M, [A]=0.25M, k=0.1 s⁻¹: t = ln(1.0/0.25)/0.1 = ln(4)/0.1 = 1.386/0.1 = 13.86 seconds. Note that 0.25/1.0 = 0.25 = (1/2)², so this is exactly 2 half-lives × 6.93 s = 13.86 s ✓