Integrated Rate Equations
Real-Life Connection: Carbon Dating and Archaeological Discoveries
How do scientists determine that a fossil is 50,000 years old? Or that a mummy is 3,000 years old? The answer lies in integrated rate equations!
Carbon-14 dating uses the first-order integrated rate equation:
$$\ln\frac{N_0}{N_t} = kt$$By measuring the ratio of C-14 to C-12 in ancient samples:
- Scientists calculate time elapsed since death
- Archaeologists date historical artifacts
- Geologists determine rock ages
- Forensic experts establish time of death
From Shroud of Turin to Egyptian pyramids, integrated rate equations help us peek into the past!
From Differential to Integrated Form
Why Integration?
Differential rate law: Tells us rate at any instant
$$-\frac{d[A]}{dt} = k[A]^n$$Integrated rate law: Tells us concentration at any time
$$[A]_t = f(t, k, [A]_0)$$Integration converts rate (derivative) into concentration (function of time).
Zero Order Reactions
Differential Form
$$\boxed{-\frac{d[A]}{dt} = k}$$Rate is independent of concentration.
Derivation of Integrated Form
$$-\frac{d[A]}{dt} = k$$Separating variables:
$$-d[A] = k \cdot dt$$Integrating both sides:
$$-\int_{[A]_0}^{[A]_t} d[A] = k\int_0^t dt$$ $$-([A]_t - [A]_0) = kt$$ $$\boxed{[A]_t = [A]_0 - kt}$$This is equation of a straight line: y = mx + c
- y = [A]ₜ
- x = t
- m = -k (slope)
- c = [A]₀ (intercept)
Graphical Representation
Plot [A] vs t:
- Straight line with negative slope
- Slope = -k
- y-intercept = [A]₀
[A]
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[A]₀|___
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0 t
Key Characteristics
- Linear decrease in concentration with time
- Constant rate throughout reaction
- Half-life depends on [A]₀: $$t_{1/2} = \frac{[A]_0}{2k}$$
- Units of k: mol L⁻¹ s⁻¹ or M s⁻¹
Examples
Photochemical reactions:
$$H_2 + Cl_2 \xrightarrow{hv} 2HCl$$Rate limited by light intensity, not concentration.
Enzyme-catalyzed at saturation: All enzyme active sites occupied; adding more substrate doesn’t increase rate.
Surface catalysis:
$$2NH_3 \xrightarrow{Pt} N_2 + 3H_2$$When catalyst surface fully covered.
First Order Reactions
Differential Form
$$\boxed{-\frac{d[A]}{dt} = k[A]}$$Rate directly proportional to concentration.
Derivation of Integrated Form
$$-\frac{d[A]}{dt} = k[A]$$Separating variables:
$$-\frac{d[A]}{[A]} = k \cdot dt$$Integrating both sides:
$$-\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = k\int_0^t dt$$ $$-[\ln[A]]_{[A]_0}^{[A]_t} = kt$$ $$-(\ln[A]_t - \ln[A]_0) = kt$$ $$\ln[A]_0 - \ln[A]_t = kt$$ $$\boxed{\ln\frac{[A]_0}{[A]_t} = kt}$$Alternative forms:
$$\boxed{\ln[A]_t = \ln[A]_0 - kt}$$(straight line form)
$$\boxed{[A]_t = [A]_0 e^{-kt}}$$(exponential form)
$$\boxed{\log\frac{[A]_0}{[A]_t} = \frac{kt}{2.303}}$$(common log form)
Graphical Representation
Plot ln[A] vs t:
- Straight line with negative slope
- Slope = -k
- y-intercept = ln[A]₀
Interactive Demo: Watch Concentration Change Over Time
Visualize how reactant concentration decreases over time for different reaction orders. Compare zero, first, and second order kinetics.
ln[A]
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ln[A]₀|___
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0 t
Plot [A] vs t:
- Exponential decay curve
- Never reaches zero (asymptotic to time axis)
Key Characteristics
- Exponential decay of concentration
- Half-life independent of [A]₀: $$t_{1/2} = \frac{0.693}{k} = \frac{\ln 2}{k}$$
- Units of k: s⁻¹ (time⁻¹)
- Most common order in nature
Examples
Radioactive decay:
$$^{14}C \rightarrow ^{14}N + \beta^-$$ $$^{238}U \rightarrow ^{234}Th + ^4He$$N₂O₅ decomposition:
$$2N_2O_5 \rightarrow 4NO_2 + O_2$$ $$\text{Rate} = k[N_2O_5]$$Drug elimination: Most pharmaceuticals follow first-order kinetics in the body.
Acid-catalyzed ester hydrolysis (water in excess):
$$RCOOR' + H_2O \xrightarrow{H^+} RCOOH + R'OH$$
Second Order Reactions
Type I: Rate = k[A]²
Differential Form
$$\boxed{-\frac{d[A]}{dt} = k[A]^2}$$Derivation of Integrated Form
$$-\frac{d[A]}{dt} = k[A]^2$$Separating variables:
$$-\frac{d[A]}{[A]^2} = k \cdot dt$$Integrating:
$$-\int_{[A]_0}^{[A]_t} [A]^{-2} d[A] = k\int_0^t dt$$ $$-\left[\frac{[A]^{-1}}{-1}\right]_{[A]_0}^{[A]_t} = kt$$ $$\left[\frac{1}{[A]}\right]_{[A]_0}^{[A]_t} = kt$$ $$\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt$$ $$\boxed{\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt}$$Graphical Representation
Plot 1/[A] vs t:
- Straight line with positive slope
- Slope = +k
- y-intercept = 1/[A]₀
1/[A]
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0 t
Key Characteristics
- Hyperbolic decrease in concentration
- Half-life depends on [A]₀: $$t_{1/2} = \frac{1}{k[A]_0}$$
- Units of k: L mol⁻¹ s⁻¹ or M⁻¹ s⁻¹
- Each successive half-life is twice the previous one
Examples
NO₂ decomposition:
$$2NO_2 \rightarrow 2NO + O_2$$ $$\text{Rate} = k[NO_2]^2$$Gaseous dimerization:
$$2C_4H_6 \rightarrow C_8H_{12}$$
Type II: Rate = k[A][B]
For
$$A + B \rightarrow \text{Products}$$Case 1: Equal Initial Concentrations [A]₀ = [B]₀
Reduces to Type I:
$$\boxed{\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt}$$Case 2: Different Initial Concentrations [A]₀ ≠ [B]₀
$$\boxed{\frac{1}{[A]_0 - [B]_0} \ln\frac{[B]_0[A]_t}{[A]_0[B]_t} = kt}$$More complex - rarely asked in JEE, but important for research.
Example
Alkaline hydrolysis of ethyl acetate:
$$CH_3COOC_2H_5 + NaOH \rightarrow CH_3COONa + C_2H_5OH$$ $$\text{Rate} = k[CH_3COOC_2H_5][NaOH]$$Summary Table: Integrated Rate Equations
| Order | Differential | Integrated | Linear Plot | Slope | Half-life | k Units |
|---|---|---|---|---|---|---|
| 0 | -d[A]/dt = k | [A]ₜ = [A]₀ - kt | [A] vs t | -k | [A]₀/2k | M s⁻¹ |
| 1 | -d[A]/dt = k[A] | ln[A]ₜ = ln[A]₀ - kt | ln[A] vs t | -k | 0.693/k | s⁻¹ |
| 2 | -d[A]/dt = k[A]² | 1/[A]ₜ = 1/[A]₀ + kt | 1/[A] vs t | +k | 1/(k[A]₀) | M⁻¹ s⁻¹ |
Determining Order from Concentration-Time Data
Method 1: Graphical Method
Plot all three graphs:
- [A] vs t
- ln[A] vs t
- 1/[A] vs t
Whichever is linear indicates the order!
Method 2: Half-Life Method
Measure multiple half-lives:
- Constant t₁/₂ → First order
- t₁/₂ increases linearly → Zero order
- t₁/₂ doubles each time → Second order
Method 3: Substitution Method
Try each integrated equation:
- Calculate k at different times
- If k is constant → correct order
- If k varies → wrong order
Memory Tricks
“LIS” for Linear Plots
Log plot for first order (ln[A] vs t) Inverse plot for second order (1/[A] vs t) Straight plot for zero order ([A] vs t)
“0-1-2 Pattern”
- 0 order: [A] to power 0 in plot = [A]¹
- 1 order: [A] to power 1 in log = ln[A]
- 2 order: [A] to power -1 (inverse) = 1/[A]
“HIL” for Half-life
Half of initial (zero order): t₁/₂ ∝ [A]₀ Independent (first order): t₁/₂ constant Less when more (second order): t₁/₂ ∝ 1/[A]₀
Common JEE Mistakes
Mistake 1: Wrong Plot for Order Determination
Wrong: Plotting only [A] vs t and concluding it’s not zero order because curve isn’t perfectly linear
Correct: Plot all three graphs ([A] vs t, ln[A] vs t, 1/[A] vs t) and identify which is most linear
Mistake 2: Sign of Slope
Wrong: “First order has positive slope in ln[A] vs t plot”
Correct: Slope is -k (negative) in ln[A] vs t plot for first order
Mistake 3: Half-Life Formulas
Wrong: Using t₁/₂ = 0.693/k for all reactions
Correct: This formula is only for first order!
- Zero order: t₁/₂ = [A]₀/(2k)
- Second order: t₁/₂ = 1/(k[A]₀)
Mistake 4: Units
Wrong: Treating k from different orders as comparable
Correct: k has different units for different orders - cannot compare directly!
Practice Problems
Level 1: JEE Main Foundation
Problem 1: For a first order reaction, initial concentration is 0.1 M. After 10 minutes, it becomes 0.05 M. Calculate the rate constant.
Solution:
First order:
$$\ln\frac{[A]_0}{[A]_t} = kt$$ $$\ln\frac{0.1}{0.05} = k \times 10$$ $$\ln 2 = 10k$$ $$k = \frac{0.693}{10} = 0.0693 \text{ min}^{-1}$$Alternative: Since concentration halved, t = t₁/₂ = 10 min
$$t_{1/2} = \frac{0.693}{k}$$ $$k = \frac{0.693}{10} = 0.0693 \text{ min}^{-1}$$Level 2: JEE Main/Advanced
Problem 2: A reaction A → Products follows second order kinetics. Initial concentration is 0.2 M and after 50 minutes it reduces to 0.05 M. Calculate: (a) Rate constant (b) Half-life at start (c) Time for 90% completion
Solution:
(a) Second order:
$$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$$ $$\frac{1}{0.05} = \frac{1}{0.2} + k \times 50$$ $$20 = 5 + 50k$$ $$k = \frac{15}{50} = 0.3 \text{ M}^{-1}\text{min}^{-1}$$(b) Half-life at start:
$$t_{1/2} = \frac{1}{k[A]_0} = \frac{1}{0.3 \times 0.2} = \frac{1}{0.06} = 16.67 \text{ min}$$(c) For 90% completion: [A]ₜ = 0.1[A]₀ = 0.02 M
$$\frac{1}{0.02} = \frac{1}{0.2} + 0.3 \times t$$ $$50 = 5 + 0.3t$$ $$t = \frac{45}{0.3} = 150 \text{ min}$$Level 3: JEE Advanced
Problem 3: The following data is obtained for decomposition of A:
| Time (s) | [A] (M) | ln[A] | 1/[A] (M⁻¹) |
|---|---|---|---|
| 0 | 0.100 | -2.303 | 10.0 |
| 100 | 0.067 | -2.708 | 14.9 |
| 200 | 0.050 | -2.996 | 20.0 |
| 300 | 0.040 | -3.219 | 25.0 |
Determine: (a) Order of reaction (b) Rate constant (c) Half-life
Solution:
(a) Check for linear relationship:
For zero order ([A] vs t):
- Slope between 0-100s: (0.067-0.100)/100 = -0.00033
- Slope between 200-300s: (0.040-0.050)/100 = -0.0001 Not constant → Not zero order
For first order (ln[A] vs t):
- Slope between 0-100s: (-2.708-(-2.303))/100 = -0.00405
- Slope between 200-300s: (-3.219-(-2.996))/100 = -0.00223 Not constant → Not first order
For second order (1/[A] vs t):
- Slope between 0-100s: (14.9-10.0)/100 = 0.049
- Slope between 200-300s: (25.0-20.0)/100 = 0.050 Nearly constant → Second order!
(b) From graph, average slope ≈ 0.05 M⁻¹s⁻¹
$$k = 0.05 \text{ M}^{-1}\text{s}^{-1}$$Verification:
$$\frac{1}{0.050} - \frac{1}{0.100} = k \times 200$$ $$20 - 10 = 200k$$ $$k = 0.05 \text{ M}^{-1}\text{s}^{-1}$$✓
(c) Half-life at [A]₀ = 0.100 M:
$$t_{1/2} = \frac{1}{k[A]_0} = \frac{1}{0.05 \times 0.100} = 200 \text{ s}$$Problem 4: ¹⁴C has half-life of 5730 years. A wooden artifact has 62.5% of original ¹⁴C. How old is it?
Solution:
First order radioactive decay:
$$\ln\frac{N_0}{N_t} = kt$$First, find k from half-life:
$$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} = 1.21 \times 10^{-4} \text{ year}^{-1}$$Given: Nₜ = 0.625N₀
$$\ln\frac{N_0}{0.625N_0} = kt$$ $$\ln\frac{1}{0.625} = 1.21 \times 10^{-4} \times t$$ $$\ln 1.6 = 1.21 \times 10^{-4} \times t$$ $$0.470 = 1.21 \times 10^{-4} \times t$$ $$t = \frac{0.470}{1.21 \times 10^{-4}} = 3884 \text{ years}$$Alternative approach using half-lives:
- 100% → 50% (1 half-life)
- 50% → 25% (2 half-lives)
- But we have 62.5% (between 100% and 50%)
Age ≈ 3885 years
Special Cases and Applications
Pseudo First Order Reactions
When one reactant is in large excess:
$$A + B \rightarrow \text{Products}$$True rate law:
$$\text{Rate} = k[A][B]$$If [B] » [A], then [B] ≈ constant
$$\text{Rate} = k[A][B] = k'[A]$$where k’ = k[B] = pseudo first order rate constant
Integrated form:
$$\ln[A]_t = \ln[A]_0 - k't$$Example: Acid hydrolysis of ester (water in excess)
Consecutive Reactions
$$A \xrightarrow{k_1} B \xrightarrow{k_2} C$$- A decreases exponentially
- B first increases, then decreases
- C increases continuously
Applications: Multi-step drug metabolism, radioactive decay series
Parallel Reactions
$$A \xrightarrow{k_1} B$$ $$A \xrightarrow{k_2} C$$ $$[A]_t = [A]_0 e^{-(k_1+k_2)t}$$Effective rate constant = k₁ + k₂
Experimental Determination
For Gaseous Reactions
Use partial pressure instead of concentration:
$$P_A = [A]RT$$For first order:
$$\ln P_t = \ln P_0 - kt$$Example: Decomposition of N₂O₅
$$2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$$Monitor total pressure change over time.
For Reactions in Solution
Methods:
- Titration - withdraw samples and titrate
- Spectroscopy - measure absorbance (colored species)
- Conductivity - for ionic reactions
- pH - for acid/base reactions
- Gas evolution - collect and measure gas volume
Connection to Other Topics
Link to Rate Law
- Integrated form derived from differential rate law
- Different orders → different integrated equations
- See: Rate Law
Link to Half-Life
- Half-life formulas derived from integrated equations
- See: Half-Life
Link to Graphical Methods
- Linear plots identify reaction order
- Slope gives rate constant k
- See: Order and Molecularity
Link to Arrhenius Equation
- Temperature dependence of k
- k values used in Arrhenius plots
- See: Arrhenius Equation
JEE Previous Year Questions
JEE Main 2021: For a first order reaction, time required for 99% completion is ‘x’ times the time required for 90% completion. The value of ‘x’ is:
Solution:
For first order:
$$t = \frac{2.303}{k}\log\frac{[A]_0}{[A]_t}$$For 90% completion: [A]ₜ = 0.1[A]₀
$$t_{90} = \frac{2.303}{k}\log\frac{[A]_0}{0.1[A]_0} = \frac{2.303}{k}\log 10 = \frac{2.303}{k}$$For 99% completion: [A]ₜ = 0.01[A]₀
$$t_{99} = \frac{2.303}{k}\log\frac{[A]_0}{0.01[A]_0} = \frac{2.303}{k}\log 100 = \frac{2 \times 2.303}{k}$$ $$x = \frac{t_{99}}{t_{90}} = \frac{2 \times 2.303/k}{2.303/k} = 2$$Answer: x = 2
JEE Advanced 2020: A substance decomposes following first order kinetics. If 50% is decomposed in 120 minutes, what % will decompose in 240 minutes?
Solution:
t₁/₂ = 120 min
After 240 min = 2 half-lives:
- After 1st half-life: 50% remains
- After 2nd half-life: 25% remains
Decomposed = 100% - 25% = 75%
Quick Revision Points
- Zero order: [A] = [A]₀ - kt (linear decrease)
- First order: ln[A] = ln[A]₀ - kt (exponential decay)
- Second order: 1/[A] = 1/[A]₀ + kt (hyperbolic decrease)
- Linear plots identify order - whichever is linear
- Half-life formulas differ for each order
- Units of k differ based on order
- First order most common in nature (radioactive decay, drug elimination)
- Graphical method most reliable for determining order
Summary
Integrated rate equations are essential tools in chemical kinetics that:
- Connect concentration to time
- Allow prediction of reaction progress
- Enable determination of reaction order
- Provide basis for half-life calculations
- Have widespread applications from archaeology to pharmacology
Mastering these equations and their graphical representations is crucial for JEE success and understanding real-world kinetic processes!