Rate Law and Rate Constant

Real-Life Connection: Antibiotic Dosage Timing

Why does your doctor prescribe antibiotics “every 6 hours” or “every 12 hours”? The answer lies in rate laws and rate constants!

When you take a pill:

• The drug enters your bloodstream at a certain rate
• Your body eliminates it at a specific rate (characterized by rate constant k)
• The rate law describes how elimination rate depends on drug concentration
• The rate constant determines how quickly the drug is cleared

A drug with:

• Large k (fast elimination) → needs frequent dosing (every 6 hrs)
• Small k (slow elimination) → longer intervals (every 24 hrs)

Understanding rate laws helps pharmaceutical scientists design optimal drug delivery systems!

What is a Rate Law?

The rate law (or rate equation) expresses the relationship between reaction rate and concentrations of reactants.

For a general reaction:

$$aA + bB \rightarrow \text{Products}$$ $$\boxed{\text{Rate} = k[A]^m[B]^n}$$

where:

• k = rate constant
• m = order with respect to A
• n = order with respect to B
• [A], [B] = molar concentrations

Critical Points

1. Rate law is determined experimentally (not from balanced equation!)
2. The exponents m and n are not necessarily the stoichiometric coefficients
3. k is specific to a particular reaction at a given temperature
4. Rate law reveals the mechanism of the reaction

The Rate Constant (k)

Definition

The rate constant (k) is the proportionality constant in the rate law that is characteristic of a particular reaction at a given temperature.

$$\boxed{k = \text{rate constant} = \text{specific rate constant}}$$

Properties of k

1. Temperature dependent - increases exponentially with temperature (Arrhenius equation)
2. Independent of concentration - constant for a given reaction at fixed temperature
3. Catalyst dependent - changes if catalyst is added
4. Unique for each reaction - different reactions have different k values
5. Has units - units depend on overall order of reaction

Physical Meaning of k

• Large k → Fast reaction (products form quickly)
• Small k → Slow reaction (products form slowly)

Example:

• Explosion: k ~ 10⁶ s⁻¹ (very large, very fast)
• Rusting of iron: k ~ 10⁻⁸ s⁻¹ (very small, very slow)
• Radioactive decay (¹⁴C): k ~ 10⁻¹² s⁻¹ (extremely slow)

Units of Rate Constant

The units of k depend on overall order of the reaction.

General Formula

$$\boxed{\text{Units of } k = (\text{concentration})^{1-n} \times (\text{time})^{-1}}$$

where n = overall order

For Different Orders

Order	Rate Law	Units of k	Common Form
0	Rate = k	mol L⁻¹ s⁻¹	M s⁻¹
1	Rate = k[A]	s⁻¹	s⁻¹, min⁻¹, hr⁻¹
2	Rate = k[A]² or k[A][B]	L mol⁻¹ s⁻¹	M⁻¹ s⁻¹
3	Rate = k[A]²[B]	L² mol⁻² s⁻¹	M⁻² s⁻¹

Derivation of Units

For second order:

$$\text{Rate} = k[A]^2$$ $$\text{M s}^{-1} = k \times \text{M}^2$$ $$k = \frac{\text{M s}^{-1}}{\text{M}^2} = \text{M}^{-1}\text{s}^{-1}$$

Types of Rate Laws

1. Differential Rate Law

Expresses rate as a function of concentration at any instant:

$$\boxed{-\frac{d[A]}{dt} = k[A]^n}$$

Uses:

• Determining order of reaction
• Finding how rate varies with concentration
• Understanding instantaneous behavior

2. Integrated Rate Law

Expresses concentration as a function of time:

$$\boxed{[A]_t = f(t)}$$

Uses:

• Calculating concentration at any time
• Finding time for specific concentration change
• Determining half-life

See detailed discussion: Integrated Rate Equations

Rate Laws for Different Orders

Zero Order Reaction

Differential form:

$$\boxed{\text{Rate} = k}$$

Integrated form:

$$\boxed{[A]_t = [A]_0 - kt}$$

Characteristics:

• Rate is constant (independent of concentration)
• Linear plot: [A] vs t (slope = -k)
• Half-life: $$t_{1/2} = \frac{[A]_0}{2k}$$ (depends on [A]₀)

Example: Photochemical reactions, enzyme-saturated reactions

First Order Reaction

Differential form:

$$\boxed{\text{Rate} = k[A]}$$

Integrated form:

$$\boxed{\ln[A]_t = \ln[A]_0 - kt}$$

or

$$\boxed{[A]_t = [A]_0 e^{-kt}}$$

Characteristics:

• Rate directly proportional to concentration
• Linear plot: ln[A] vs t (slope = -k)
• Half-life: $$t_{1/2} = \frac{0.693}{k}$$ (independent of [A]₀)

Example: Radioactive decay, drug elimination, N₂O₅ decomposition

Second Order Reaction

Differential form:

$$\boxed{\text{Rate} = k[A]^2}$$

(Type I)

or

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

(Type II)

Integrated form (Type I):

$$\boxed{\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt}$$

Characteristics:

• Rate proportional to square of concentration
• Linear plot: 1/[A] vs t (slope = +k)
• Half-life: $$t_{1/2} = \frac{1}{k[A]_0}$$ (inversely proportional to [A]₀)

Example: NO₂ decomposition, gaseous dimerization

Determining Rate Law Experimentally

Method 1: Initial Rate Method

Procedure:

1. Perform multiple experiments with different initial concentrations
2. Measure initial rate in each experiment
3. Compare how rate changes with concentration
4. Determine order and k

Example:

For

$$A + B \rightarrow \text{Products}$$

Exp	[A]₀ (M)	[B]₀ (M)	Initial Rate (M/s)
1	0.10	0.10	1.0 × 10⁻³
2	0.20	0.10	4.0 × 10⁻³
3	0.10	0.20	2.0 × 10⁻³

Step 1: Find order w.r.t. A (compare exp 1 & 2):

$$\frac{r_2}{r_1} = \frac{k(0.20)^m(0.10)^n}{k(0.10)^m(0.10)^n} = \frac{4.0 \times 10^{-3}}{1.0 \times 10^{-3}}$$ $$2^m = 4 \implies m = 2$$

Step 2: Find order w.r.t. B (compare exp 1 & 3):

$$\frac{r_3}{r_1} = \frac{k(0.10)^m(0.20)^n}{k(0.10)^m(0.10)^n} = \frac{2.0 \times 10^{-3}}{1.0 \times 10^{-3}}$$ $$2^n = 2 \implies n = 1$$

Step 3: Write rate law:

$$\boxed{\text{Rate} = k[A]^2[B]}$$

Step 4: Calculate k using any experiment:

$$1.0 \times 10^{-3} = k(0.10)^2(0.10)$$ $$k = \frac{1.0 \times 10^{-3}}{0.001} = 1.0 \text{ M}^{-2}\text{s}^{-1}$$

Method 2: Graphical Method

Plot different graphs to identify linear relationship:

Order	Linear Plot	Equation	Slope
0	[A] vs t	y = -kt + [A]₀	-k
1	ln[A] vs t	y = -kt + ln[A]₀	-k
2	1/[A] vs t	y = kt + 1/[A]₀	+k

Whichever plot is linear indicates the order!

Method 3: Half-Life Method

Measure half-lives at different initial concentrations:

• If t₁/₂ is constant → First order
• If t₁/₂ ∝ [A]₀ → Zero order
• If t₁/₂ ∝ 1/[A]₀ → Second order

Memory Tricks

“RUCK” for Rate Law Components

Rate equals Unique constant k Concentrations raised to powers Kinetically determined (not from stoichiometry)

“ZOT” for Zero Order

Zero dependence on concentration Observed in saturated reactions Time gives linear [A] plot

“FELINE” for First Order

First power of concentration Exponential decay Ln[A] gives linear plot Independent half-life (constant t₁/₂) Nuclear decay example Elimination follows this often

Units Memory

“Order increases, k units get messier”

• 0 order: M/s (simple)
• 1st order: 1/s (simpler!)
• 2nd order: 1/(M·s) (complex)
• 3rd order: 1/(M²·s) (very complex)

Common JEE Mistakes

Mistake 1: Order from Stoichiometry

Wrong: For

$$2NO + O_2 \rightarrow 2NO_2$$

, assuming Rate = k[NO]²[O₂]

Correct: Must determine experimentally! Actual rate law is Rate = k[NO]²[O₂], but this is coincidental - not guaranteed from equation.

Mistake 2: Units of k

Wrong: “k always has units of s⁻¹”

Correct: Units of k depend on overall order:

• Zero order: M s⁻¹
• First order: s⁻¹
• Second order: M⁻¹ s⁻¹

Mistake 3: Temperature and k

Wrong: “k is constant for a reaction”

Correct: k is constant at fixed temperature. k changes significantly with temperature!

Mistake 4: Confusing Differential and Integrated Forms

Wrong: Using [A] = [A]₀ - kt for first order reaction

Correct:

• Zero order: [A] = [A]₀ - kt
• First order: ln[A] = ln[A]₀ - kt

Practice Problems

Level 1: JEE Main Foundation

Problem 1: For a first order reaction, rate constant k = 0.693 hr⁻¹. Calculate the half-life.

Solution:

For first order:

$$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.693} = 1 \text{ hour}$$

Answer: 1 hour

Level 2: JEE Main/Advanced

Problem 2: For the reaction

$$2A + B \rightarrow C$$

, following data is obtained:

[A] (M)	[B] (M)	Rate (M/min)
0.1	0.1	0.004
0.2	0.2	0.032
0.2	0.1	0.016

Determine: (a) Rate law (b) Order of reaction (c) Rate constant with units

Solution:

Let Rate = k[A]^m[B]^n

Finding m: Compare experiments where only [A] changes:

From exp 1 and 3:

$$\frac{0.016}{0.004} = \left(\frac{0.2}{0.1}\right)^m \times \left(\frac{0.1}{0.1}\right)^n$$ $$4 = 2^m \implies m = 2$$

Finding n: From exp 3 and 2:

$$\frac{0.032}{0.016} = \left(\frac{0.2}{0.2}\right)^m \times \left(\frac{0.2}{0.1}\right)^n$$ $$2 = 2^n \implies n = 1$$

(a) Rate law:

$$\text{Rate} = k[A]^2[B]$$

(b) Order = 2 + 1 = 3 (third order)

(c) From experiment 1:

$$0.004 = k(0.1)^2(0.1)$$ $$k = \frac{0.004}{0.001} = 4 \text{ M}^{-2}\text{min}^{-1}$$

Units: L² mol⁻² min⁻¹

Level 3: JEE Advanced

Problem 3: A reaction is second order in A. When [A]₀ = 0.10 M, the half-life is 100 seconds. Calculate: (a) Rate constant (b) Time for [A] to decrease to 0.025 M (c) Half-life when [A]₀ = 0.20 M

Solution:

(a) For second order:

$$t_{1/2} = \frac{1}{k[A]_0}$$ $$100 = \frac{1}{k \times 0.10}$$ $$k = \frac{1}{100 \times 0.10} = 0.1 \text{ M}^{-1}\text{s}^{-1}$$

(b) Second order integrated equation:

$$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$$ $$\frac{1}{0.025} = \frac{1}{0.10} + 0.1 \times t$$ $$40 = 10 + 0.1t$$ $$t = \frac{30}{0.1} = 300 \text{ s}$$

Alternative approach: From 0.10 M → 0.05 M: 100 s (first half-life) From 0.05 M → 0.025 M: Need to calculate new t₁/₂

$$t_{1/2,new} = \frac{1}{k \times 0.05} = \frac{1}{0.1 \times 0.05} = 200 \text{ s}$$

Total time = 100 + 200 = 300 s

(c) For [A]₀ = 0.20 M:

$$t_{1/2} = \frac{1}{k[A]_0} = \frac{1}{0.1 \times 0.20} = 50 \text{ s}$$

Key insight: When [A]₀ doubles, t₁/₂ halves (for second order)!

Problem 4: The decomposition of N₂O₅ follows first order kinetics:

$$2N_2O_5 \rightarrow 4NO_2 + O_2$$

At 300 K, k = 2.5 × 10⁻⁴ s⁻¹

(a) What % of N₂O₅ decomposes in 1 hour? (b) How long for 90% decomposition?

Solution:

(a) First order:

$$\ln\frac{[A]_0}{[A]_t} = kt$$

t = 1 hour = 3600 s

$$\ln\frac{[A]_0}{[A]_t} = 2.5 \times 10^{-4} \times 3600 = 0.90$$ $$\frac{[A]_0}{[A]_t} = e^{0.90} = 2.46$$ $$[A]_t = \frac{[A]_0}{2.46} = 0.407[A]_0$$

Decomposed = [A]₀ - [A]ₜ = [A]₀ - 0.407[A]₀ = 0.593[A]₀

% decomposed = 59.3%

(b) For 90% decomposition: [A]ₜ = 0.10[A]₀

$$\ln\frac{[A]_0}{0.10[A]_0} = kt$$ $$\ln(10) = 2.5 \times 10^{-4} \times t$$ $$t = \frac{2.303}{2.5 \times 10^{-4}} = 9212 \text{ s} = 153.5 \text{ min} \approx 2.56 \text{ hours}$$

Shortcut: For 90% completion (10% remaining):

$$t = \frac{2.303}{k} = \frac{2.303}{2.5 \times 10^{-4}} = 9212 \text{ s}$$

Relationship: Rate Law and Mechanism

The rate law provides clues about reaction mechanism!

Elementary Reactions

For elementary reactions, rate law can be written from stoichiometry:

$$A + B \rightarrow \text{Products}$$

(elementary)

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

✓

Complex Reactions

For complex reactions, rate law differs from stoichiometry:

$$2N_2O_5 \rightarrow 4NO_2 + O_2$$

(overall)

$$\text{Rate} = k[N_2O_5]$$

(NOT k[N₂O₅]²!)

Why? Multiple elementary steps involved.

Rate-Determining Step (RDS)

The slowest step in a mechanism determines the overall rate.

Example:

$$NO_2 + CO \rightarrow NO + CO_2$$

Mechanism:

• Step 1: $$2NO_2 \xrightarrow{slow} NO_3 + NO$$ (RDS)
• Step 2: $$NO_3 + CO \xrightarrow{fast} NO_2 + CO_2$$

Rate = rate of step 1 = k[NO₂]²

This explains why CO doesn’t appear in rate law!

Temperature Dependence of k

Arrhenius Equation

$$\boxed{k = Ae^{-E_a/RT}}$$

Logarithmic form:

$$\boxed{\ln k = \ln A - \frac{E_a}{RT}}$$

Two-temperature form:

$$\boxed{\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)}$$

See detailed discussion: Arrhenius Equation

Effect of Temperature

• Increase in temperature → k increases exponentially
• Typical: k doubles for every 10°C rise (depends on Ea)
• Large Ea → strong temperature dependence
• Small Ea → weak temperature dependence

Connection to Other Topics

Link to Order and Molecularity

• Order appears as exponents in rate law
• For elementary reactions: order = molecularity
• See: Order and Molecularity

Link to Integrated Rate Equations

• Integration of differential rate law gives integrated form
• Different orders → different integrated equations
• See: Integrated Rate Equations

Link to Half-Life

• Half-life derived from integrated rate equations
• Dependence on [A]₀ varies with order
• See: Half-Life

Link to Catalysis

• Catalyst increases k (by lowering Ea)
• Does not change order or form of rate law
• See: Catalysis

Link to Equilibrium

• At equilibrium: k_forward[reactants] = k_backward[products]
• Equilibrium constant K = k_f/k_b
• See: Chemical Equilibrium

JEE Previous Year Questions

JEE Main 2020: The rate constant of a reaction is 1.5 × 10⁻³ s⁻¹. What is the order of the reaction?

Answer: First order

Explanation: Units of k are s⁻¹, which is characteristic of first order reactions.

JEE Advanced 2019: For a reaction A → Products, the following data is obtained:

Time (min)	[A] (M)
0	0.10
10	0.05
20	0.025

What is the order?

Solution:

• At t = 10 min: [A] = 0.05 = 0.10/2 (half of initial)
• At t = 20 min: [A] = 0.025 = 0.05/2 (half of previous)

Half-life is constant (10 min) → First order

Alternatively:

$$t_{1/2} = \frac{0.693}{k} = 10 \text{ min}$$ $$k = 0.0693 \text{ min}^{-1}$$

Comparison Table: Rate Laws

Order	Differential Form	Integrated Form	Linear Plot	Slope	Half-life
0	Rate = k	[A] = [A]₀ - kt	[A] vs t	-k	[A]₀/2k
1	Rate = k[A]	ln[A] = ln[A]₀ - kt	ln[A] vs t	-k	0.693/k
2	Rate = k[A]²	1/[A] = 1/[A]₀ + kt	1/[A] vs t	+k	1/(k[A]₀)

Quick Revision Points

1. Rate law determined experimentally, not from balanced equation
2. k is temperature dependent (increases with T)
3. Units of k depend on overall order
4. Order = sum of exponents in rate law
5. Different orders → different integrated equations
6. Half-life dependence on [A]₀ varies with order
7. RDS (rate-determining step) controls overall rate
8. For elementary reactions: rate law from stoichiometry
9. For complex reactions: rate law from RDS

Summary

Rate laws are the mathematical expressions that connect reaction rates to reactant concentrations. The rate constant k is a fundamental parameter that:

• Characterizes reaction speed
• Depends on temperature (Arrhenius equation)
• Has units determined by reaction order
• Provides insights into reaction mechanisms

Mastering rate laws is essential for:

• Predicting reaction behavior
• Designing industrial processes
• Understanding drug kinetics
• Solving JEE problems effectively

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