Rate law — order vs molecularity — Chemistry notes

1. Rate Law and Order

Derivation from first principles

WHY do we write it this way?
Because in the lab, we change [A] and [B] systematically and measure how Rate responds. If doubling [A] doubles the rate (keeping [B] fixed), then Rate ∝ [A]¹. If doubling [B] quadruples the rate, then Rate ∝ [B]². The exponents $m, n$ are pure empirical fits to data, not guesses from the balanced equation.

HOW to determine order experimentally:

Initial rates method: Run multiple trials with different starting concentrations, measure initial rate each time.
Compare trials where only one concentration changes: $\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[\text{A}]_2^m[\text{B}]_2^n}{k[\text{A}]_1^m[\text{B}]_1^n}$ If [B] is constant: $\frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[\text{A}]_2}{[\text{A}]_1}\right)^m$
Solve for $m$ using logarithms: $m = \frac{\ln(\text{Rate}_2/\text{Rate}_1)}{\ln([\text{A}]_2/[\text{A}]_1)}$

WHY this works: We're isolating the effect of one reactant at a time, holding others constant. The exponent that makes the ratio match is the order.

WHY these graphs? Integrate the rate equation to get A. For first order: $\frac{d[\text{A}]}{dt} = -k[\text{A}] \implies \int_{[\text{A}]_0}^{[\text{A}]} \frac{d[\text{A}]}{[\text{A}]} = -k\int_0^t dt \implies \ln[\text{A}] - \ln[\text{A}]_0 = -kt$ So $\ln[\text{A}] = \ln[\text{A}]_0 - kt$ (linear in $t$ ).

Figure — Rate law — order vs molecularity

Worked example Determining order from data

Given: Reaction A + B → products

Trial	[A]₀ (M)	[B]₀ (M)	Initial Rate (M/s)
1	0.10	0.10	0.020
2	0.20	0.10	0.080
3	0.10	0.20	0.020

Find: Order w.r.t. A, B, and overall order.

Solution:
Compare trials 1 and 2 ([B] constant): $\frac{0.080}{0.020} = \left(\frac{0.20}{0.10}\right)^m \implies 4 = 2^m \implies m = 2$ Why this step? Doubling [A] quadrupled the rate, so Rate ∝ [A]².

Compare trials 1 and 3 ([A] constant): $\frac{0.020}{0.020} = \left(\frac{0.20}{0.10}\right)^n \implies 1 = 2^n \implies n = 0$ Why this step? Doubling [B] had zero effect, so the rate doesn't depend on [B].

Rate law: Rate = $k[\text{A}]^2[\text{B}]^0 = k[\text{A}]^2$
Overall order: 2 + 0 = 2

Calculate $k$ from trial 1: $0.020 = k(0.10)^2 \implies k = \frac{0.020}{0.01} = 2.0 \, \text{M}^{-1}\text{s}^{-1}$

2. Molecularity

WHY rare beyond 3? Probability of simultaneous collision drops exponentially. Even termolecular steps are uncomon; reactions proceed through sequences of uni- and bimolecular steps instead.

For elementary steps ONLY

CRITICAL: This is NOT true for overall multi-step reactions. The stoichiometric equation $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$ does not imply Rate = $k[\text{H}_2]^2[\text{O}_2]$ because the mechanism has many elementary steps.

Worked example Elementary vs overall reaction

Reaction: $2\text{NO}_2 + \text{F}_2 \to 2\text{NO}_2\text{F}$

Case 1: Elementary (one step)
If this happens in a single termolecular collision (molecularity = 3):
Rate = $k[\text{NO}_2]^2[\text{F}_2]$
Order w.r.t. NO₂ = 2, F₂ = 1, overall = 3. Here, order = molecularity (both = 3).

Case 2: Multi-step mechanism (actual)
Experiments show the real mechanism is:

$\text{NO}_2 + \text{F}_2 \to \text{NO}_2\text{F} + \text{F}$ (slow, bimolecular)
$\text{NO}_2 + \text{F} \to \text{NO}_2\text{F}$ (fast, bimolecular)

The slow step determines the overall rate:
Rate = $k_1[\text{NO}_2][\text{F}_2]$
Order w.r.t. NO₂ = 1, F₂ = 1, overall = 2.
Molecularity of step 1 = 2, step 2 = 2.
Overall order (2) ≠ sum of stoichiometric coefficients (3). The molecularity of individual steps (2 each) is not the overall order.

Why this step? The slow step is the bottleneck. Fast step 2 consumes F as soon as it forms, so [F] doesn't build up and doesn't appear in the rate law.

3. Order vs Molecularity — Key Differences

Property	Order	Molecularity
Source	Experimental (rate law from lab data)	Theoretical (from proposed mechanism)
Applies to	Elementary steps AND overall reactions	Only elementary steps
Values	0, 1, 2, or even fractional/negative (rare)	1, 2, or 3 (positive integers only)
Depends on	Which reactants appear in rate law, their exponents	How many molecules collide in one step
Relation to stoichiometry	NO relation for overall reactions	Equals stoichiometric coefficients for elementary steps

Common mistake Assigning molecularity to overall reactions

Wrong idea: "The reaction $\text{H}_2 + \text{I}_2 \to 2\text{HI}$ has molecularity 2."
Why it feels right: Two molecules appear as reactants.
Fix: Molecularity is only defined for elementary steps. The overall reaction proceeds through a multi-step mechanism (e.g., I₂ dissociation, then H₂ + I → HI + H, then H + I₂ → HI + I). Each elementary step has its own molecularity (1 for dissociation, 2 for each bimolecular step), but the overall reaction has no molecularity, only an overall order (which is experimentally 2 for this reaction, matching the rate law Rate = $k[\text{H}_2][\text{I}_2]$ from the slow bimolecular step in modern understanding, but that's coincidence for this mechanism, not a rule).

4. Connecting order and molecularity via mechanism

HOW they relate:

Write the proposed mechanism (sequence of elementary steps).
Each elementary step's molecularity tells you its rate law (exponents = coefficients).
Identify the rate-determining step (RDS, slowest).
The RDS rate law (possibly with fast-equilibrium pre-steps substituted) becomes the overall rate law.
The exponents in the overall rate law are the orders.

WHY this works: The slowest step is the bottleneck. Faster steps before it reach equilibrium (forward/reverse rates equal), and faster steps after it clear intermediates instantly. So the overall rate ≈ rate of RDS.

Worked example Mechanism with intermediate

Reaction: $2\text{NO} + \text{O}_2 \to 2\text{NO}_2$
Mechanism:

$2\text{NO} \rightleftharpoons \text{N}_2\text{O}_2$ (fast equilibrium, $K_{\text{eq}} = k_1/k_{-1}$ )
$\text{N}_2\text{O}_2 + \text{O}_2 \to 2\text{NO}_2$ (slow, bimolecular)

Step-by-step derivation:
Rate of overall reaction = rate of step 2 (RDS): $\text{Rate} = k_2[\text{N}_2\text{O}_2][\text{O}_2]$ Why? Step 2 is slow, so its rate controls the overall rate.

But N₂O₂ is an intermediate (not measurable). Use equilibrium from step 1: $K_{\text{eq}} = \frac{[\text{N}_2\text{O}_2]}{[\text{NO}]^2} \implies [\text{N}_2\text{O}_2] = K_{\text{eq}}[\text{NO}]^2$ Why? Fast equilibrium means $k_1[\text{NO}]^2 = k_{-1}[\text{N}_2\text{O}_2]$ , so $[\text{N}_2\text{O}_2] = \frac{k_1}{k_{-1}}[\text{NO}]^2 = K_{\text{eq}}[\text{NO}]^2$ .

Substitute into rate equation: $\text{Rate} = k_2 K_{\text{eq}}[\text{NO}]^2[\text{O}_2] = k_{\text{obs}}[\text{NO}]^2[\text{O}_2]$ where $k_{\text{obs}} = k_2 K_{\text{eq}}$ .

Result:
Order w.r.t. NO = 2, O₂ = 1, overall = 3.
Molecularity of step 1 (forward) = 2, step 2 = 2.
The overall order (3) equals the sum of coefficients in the balanced equation BY COINCIDENCE because this specific mechanism's RDS involves all stoichiometric species. Don't generalize this!

Recall Explain to a 12-year-old

Imagine you're making sandwiches

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, yahan pe sabse important baat samajhne wali yeh hai ki order aur molecularity do bilkul alag cheezein hain, chahe naam sunne mein similar lagte hon. Order ek experimental quantity hai — matlab lab mein hum concentration ko change karte hain aur dekhte hain ki reaction ki speed kaise badalti hai. Jaise agar [A] double karne se rate double ho gayi, toh order w.r.t. A is 1; agar rate 4 guna ho gayi, toh order 2 hai. Yeh exponents ( $m, n$ ) balanced equation ke coefficients se nahi aate — yeh sirf aur sirf data se milte hain. Isliye rate law hamesha experiment se determine hota hai, guess se nahi.

Dusri taraf, molecularity ek theoretical concept hai jo reaction ke mechanism ke slowest step (elementary step) mein kitne molecules physically collide kar rahe hain, usko batata hai. Yeh hamesha ek whole number hota hai (1, 2, ya 3), aur mechanism se aata hai, na ki experiment se. Isiliye complex reactions mein overall order ka molecularity ya stoichiometry se koi seedha relation nahi hota — yeh point exam mein bahut confuse karta hai, isliye clear rakhna zaroori hai.

Yeh concepts kyun matter karte hain? Kyunki agar tumhe kisi reaction ki speed control karni hai — chahe industry mein product banana ho ya medicine ke degradation ko slow karna ho — toh tumhe pata hona chahiye ki kaunse reactant ka concentration badhane se kya asar padega. "Initial rates method" isi kaam aata hai: ek-ek karke sirf ek concentration change karo, baaki constant rakho, aur logarithm use karke order nikaalo. Iske baad graphs (jaise first order ke liye $\ln[A]$ vs time straight line dega) se tum easily identify kar sakte ho ki reaction kaunse order ka hai. Yeh practical skill hai jo kinetics ke poore chapter ki foundation banati hai.

Rate law — order vs molecularity