Rate of Reaction, Order & Molecularity
Master the speed of chemistry. Decode the stoichiometry of rate expressions, calculate the universal units of 'k', and distinguish Order from Molecularity.
Module Focus: The Dimension of Time
Thermodynamics tells us if a reaction will happen. Chemical Kinetics tells us how fast it will happen. Understanding the rate at which reactants vanish and products appear requires a strict mathematical framework. In NEET, you must master the difference between the experimental Rate Law (Order) and the theoretical collision model (Molecularity).
1. Rate of Reaction & Stoichiometry
The rate of a reaction is the change in concentration of a reactant or product per unit time. Because reactants disappear, we use a negative sign to ensure the rate itself is a positive value.
Consider a general reaction: $\mathbf{aA + bB \rightarrow cC + dD}$
To express a single, universal "Rate of Reaction", we must divide the individual rates of appearance/disappearance by their respective stoichiometric coefficients:
- Rate of Disappearance of A = $-\frac{d[A]}{dt}$ (No stoichiometric coefficient!)
- Rate of Appearance of C = $+\frac{d[C]}{dt}$ (No stoichiometric coefficient!)
- Rate of Reaction = Uses the stoichiometric coefficients ($\frac{1}{a}, \frac{1}{c}$, etc.)
Because the rate of reaction constantly changes as reactants are consumed, we define it in two specific ways for calculations:
The change in concentration over a macroscopic, measurable time interval ($\Delta t$). On a graph, it represents the slope of the secant line connecting two points.
The true, exact rate at a specific moment in time. On a graph, it is exactly equal to the slope of the tangent line drawn at time $t$.
Concept: For any given reaction, as the time interval approaches zero ($\Delta t \to 0$), the Average Rate becomes equal to the Instantaneous Rate.
2. Rate Law and Rate Constant ($k$)
The Rate Law expresses the rate of a reaction in terms of the molar concentrations of reactants, raised to some power. These powers are determined experimentally and may or may not equal the stoichiometric coefficients.
For $aA + bB \rightarrow \text{Products}$:
- $x$ and $y$ are the partial orders.
- Overall Order (n) = $x + y$.
- $k$ is the Rate Constant (Specific Reaction Rate).
- It is independent of concentration.
- It depends strictly on Temperature and the presence of a Catalyst.
- A larger $k$ means a faster reaction.
- Specific Reaction Rate: It is the rate of reaction when the concentration of all reactants is exactly 1 M.
The unit of the rate constant $k$ changes depending on the order of the reaction ($n$). You don't need to memorize each one, just memorize this single master formula:
mol L⁻¹ s⁻¹
s⁻¹
L mol⁻¹ s⁻¹
3. Order vs. Molecularity
This is a classic theoretical distinction heavily tested in NEET. They sound similar but describe entirely different aspects of kinetics.
| Order of Reaction | Molecularity |
|---|---|
| The sum of powers of the concentration terms in the experimental rate law. | The number of reacting species (atoms/ions/molecules) colliding simultaneously in an elementary reaction. |
| Strictly an Experimental quantity. | Strictly a Theoretical concept. |
| Can be Zero, Fractional, or a Whole Number. | Must be a Whole Number (1, 2, or 3). It cannot be zero, negative, or fractional. |
| Applicable to both elementary and complex reactions. | Applicable only to elementary steps. It has no meaning for a complex overall reaction. |
Most chemical reactions do not occur in a single, simple collision. They proceed through a series of basic elementary steps, known collectively as the reaction mechanism. The overall Differential Rate Equation is dictated strictly by the slowest step in this sequence.
Consider the overall reaction: $NO_2(g) + CO(g) \rightarrow NO(g) + CO_2(g)$
- Step 1: $NO_2 + NO_2 \xrightarrow{k_1} NO + NO_3$ SLOW (RDS)
- Step 2: $NO_3 + CO \xrightarrow{k_2} NO_2 + CO_2$ FAST
Because Step 1 has the highest activation energy barrier, it acts as a bottleneck for the entire process. Therefore, we write the Differential Rate Equation using ONLY the reactants present in the Slow Step:
4. Pseudo First-Order Reactions
Some reactions are naturally second order (or higher) because two different molecules participate in the rate-determining step. However, if one reactant is present in massive excess (like water acting as a solvent), its concentration remains practically unchanged during the reaction. The rate law simplifies, and the reaction behaves as if it were First Order.
$CH_3COOH + C_2H_5OH$
True Rate = $k'[Ester][H_2O]$
Pseudo Rate = $k[Ester]$
(Where $k = k'[H_2O]$)
$C_6H_{12}O_6 \ (\text{Glu}) + C_6H_{12}O_6 \ (\text{Fru})$
True Rate = $k'[Sucrose][H_2O]$
Pseudo Rate = $k[Sucrose]$
(Where $k = k'[H_2O]$)
NEET Grand Test: Kinetics Basics
15 High-Yield Questions testing rate stoichiometry, units of $k$, RDS logic, and molecularity limits.
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