Q.Why does the rate of any reaction generally decreases during the course of the reaction?

The key idea is that reaction rate depends on the concentration of reactants. As the reaction proceeds, reactants are consumed, so their concentrations fall.

1. For most reactions, the rate law is of the form: $\text{Rate} = k[\text{Reactant}]^n$, where $n$ is the order.
2. As reactants are used up, $[\text{Reactant}]$ decreases. …

The rate of a chemical reaction is not constant — it changes as the reaction proceeds. Most students memorise that "rate decreases with time", but the real insight lies in collision theory and the rate law.

Think about what a reaction needs to happen: reactant particles must collide with sufficient energy and proper orientation. The number of such collisions per second depends directly on how many reactant particles are present in a given volume — that is, their concentration. As the reaction runs, reactants are converted into products, so their concentrations drop. Fewer particles mean fewer effective collisions per second, so the rate falls.

This is not a guess — it is built into the mathematical definition of rate. For a general reaction $aA + bB \rightarrow \text{products}$, the instantaneous rate is given by:

where $k$ is the rate constant (temperature-dependent, but constant at a fixed $T$), and $m$, $n$ are the orders with respect to A and B. Unless the reaction is zero-order ($m=n=0$), the rate depends on $[A]$ and $[B]$. As these concentrations decrease, the product $[A]^m[B]^n$ shrinks, so the rate drops.

Let us walk through the reasoning step by step.

1. The rate law ties rate to concentration. For any reaction (except zero-order), the rate is proportional to the concentration of reactants raised to some power. For a simple first-order reaction $A \rightarrow \text{products}$, the rate law is:

If $[A]$ halves, the rate halves. This is the direct mathematical link.

2. Reactant concentration falls as reaction proceeds.

   In a closed system, the total amount of reactant is finite. Every time a reaction event occurs, a reactant molecule is consumed. So $[A]$ decreases monotonically from its initial value $[A]_0$ toward zero (or toward equilibrium, if reversible).

   Watch out

   A common mistake is to think the rate constant $k$ changes during the reaction. It does not — $k$ depends only on temperature and the nature of the reaction, not on concentration. The rate changes because $[A]$ changes, not because $k$ changes.

3. Collision frequency drops with concentration.

   From kinetic molecular theory, the frequency of collisions between reactant molecules is proportional to the product of their concentrations. For a bimolecular step $A + B \rightarrow$ products, the collision frequency $\propto [A][B]$. Fewer molecules → fewer collisions → fewer successful reactions per second → lower rate.

4. Even for complex reactions, the same logic holds.

   Consider a second-order reaction $2A \rightarrow$ products with rate $= k[A]^2$. As $[A]$ drops, $[A]^2$ drops even faster. For a zero-order reaction (rate $= k$), the rate is constant — but this is rare and usually occurs only when a catalyst surface is saturated, so concentration of reactant in solution does not affect the rate. For the vast majority of reactions you will encounter in exams, the rate decreases.

5. Graphical evidence confirms this.

   If you plot concentration of reactant vs. time, you get a curve that slopes downward. The slope at any point is $-\frac{d[A]}{dt}$, which is the rate. This slope becomes less steep as time increases — exactly what we expect. …

Concept: Rate Law and Concentration Dependence

The rate of a reaction depends on the concentration of reactants. As the reaction proceeds, reactants are consumed — their concentrations fall — so the rate naturally decreases.

Method: Collision Theory & Rate Law Analysis

Steps:

1. Recall the general rate law For a reaction $\text{A} \to \text{Products}$, the rate is:

where $k$ is the rate constant and $n$ is the order of reaction.

2. Identify the key factor

   The rate is directly proportional to $[\text{A}]^n$.

   • As the reaction proceeds, $[\text{A}]$ decreases.
   • Therefore, the rate decreases (unless $n = 0$, which is rare).

3. Connect to collision theory

   • Reactions occur when reactant particles collide with sufficient energy and proper orientation.
   • Fewer reactant molecules → fewer collisions per second → lower reaction rate.

4. Conclusion …

Here are the common mistakes students make when answering “Why does the rate of any reaction generally decrease during the course of the reaction?” and how to avoid each.

✗ Mistake 1: Confusing “rate” with “speed” or “time”

What students write:

“The reaction slows down because the reaction is taking longer.”

Why it’s wrong:

This is circular reasoning. The rate is defined as change in concentration per unit time. Saying “it slows down because it takes longer” explains nothing.

How to avoid:

Always anchor your answer in the definition of rate:

Rate $= -\frac{d[\text{reactant}]}{dt}$

Then explain why this derivative decreases — not just that it does.

✗ Mistake 2: Saying “the reaction stops” or “rate becomes zero”

What students write:

“The rate decreases and finally becomes zero.”

Why it’s wrong:

For most reactions, the rate approaches zero asymptotically as reactants are consumed, but it never truly becomes zero in finite time (unless the reaction goes to completion and one reactant is completely exhausted). In many equilibrium reactions, the net rate becomes zero only at equilibrium, but the forward and backward rates are equal — not zero.

How to avoid:

Use precise language:

The rate decreases gradually and tends toward zero as the reactant concentration approaches zero (or as equilibrium is reached).

✗ Mistake 3: Blaming “catalyst deactivation” or “temperature change” without evidence

What students write:

“The catalyst gets used up, so the rate falls.”

Why it’s wrong:

A catalyst is not consumed in the reaction. Unless the problem explicitly mentions catalyst poisoning or decomposition, this is an incorrect assumption. Similarly, unless the reaction is exothermic/endothermic in a non-thermostatted system, temperature is assumed constant.

How to avoid:

Stick to the universal reason that applies to any reaction under constant conditions:

The rate decreases because the concentration of reactants decreases with time, reducing the frequency of effective collisions.

✗ Mistake 4: Forgetting the collision theory link

What students write:

“There are fewer reactant particles, so the rate is lower.”

Why it’s incomplete:

This is correct but lacks the mechanism. Examiners expect you to connect concentration → collision frequency → rate.

How to avoid:

Write a two-step logical chain:

1. As the reaction proceeds, reactant concentration decreases.
2. According to collision theory, lower concentration means fewer effective collisions per unit time.
3. Hence, the rate of reaction decreases.

✗ Mistake 5: Using the wrong rate law or assuming order = 1 …