Which of the following statements is correct? (i) The rate of a reaction decreases with passage of time as the

NCERT Exemplar · Q11

Q. Which of the following statements is correct?

(i) The rate of a reaction decreases with passage of time as the concentration of reactants decreases.

(ii) The rate of a reaction is same at any time during the reaction.

(iii) The rate of a reaction is independent of temperature change.

(iv) The rate of a reaction decreases with increase in concentration of reactant(s).

Punjab PsebMCQ· 1mImportance★★★★★

51% · 60/117 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Factors Affecting Reaction Rate

Reaction Rate Stoichiometry – From Intuition to Precision

Imagine you are watching a simple reaction:

$2 \text{NO}_2 \rightarrow 2 \text{NO} + \text{O}_2$

As NO₂ disappears, NO appears twice as fast as O₂ appears. Why? Because the balanced equation says: for every 2 molecules of NO₂ that break apart, you get 2 molecules of NO and 1 molecule of O₂. The numbers in front of the species — the stoichiometric coefficients — tell you the relative speeds at which reactants vanish and products appear.

That is the core idea: reaction rate stoichiometry is the relationship between the rates of change of different species in a chemical reaction, dictated by their coefficients in the balanced equation.

The Intuitive Picture

Think of a factory assembly line. The balanced equation is like a recipe:

2 units of raw material A → 2 units of product B + 1 unit of byproduct C

If the line runs steadily, every time 2 units of A are consumed, 2 units of B are produced and 1 unit of C is produced. So the rate at which A disappears must be twice the rate at which C appears. The rate at which B appears equals the rate at which A disappears (both have coefficient 2).

Note

The stoichiometric coefficients are not speeds themselves — they are scaling factors that connect the speeds of different species.

The Precise Statement

For a general reaction:

aA + bB \rightarrow cC + dD

The rate of reaction (often called the rate of the process, $r$ ) is defined as:

r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

Here:

$\frac{d[X]}{dt}$ is the instantaneous rate of change of concentration of species X (in mol L⁻¹ s⁻¹).
The minus sign is used for reactants (their concentration decreases with time).
The plus sign is used for products (their concentration increases with time).
Dividing by the coefficient normalises the rate — so $r$ is the same number for every species.

r = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}

This single value $r$ is the intrinsic rate of the reaction, independent of which species you measure.

How to Use It (Step by Step)

Suppose you are given:

$2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}$

And you measure that $\frac{d[\text{H}_2]}{dt} = -0.040 \text{ M/s}$ (negative because H₂ is being consumed).

Step 1: Write the rate relation:

r = -\frac{1}{2}\frac{d[\text{H}_2]}{dt} = -\frac{1}{1}\frac{d[\text{O}_2]}{dt} = \frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt}

Step 2: Plug in the known value:

r = -\frac{1}{2}(-0.040) = +0.020 \text{ M/s}

Step 3: Find the rate for O₂:

-\frac{d[\text{O}_2]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{O}_2]}{dt} = -0.020 \text{ M/s}

Step 4: Find the rate for H₂O:

\frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{H}_2\text{O}]}{dt} = 2r = 0.040 \text{ M/s}

Tip

A quick check: the coefficients tell you the relative rates. Here, H₂ disappears twice as fast as O₂, and H₂O appears at the same rate as H₂ disappears (both coefficient 2). Always verify your numbers match the coefficient ratios.

Common Pitfall to Avoid …

Why this formula?

Factors Affecting Reaction Rate: Understanding the Why Behind the Formulas

In chemical kinetics, the rate of a reaction depends on several factors. The key formulas are not arbitrary — they arise from collision theory and transition state theory. Let’s break down each factor and derive the reasoning step-by-step.

1. Effect of Concentration: The Rate Law

The Formula

For a reaction $\text{aA} + \text{bB} \rightarrow \text{products}$ , the rate is often:

\text{Rate} = k [\text{A}]^m [\text{B}]^n

where $m$ and $n$ are orders (not necessarily equal to $a$ and $b$ ).

Why This Holds

Collision theory: For a reaction to occur, reactant particles must collide with sufficient energy and correct orientation.
Doubling concentration of A doubles the number of A particles per unit volume. This doubles the collision frequency between A and B (if B is constant). Hence, rate ∝ [A].
But the order $m$ $m$ is determined experimentally because:
- Not all collisions are effective.
- Some reactions involve multiple steps (mechanism). The rate depends on the slowest step (rate-determining step), which may involve only some reactants.

Example: For $\text{NO}_2 + \text{CO} \rightarrow \text{NO} + \text{CO}_2$ , the rate law is $\text{Rate} = k[\text{NO}_2]^2$ — because the slow step involves two NO₂ molecules colliding.

2. Effect of Temperature: The Arrhenius Equation

The Formula

k = A e^{-E_a / RT}

where:

$k$ = rate constant
$A$ = frequency factor (collision frequency × orientation factor)
$E_a$ = activation energy (J/mol)
$R$ = gas constant (8.314 J/mol·K)
$T$ = temperature (K)

Why This Holds

Boltzmann distribution: At a given temperature, molecules have a range of kinetic energies. Only those with energy ≥ $E_a$ can react.
The fraction of molecules with energy ≥ $E_a$ is given by the Maxwell–Boltzmann distribution:

\text{Fraction} = e^{-E_a / RT}

This is derived from integrating the energy distribution function.

The frequency factor $A$ accounts for how often collisions occur and the fraction with correct orientation.
Temperature increase shifts the distribution to higher energies, exponentially increasing the fraction of molecules that can overcome $E_a$ .

Key insight: A 10°C rise near room temperature roughly doubles the rate — because $e^{-E_a/RT}$ changes significantly.

3. Effect of a Catalyst

No Simple Formula, But a Key Principle

A catalyst lowers the activation energy $E_a$ by providing an alternative reaction pathway (e.g., forming an intermediate).

Why This Works

In the Arrhenius equation, $k = A e^{-E_a/RT}$ .
Lowering $E_a$ increases the exponential factor $e^{-E_a/RT}$ dramatically.
Example: If $E_a$ drops from 100 kJ/mol to 80 kJ/mol at 300 K, the rate constant increases by a factor of:

\frac{e^{-80,000/(8.314 \times 300)}}{e^{-100,000/(8.314 \times 300)}} = e^{(20,000)/(2494)} \approx e^{8.02} \approx 3000

The catalyst itself is not consumed — it participates in the mechanism but is regenerated.

4. Effect of Surface Area (for Heterogeneous Reactions) …

Start your 14-day free trial to unlock the full solution →

The rate of a reaction depends on the concentration of reactants — as reactants are consumed, concentration falls, and the rate decreases over time. Therefore, statement (i) is correct.

Let’s first understand what “rate of reaction” really means. The rate tells us how fast the concentration of a reactant (or product) changes per unit time. For most reactions, the rate is not constant — it changes as the reaction proceeds. Why? Because the rate depends on how frequently reactant particles collide, and that frequency depends on how many particles are present (i.e., their concentration). As reactants get used up, their concentration drops, collisions become less frequent, and the rate slows down. This is the core idea behind all the options.

Now examine each statement one by one.

Statement (i): “The rate of a reaction decreases with passage of time as the concentration of reactants decreases.”

This is exactly what we just described. For a typical reaction, say $A \rightarrow B$ , the rate law is often $\text{rate} = k[A]^n$ (where $n$ is the order). As $[A]$ falls, the rate falls. Even for zero-order reactions (where rate is constant until the reactant is nearly gone), the rate eventually drops to zero when the reactant is exhausted. So this statement is true for the vast majority of reactions.
Statement (ii): “The rate of a reaction is same at any time during the reaction.”

This would only be true for a zero-order reaction with a constant supply of reactant — but even then, the rate is constant only while reactant is present, not “at any time” (it becomes zero when reactant runs out). For first-order, second-order, etc., the rate clearly changes. So this is false.
Statement (iii): “The rate of a reaction is independent of temperature change.”

This is completely false. The rate constant $k$ depends strongly on temperature, as given by the Arrhenius equation: $k = A e^{-E_a/(RT)}$ . Raising temperature increases $k$ dramatically, and thus the rate increases. Temperature is one of the most important factors affecting reaction rate. …

Start your 14-day free trial to unlock the full solution →

Here are the common mistakes students make on this question, along with how to avoid each.

Mistake 1: Choosing (ii) — "The rate is the same at any time"

Why students pick it:

They confuse average rate (over a time interval) with instantaneous rate (at a specific moment). They think "rate" is constant because they memorise the formula $\frac{\Delta [R]}{\Delta t}$ without thinking about the slope.

How to avoid:

Draw a concentration vs. time graph. The slope at any point is the instantaneous rate.
As reactants are used up, the slope becomes less steep → rate decreases with time.
Key fact: Rate is not constant — it depends on concentration at that instant.

Mistake 2: Choosing (iv) — "Rate decreases with increase in concentration"

Why students pick it:

They misread "decreases" as "increases" or confuse the direction. Some think "more reactants means slower reaction" because they imagine crowding.

How to avoid:

Recall the collision theory: more reactant molecules per unit volume → more frequent collisions → higher rate.
For most reactions, rate $\propto$ [reactant] $^n$ (where $n > 0$ ).
Rule of thumb: Increasing concentration increases rate (unless it’s a zero-order reaction, but that’s rare at this level).

Mistake 3: Choosing (iii) — "Rate is independent of temperature"

Why students pick it:

They think "if the reaction happens at room temperature, temperature doesn’t matter." Or they confuse temperature with catalyst.

How to avoid:

Remember the Arrhenius equation: $k = A e^{-E_a/RT}$ .
Even a small temperature rise (e.g., 10 °C) can double or triple the rate.
Exam tip: Temperature always affects rate — it’s a fundamental factor.

Mistake 4: Not reading "correct" vs. "incorrect" carefully

Why students pick it:

They see a statement that sounds familiar (like (i)) but second-guess themselves because (ii) or (iv) also look plausible.

How to avoid:

Read all four options before deciding.
Eliminate clearly wrong ones first:
- (ii) is wrong — rate changes. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ10 — For the same plot of the volume of hydrogen collected against time in …→Next questionQ12 — Which of the following expressions is correct for the rate of reaction…

Q.Which of the following statements is correct?

Concept understanding — Factors Affecting Reaction Rate

Reaction Rate Stoichiometry – From Intuition to Precision

The Intuitive Picture

The Precise Statement

How to Use It (Step by Step)

Common Pitfall to Avoid …

Why this formula?

Factors Affecting Reaction Rate: Understanding the Why Behind the Formulas

1. Effect of Concentration: The Rate Law

The Formula

Why This Holds

2. Effect of Temperature: The Arrhenius Equation

The Formula

Why This Holds

3. Effect of a Catalyst

No Simple Formula, But a Key Principle

Why This Works

4. Effect of Surface Area (for Heterogeneous Reactions) …

Method: Conceptual Elimination Using the Definition of Reaction Rate

Step-by-step reasoning:

Mistake 1: Choosing (ii) — "The rate is the same at any time"

Mistake 2: Choosing (iv) — "Rate decreases with increase in concentration"

Mistake 3: Choosing (iii) — "Rate is independent of temperature"

Mistake 4: Not reading "correct" vs. "incorrect" carefully

Q.Which of the following statements is correct?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days