Match the items of Column I and Column II. Column I: (i) Mathematical expression for rate of reaction (ii) Rat

NCERT Exemplar · Q61

Q. Match the items of Column I and Column II. Column I:

(i) Mathematical expression for rate of reaction

(ii) Rate of reaction for zero order reaction is equal to

(iii) Units of rate constant for zero order reaction is same as that of

(iv) Order of a complex reaction is determined by
Column II:

(a) rate constant

(b) rate law

(d) rate of a reaction

Nagaland NbseShort· 2mImportance★★★★★

94% · 110/117 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 — Average Rate Of Reaction

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.

--- …

Why this formula?

Average Rate of Reaction — Why the Formula Holds

Let’s build this from the ground up. The goal is to understand why the average rate formula looks the way it does — not just memorise it.

1. What does "rate of reaction" mean physically?

A chemical reaction changes the concentration of reactants (decreasing) and products (increasing) over time.

Rate = how fast this change happens.
If you measure the change over a finite time interval, you get the average rate.

2. The core idea: change per unit time

For any quantity $X$ that changes from $X_1$ to $X_2$ over time $t_1$ to $t_2$ :

\text{Average rate of change of } X = \frac{\Delta X}{\Delta t} = \frac{X_2 - X_1}{t_2 - t_1}

This is just the slope of the straight line connecting the two points on a concentration vs. time graph.

3. Applying this to a reaction

Consider a simple reaction:

A \rightarrow B

Reactant A is consumed: $[A]$ decreases.
Product B is formed: $[B]$ increases.

For reactant A (disappearing):

\text{Average rate} = -\frac{\Delta [A]}{\Delta t}

Why the minus sign?

Because $\Delta [A] = [A]_2 - [A]_1$ is negative (concentration drops). The rate itself must be positive (speed is never negative). So we multiply by $-1$ .

For product B (appearing):

\text{Average rate} = +\frac{\Delta [B]}{\Delta t}

Here $\Delta [B]$ is positive, so no minus sign needed.

4. The general formula for any reaction

For a balanced reaction:

aA + bB \rightarrow cC + dD

The average rate is defined per mole of reaction — so it’s the same number regardless of which species you track.

We divide each $\frac{\Delta [\text{species}]}{\Delta t}$ by its stoichiometric coefficient:

\text{Average rate} = -\frac{1}{a} \frac{\Delta [A]}{\Delta t} = -\frac{1}{b} \frac{\Delta [B]}{\Delta t} = +\frac{1}{c} \frac{\Delta [C]}{\Delta t} = +\frac{1}{d} \frac{\Delta [D]}{\Delta t}

Why divide by the coefficient?

If 2 moles of A disappear for every 1 mole of C formed, then $\frac{\Delta [A]}{\Delta t}$ is twice as large as $\frac{\Delta [C]}{\Delta t}$ . Dividing by the coefficient normalises them to the same "per mole of reaction" rate.

5. Key exam point: the formula in one line

For any species $X$ with stoichiometric coefficient $\nu_X$ (negative for reactants, positive for products):

$\text{Average rate} = \frac{1}{\nu_X} \frac{\Delta [X]}{\Delta t}$ …

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

The key idea is to match each term in Column I with its correct conceptual counterpart in Column II by understanding the definitions of rate of reaction, zero‑order kinetics, and the rate‑determining step. The correct matches are: (i)→(b), (ii)→(a), (iii)→(d), (iv)→(c).

The Concept: What Each Term Really Means

Before we match, let’s build a clear mental picture. The rate of reaction tells us how fast reactants are used up or products appear. Its mathematical expression is the rate law — an equation linking the rate to concentrations raised to some powers (the orders). For a zero‑order reaction, the rate is constant and equals the rate constant itself. That constant has units of concentration per time (e.g., mol L⁻¹ s⁻¹), which is exactly the same as the rate of reaction (since rate is also concentration per time). Finally, for a complex reaction (one with multiple steps), the overall order is determined by the slowest step — the bottleneck that controls the whole process.

Now let’s match each item step by step.

Step‑by‑Step Matching

1. (i) Mathematical expression for rate of reaction

The mathematical expression is the rate law, which is written as:

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

where $k$ is the rate constant and $m$ , $n$ are orders. This is not the rate constant itself, nor the rate of a reaction (which is a number, not an expression). So (i) matches with (b) rate law.

2. (ii) Rate of reaction for zero order reaction is equal to

For a zero‑order reaction, the rate law is:

\text{Rate} = k[\text{A}]^0 = k

The rate is constant and numerically equal to the rate constant $k$ . It is not equal to the rate of a reaction (that would be circular) nor to the order of a step. So (ii) matches with (a) rate constant.

Tip

A zero‑order reaction is the only case where the rate and the rate constant have the same numerical value (and same units). This is a common exam favourite.

3. (iii) Units of rate constant for zero order reaction is same as that of …

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

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

Mistake 1: Confusing "Rate of Reaction" with "Rate Constant"

The Error: Students match (i) "Mathematical expression for rate of reaction" with (a) "rate constant." They think the expression is the constant.
Why it happens: The rate constant ( $k$ ) appears inside the rate expression, so students blur the two.
How to avoid: Remember the rate law:

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

The rate law is the full mathematical expression (including $k$ and concentrations). The rate constant ( $k$ ) is just one number in that expression.

Correct match: (i) → (b) rate law.

Mistake 2: Forgetting the Definition of Zero Order Rate

The Error: Students match (ii) "Rate of reaction for zero order reaction is equal to" with (d) "rate of a reaction" (a tautology) or with (a) "rate constant" but without understanding why.
Why it happens: For zero order, $\text{Rate} = k[\text{A}]^0 = k$ . Students memorize this but don't connect it to the fact that the rate is the constant.
How to avoid: Write the general rate law for zero order:

\text{Rate} = k[\text{A}]^0 = k \times 1 = k

The rate does not depend on concentration; it is equal to the rate constant itself.

Correct match: (ii) → (a) rate constant.

Mistake 3: Mixing Up Units of Rate and Rate Constant

The Error: Students match (iii) "Units of rate constant for zero order reaction is same as that of" with (b) "rate law" or (c) "order of slowest step."
Why it happens: They know units of $k$ vary with order but forget that for zero order, $k$ has units of concentration/time (e.g., $\text{mol L}^{-1} \text{s}^{-1}$ ), which is exactly the unit of rate of reaction.
How to avoid: Recall the unit formula:

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

For $n=0$ :

\text{Units of } k = (\text{concentration})^{1} \times \text{time}^{-1} = \text{mol L}^{-1} \text{s}^{-1}

This is identical to the unit of rate of reaction.

Correct match: (iii) → (d) rate of a reaction.

Mistake 4: Assuming "Order of Complex Reaction" is Always the Sum of Stoichiometric Coefficients

The Error: Students match (iv) "Order of a complex reaction is determined by" with (b) "rate law" (which is true in general) but miss the specific exam nuance: for a complex (multi-step) reaction, the order is governed by the slowest step. …

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

←Previous questionQ60 — Match the items of Column I and Column II. Column I: (i) Diamond (ii) …→Next questionQ62 — Assertion: Order of the reaction can be zero or fractional. Reason: We…

Q.Match the items of Column I and Column II.
Column I:

Concept understanding — Average Rate Of Reaction

Reaction Rate Stoichiometry – From Intuition to Precision

The Intuitive Picture

The Precise Statement

How to Use It (Step by Step)

Why this formula?

Average Rate of Reaction — Why the Formula Holds

1. What does "rate of reaction" mean physically?

2. The core idea: change per unit time

3. Applying this to a reaction

For reactant A (disappearing):

For product B (appearing):

4. The general formula for any reaction

5. Key exam point: the formula in one line

The Concept: What Each Term Really Means

Step‑by‑Step Matching

Method: Concept Matching by Definition & Formula

Step-by-step solution

Mistake 1: Confusing "Rate of Reaction" with "Rate Constant"

Mistake 2: Forgetting the Definition of Zero Order Rate

Mistake 3: Mixing Up Units of Rate and Rate Constant

Mistake 4: Assuming "Order of Complex Reaction" is Always the Sum of Stoichiometric Coefficients

Q.Match the items of Column I and Column II. Column I:

The Concept: What Each Term Really Means

Step‑by‑Step Matching

🎓Unlock everything free for 14 days

Q.Match the items of Column I and Column II.
Column I:

Unlock everything free for 14 days