A chemical reaction 2N₂O₅ (g) arrow 4NO₂ (g) + O₂ (g) in gas phase was carried out in a closed vessel. The con

Question

Q. A chemical reaction $2N_2O_5

(g) \rightarrow 4NO_2

(g) + O_2 (g)

in gas phase was carried out in a closed vessel. The concentration of

NO_2

was found to increase by

5 \times 10^{-3}

mol

L^{-1}$ in 10 seconds. Calculate:

(a) the rate of formation of

NO_2

, and

(b) the rate of consumption of

N_2O_5

CBSECBSE Class XII Board 2023Subjective· 2mImportance★★★★★

🔒 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 — Reaction Rate Stoichiometry

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:

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

Why this formula?

Reaction Rate Stoichiometry: Why the Formula Holds

Let’s start with the core idea: In a chemical reaction, the rate at which reactants disappear and products appear is not arbitrary — it is tied directly to the stoichiometric coefficients in the balanced equation.

The Key Formula

For a general reaction:

aA + bB \rightarrow cC + dD

The rate of reaction ( $R$ ) 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}

Where:

$[A], [B], [C], [D]$ are concentrations (in mol/L)
$t$ is time
$a, b, c, d$ are stoichiometric coefficients

Why This Formula Holds: The Reasoning

1. The Physical Meaning of Stoichiometric Coefficients

The coefficients tell us the mole ratio in which substances react or are produced. For example:

2H_2 + O_2 \rightarrow 2H_2O

2 moles of $H_2$ react with 1 mole of $O_2$ to produce 2 moles of $H_2O$ .
This means: for every 2 molecules of $H_2$ that disappear, only 1 molecule of $O_2$ disappears, and 2 molecules of $H_2O$ appear.

Key insight: The number of moles changing per unit time is different for each substance, but the reaction event is the same.

2. The Problem with Raw Rates

If we simply wrote:

\text{Rate} = -\frac{d[H_2]}{dt}

This would be twice the rate of disappearance of $O_2$ (since $H_2$ disappears twice as fast). That’s inconsistent — the same reaction shouldn’t have two different numerical rates.

We need a single, unique rate that describes the reaction itself, not just one substance.

3. The Solution: Normalize by Stoichiometric Coefficients

To get a reaction rate that is the same regardless of which substance we track, we divide each substance’s rate of change by its stoichiometric coefficient.

Why division works:

If $A$ disappears at rate $-\frac{d[A]}{dt}$ , and $a$ moles of $A$ are consumed per reaction event, then the number of reaction events per unit time is:

\text{Reaction events per second} = \frac{-\frac{d[A]}{dt}}{a}

Similarly, for product $C$ appearing at rate $+\frac{d[C]}{dt}$ , with $c$ moles produced per event:

\text{Reaction events per second} = \frac{+\frac{d[C]}{dt}}{c}

Since the same reaction is happening, these must be equal. Hence:

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

4. The Sign Convention

Reactants decrease over time → $\frac{d[\text{reactant}]}{dt} < 0$ → we add a negative sign to make the rate positive.
Products increase over time → $\frac{d[\text{product}]}{dt} > 0$ → we use a positive sign. …

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

The rate of a reaction is linked to the stoichiometric coefficients. For the given reaction, the rate of formation of $NO_2$ is $5 \times 10^{-4} \text{ mol L}^{-1} \text{s}^{-1}$ , and the rate of consumption of $N_2O_5$ is $2.5 \times 10^{-4} \text{ mol L}^{-1} \text{s}^{-1}$ .

The key idea here is that in a chemical reaction, the rate at which a product appears or a reactant disappears is not arbitrary — it is strictly tied to the stoichiometric coefficients. This is because the reaction consumes and produces molecules in fixed ratios. If you know how fast one species changes, you can calculate the rate for any other species using those coefficients.

For the reaction:

2N_2O_5 (g) \rightarrow 4NO_2 (g) + O_2 (g)

the rate of reaction is defined as:

\text{Rate} = -\frac{1}{2}\frac{d[N_2O_5]}{dt} = \frac{1}{4}\frac{d[NO_2]}{dt} = \frac{1}{1}\frac{d[O_2]}{dt}

This single rate value applies to the entire reaction. The negative sign for reactants indicates they are being consumed (concentration decreasing), while products have a positive sign.

Now, let's apply this step by step.

Rate of formation of $NO_2$ The problem tells us that the concentration of $NO_2$ increases by $5 \times 10^{-3} \text{ mol L}^{-1}$ in 10 seconds. The rate of formation is simply the change in concentration per unit time:

\text{Rate of formation of } NO_2 = \frac{\Delta [NO_2]}{\Delta t} = \frac{5 \times 10^{-3}}{10} = 5 \times 10^{-4} \text{ mol L}^{-1} \text{s}^{-1}

This is a direct calculation — no stoichiometry needed yet.

Relating the rates using stoichiometry From the balanced equation, for every 4 moles of $NO_2$ produced, 2 moles of $N_2O_5$ are consumed. So the rate of consumption of $N_2O_5$ is half the rate of formation of $NO_2$ (since the coefficient ratio is $2:4 = 1:2$ ). More formally:

\frac{d[N_2O_5]}{dt} = -\frac{2}{4} \frac{d[NO_2]}{dt} = -\frac{1}{2} \frac{d[NO_2]}{dt}

The negative sign reminds us that $N_2O_5$ is being consumed.

Rate of consumption of $N_2O_5$ Plugging in the rate of formation of $NO_2$ : …

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

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 56/3/11 markMCQ
$3A \rightarrow 2B$
›Reveal solutionSolution
The rate of change of concentration of a species is related to the overall reaction rate by its stoichiometric coefficient; for $3A \rightarrow 2B$ , the rate of formation of B is $\boxed{-\dfrac{2}{3}\dfrac{d[A]}{dt}}$ .

In chemical kinetics, the rate of a reaction can be expressed in terms of the rate of disappearance of reactants or the rate of formation of products. However, these individual rates are not always equal to each other because of the stoichiometry of the reaction. For example, if 3 moles of A disappear for every 2 moles of B formed, then A is disappearing faster than B is forming (in terms of moles).

To define a single, unambiguous "rate of reaction" for the entire process, we normalize the rate of change of concentration of each species by its stoichiometric coefficient. This ensures that the overall reaction rate is independent of which reactant or product we choose to monitor.
$aA + bB \rightarrow cC + dD$
Let's apply this concept to the given reaction.

Write the general rate expression for the given reaction. The reaction is $3A \rightarrow 2B$ . Using the general formula, the rate of reaction can be expressed in terms of the disappearance of A and the formation of B:

$\text{Rate} = -\frac{1}{3}\frac{d[A]}{dt} = +\frac{1}{2}\frac{d[B]}{dt}$
Here, $\frac{d[A]}{dt}$ represents the instantaneous rate of change of concentration of A, and $\frac{d[B]}{dt}$ represents the instantaneous rate of change of concentration of B. The negative sign for A indicates it is a reactant and its concentration is decreasing. The positive sign for B indicates it is a product and its concentration is increasing.

2. Equate the relevant parts of the rate expression and solve for the desired term. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set 56/2/11 markMCQ
$A + 2B \rightarrow$
›Reveal solutionSolution
The order of a reaction is the sum of all exponents in the rate law. Here, $\frac{1}{2} + 1 = \frac{3}{2} = 1.5$ , so the answer is (A).

Understanding Reaction Order

The order of a reaction tells us how the rate depends on reactant concentrations. It's a purely experimental quantity—you cannot deduce it from the stoichiometric equation alone. The stoichiometric coefficients (1 for A, 2 for B) are irrelevant here; only the exponents in the experimentally determined rate law matter.

When we write a rate law like $\text{Rate} = k[A]^m[B]^n$ , the exponents $m$ and $n$ describe how sensitive the rate is to changes in each reactant's concentration. The overall order is simply the sum of these individual orders.

Finding the Overall Order

The rate law given is:
$\text{Rate} = k[A]^{1/2}[B]^1$
Let's identify each component:

Order with respect to A: The exponent on $[A]$ is $\frac{1}{2}$ . This means the reaction is half-order in A—if you quadruple the concentration of A (keeping B constant), the rate only doubles.

Order with respect to B: The exponent on $[B]$ is $1$ . This means the reaction is first-order in B—doubling B doubles the rate (keeping A constant).

Overall order: Add the individual orders:

$\text{Overall order} = \frac{1}{2} + 1 = \frac{3}{2} = 1.5$ …
Start your 14-day free trial to see the full solution →
CBSE 2026Set A1 markMCQ
$Q. For a reaction, aA + bB -> products, -d[A]/dt is equal to$
(a) -d[B]/dt
(b) (-b/d) d[B]/dt
(c) (-a/b) d[B]/dt
(d) (-b/a) d[B]/dt
›Reveal solutionSolution

Rate = -(1/a) d[A]/dt = -(1/b) d[B]/dt, so -d[A]/dt = (a/b)(-d[B]/dt) = (-a/b) d[B]/dt, which is option (c).

For aA + bB -> products, the rate of reaction is defined so that

Rate = -(1/a) d[A]/dt = -(1/b) d[B]/dt.

Equating the two: (1/a)(-d[A]/dt) = (1/b)(-d[B]/dt), hence

-d[A]/dt = (a/b)(-d[B]/dt) = (-a/b) d[B]/dt. …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$BrO_3^-(aq)+5Br^-(aq)+6H^+ \rightarrow 3Br_2(l)+3H_2O(l)$
›Reveal solutionSolution

Dividing each species' rate of change by its own stoichiometric coefficient gives a common "rate of reaction" expression, which links the rates of Br₂ formation and Br⁻ consumption via a 3:5 ratio.

For the balanced reaction:

$BrO_3^-(aq) + 5Br^-(aq) + 6H^+ \rightarrow 3Br_2(l) + 3H_2O(l)$

The rate of reaction, expressed consistently for every species, divides each species' rate of change by its own stoichiometric coefficient:

$Rate = \frac{1}{3}\frac{d[Br_2]}{dt} = -\frac{1}{5}\frac{d[Br^-]}{dt}$

Rearranging to relate the two rates directly:

$\frac{d[Br_2]}{dt} = -\frac{3}{5}\frac{d[Br^-]}{dt}$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set 56/1/11 markMCQ
$X + 2Y \rightarrow P$
›Reveal solutionSolution
The stoichiometric coefficients relate the rates of change of all species in a reaction; for $X + 2Y \rightarrow P$ , the rate of consumption of $Y$ is twice the rate of formation of $P$ , giving $-2\frac{d[Y]}{dt} = +\frac{d[P]}{dt}$ .

Understanding Reaction Rate Stoichiometry

When a chemical reaction proceeds, the concentrations of reactants decrease while products increase. The rate at which each species changes is directly tied to the stoichiometric coefficients in the balanced equation. The key insight is that these rates must be proportional to maintain the stoichiometry throughout the reaction.

For a general reaction $aA + bB \rightarrow cC + dD$ , the relationship between the rates of change is:
$-\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}$
The negative signs appear for reactants (concentrations decreasing), positive for products (concentrations increasing). Dividing by the stoichiometric coefficient normalizes the rate to a common "reaction rate."

Step-by-Step Analysis

Identify the stoichiometry

For $X + 2Y \rightarrow P$ , we have:

1 mole of $X$ consumed

2 moles of $Y$ consumed

1 mole of $P$ formed

Write the general rate relationship

Applying the stoichiometric rate law:

$-\frac{d[X]}{dt} = -\frac{1}{2}\frac{d[Y]}{dt} = +\frac{d[P]}{dt}$
Each term equals the overall reaction rate. Notice $Y$ has coefficient 2, so its rate of change is divided by 2.

Extract pairwise relationships

From the equality above, we can derive several equivalent statements:

Between $Y$ and $P$ : $-\frac{1}{2}\frac{d[Y]}{dt} = +\frac{d[P]}{dt}$

Multiplying both sides by $-2$ :

$-2\frac{d[Y]}{dt} = +\frac{d[P]}{dt}$

Check each option …
Start your 14-day free trial to see the full solution →
CBSE 2024Set 56/2/11 markMCQ
$P \rightarrow Q$
›Reveal solutionSolution
When the concentration of $P$ doubles and the rate increases by a factor of 8, the reaction is third order with respect to $P$ . The answer is (A) 3.

Understanding Rate Laws and Reaction Order

The rate of a reaction tells us how fast a reactant disappears or a product forms. For an elementary reaction, the rate law can be written directly from the stoichiometry. For $P \rightarrow Q$ , the rate of disappearance of $P$ is:
$\text{Rate} = k[P]^n$
where $k$ is the rate constant and $n$ is the order of the reaction with respect to $P$ . The order tells us how sensitive the rate is to changes in concentration.

The key insight here is that if we know how the rate changes when concentration changes, we can deduce the order. Specifically, if concentration is multiplied by some factor, the rate is multiplied by that factor raised to the power of the order.

Finding the Order Step by Step

Write the rate law at the initial concentration Let the initial concentration of $P$ be $[P]_0$ . The initial rate is:

$\text{Rate}_1 = k[P]_0^n$

Write the rate law at the doubled concentration When we double the concentration, $[P] = 2[P]_0$ . The new rate is:

$\text{Rate}_2 = k(2[P]_0)^n = k \cdot 2^n \cdot [P]_0^n$

Set up the ratio of rates We're told that the rate increases by a factor of 8, meaning $\text{Rate}_2 = 8 \times \text{Rate}_1$ . Taking the ratio:

$\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k \cdot 2^n \cdot [P]_0^n}{k[P]_0^n} = 2^n$

Solve for the order …
Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. For the reaction : 4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g), the rate of formation of NO is 3.6 x 10-3 mol L-1 S-1. The rate of disapperance of ammonia would be -$
(a) 5.4 x 10-3 mol L-1 S-1
(b) 2.4 x 10-3 mol L-1 S-1
(c) 3.6 x 10-3 mol L-1 S-1
(d) 4.6 x 10-3 mol L-1 S-1
›Reveal solutionSolution

Because NH $_3$ and NO both have coefficient 4 in the balanced equation, their rates of change are numerically equal.

For the reaction $4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)$ , the rate of reaction is written in terms of each species divided by its stoichiometric coefficient:

$\text{Rate} = -\frac{1}{4}\frac{d[NH_3]}{dt} = -\frac{1}{5}\frac{d[O_2]}{dt} = \frac{1}{4}\frac{d[NO]}{dt} = \frac{1}{6}\frac{d[H_2O]}{dt}$

Since NH $_3$ (a reactant) and NO (a product) both carry the coefficient 4: …

Start your 14-day free trial to see the full solution →
CBSE 2023Set 56/2/11 markMCQ
$3A \rightarrow 2B$
›Reveal solutionSolution
The rate of formation of $B$ is linked to the rate of consumption of $A$ through their stoichiometric coefficients: $\frac{d[B]}{dt} = -\frac{2}{3}\frac{d[A]}{dt}$ . The answer is (B).

Understanding Reaction Rate Stoichiometry

When a chemical reaction proceeds, different species appear or disappear at rates determined by the balanced equation. The stoichiometric coefficients tell us the ratio in which molecules react and form, but the actual rates of change of concentration must account for these ratios.

For the reaction $3A \rightarrow 2B$ , three molecules of $A$ disappear for every two molecules of $B$ that form. This means $A$ is consumed $\frac{3}{2}$ times faster than $B$ is produced. The key insight: to define a single "rate of reaction" that's the same no matter which species we measure, we divide each species' rate by its stoichiometric coefficient.

The general relationship is:
$\text{Rate} = -\frac{1}{3}\frac{d[A]}{dt} = +\frac{1}{2}\frac{d[B]}{dt}$
The negative sign appears for reactants (concentration decreasing) and positive for products (concentration increasing).

Step-by-Step Solution

Write the stoichiometric relationship From $3A \rightarrow 2B$ , the stoichiometric coefficients are 3 for $A$ and 2 for $B$ . The rate of reaction is defined as:

$\text{Rate} = -\frac{1}{3}\frac{d[A]}{dt} = +\frac{1}{2}\frac{d[B]}{dt}$

Isolate $\frac{d[B]}{dt}$ We want to express $\frac{d[B]}{dt}$ in terms of $\frac{d[A]}{dt}$ . From the equality above:

$\frac{1}{2}\frac{d[B]}{dt} = -\frac{1}{3}\frac{d[A]}{dt}$

Solve for $\frac{d[B]}{dt}$ Multiply both sides by 2: …
Start your 14-day free trial to see the full solution →
CBSE 2023Set F1 markMCQ
$Q. The rate of chemical reaction, 2A+B\toC can be represented by which of the following?$
(a) -1/2 d[A]/dt
(b) -d[B]/dt
(c) +d[C]/dt
(d) All of these
›Reveal solutionSolution
Rate is defined so that each species term is divided by its stoichiometric coefficient; all three given forms are equal.

For the reaction 2A + B → C, the unique rate of reaction is written by dividing each concentration change by the corresponding stoichiometric coefficient (with a minus sign for reactants):

Rate = -(1/2) d[A]/dt = -d[B]/dt = +d[C]/dt

A has coefficient 2, so its term is -(1/2) d[A]/dt. …
Start your 14-day free trial to see the full solution →
CBSE 2020Set ANNUAL1 mark
$Q. Clculate the rate of the reaction, A + 2B -> 2C + D.$
›Reveal solutionSolution

For a general reaction, the rate is defined as the rate of change of concentration of any species divided by its stoichiometric coefficient, with reactants taken as negative (since they are consumed) and products as positive (since they are formed).

For the reaction A + 2B -> 2C + D, the stoichiometric coefficients are 1 (A), 2 (B), 2 (C), and 1 (D). By definition, the rate of reaction is the same value whichever species you track, once each rate-of-change term is divided by its own coefficient:

Rate = -d[A]/dt = -(1/2) d[B]/dt = (1/2) d[C]/dt = d[D]/dt

…

Start your 14-day free trial to see the full solution →
CBSE 2018Set ANNUAL1 markMCQ
$2A + B \longrightarrow 3C + D$
(a) $\dfrac{-d[A]}{2\,dt}$
(b) $\dfrac{-d[B]}{dt}$
(c) $\dfrac{-d[C]}{3\,dt}$
(d) $\dfrac{-d[D]}{3\,dt}$
›Reveal solutionSolution

The correct rate expressions divide by each species' own stoichiometric coefficient — $D$ 's coefficient is 1, so dividing by 3 is wrong.

For $2A + B \rightarrow 3C + D$ , the rate of reaction is expressed as:

$Rate = -\frac{1}{2}\frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{3}\frac{d[C]}{dt} = \frac{d[D]}{dt}$

each concentration-change term divided by its own stoichiometric coefficient (2 for A, 1 for B, 3 for C, 1 for D), with a negative sign for reactants being consumed and a positive sign for products being formed.

…

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

Q.A chemical reaction $2N_2O_5

🎓Unlock everything free for 14 days

Unlock everything free for 14 days