The experimental data for decomposition of $N_2O_5$ $[2N_2O_5 \rightarrow 4NO_2 + O_2]$ in gas phase at 318 K are given below: | $t$/s | 0 | 400 | 800 | 1200 | 1600 | 2000 | 2400 | 2800 | 3200 | |---|---|---|---|---|---|---|---|---|---| | $10^2\times[N_2O_5]$/$\text{mol L}^{-1}$ | 1.63 | 1.36 | 1.14 | 0.93 | 0.78 | 0.64 | 0.53 | 0.43 | 0.35 | (i) Plot $[N_2O_5]$ against $t$. (ii) Find the half-life period for the reaction. (iii) Draw a graph between $\log[N_2O_5]$ and $t$. (iv) What is the rate law? (v) Calculate the rate constant. (vi) Calculate the half-life period from $k$ and compare it with (ii).

The experimental data for decomposition of N₂O₅ [2N₂O₅ arrow 4NO₂ + O₂] in gas phase at 318 K are given below:

Exercises · 3.15

The experimental data for decomposition of $N_2O_5$

$[2N_2O_5 \rightarrow 4NO_2 + O_2]$

in gas phase at 318 K are given below:

$t$ /s	0	400	800	1200	1600	2000	2400	2800	3200
$10^2\times[N_2O_5]$ / $\text{mol L}^{-1}$	1.63	1.36	1.14	0.93	0.78	0.64	0.53	0.43	0.35

Plot $[N_2O_5]$ against $t$ .
Find the half-life period for the reaction.
Draw a graph between $\log[N_2O_5]$ and $t$ .
What is the rate law?
Calculate the rate constant.
Calculate the half-life period from $k$ and compare it with (ii).

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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):

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

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The plot of $\log[N_2O_5]$ against $t$ is a straight line, so the reaction is first order: $\text{Rate} = k[N_2O_5]$ . The rate constant is $k \approx 4.81 \times 10^{-4}\ \text{s}^{-1}$ , giving $t_{1/2} \approx 1441\ \text{s}$ from $k$ , in good agreement with $\approx 1450\ \text{s}$ read from the graph.

(i) Plot $[N_2O_5]$ vs $t$ . Plotting the concentration against time gives a smooth downward curve (steep at first, then flattening). It is not a straight line, so the reaction is not zero order.

(ii) Half-life from the graph. The initial concentration is $[N_2O_5]_0 = 1.63 \times 10^{-2}\ \text{mol L}^{-1}$ , so half of it is $0.815 \times 10^{-2}\ \text{mol L}^{-1}$ . Reading the curve, this value is reached at about $t_{1/2} \approx 1450\ \text{s}$ .

(iii) Plot $\log[N_2O_5]$ vs $t$ . Computing $\log[N_2O_5]$ (with $[N_2O_5]$ in $\text{mol L}^{-1}$ ):

$t$ /s	$[N_2O_5]$ / $\text{mol L}^{-1}$	$\log[N_2O_5]$
0	$1.63 \times 10^{-2}$	$-1.788$
400	$1.36 \times 10^{-2}$	$-1.866$
800	$1.14 \times 10^{-2}$	$-1.943$
1200	$0.93 \times 10^{-2}$	$-2.031$
1600	$0.78 \times 10^{-2}$	$-2.108$
2000	$0.64 \times 10^{-2}$	$-2.194$
2400	$0.53 \times 10^{-2}$	$-2.276$
2800	$0.43 \times 10^{-2}$	$-2.367$
3200	$0.35 \times 10^{-2}$	$-2.456$

These points fall on a straight line, which confirms first-order kinetics.

(iv) Rate law. A linear $\log[N_2O_5]$ vs $t$ plot means the reaction is first order in $N_2O_5$ :

\text{Rate} = k[N_2O_5]

(v) Rate constant. For a first-order reaction, $\log[N_2O_5] = \log[N_2O_5]_0 - \dfrac{k}{2.303}\,t$ , so the slope is $-\dfrac{k}{2.303}$ . Using the endpoints $t = 0$ and $t = 3200\ \text{s}$ : …

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Method: Integrated Rate Law Analysis for First-Order Reactions

This method uses the integrated rate equation for a first-order reaction to determine the rate law, rate constant, and half-life from concentration–time data.

Steps

1. Plot $[N_2O_5]$ vs $t$ (part i)

Plot the given data with time $t$ (s) on the x-axis and $[N_2O_5]$ (mol L⁻¹) on the y-axis.
The curve shows a continuous decrease in concentration, typical of a first-order decay.

2. Test for first-order kinetics — plot $\log[N_2O_5]$ vs $t$ (part iii)

For a first-order reaction:

\log[N_2O_5] = \log[N_2O_5]_0 - \frac{k}{2.303} t

Calculate $\log_{10}[N_2O_5]$ for each time point.
Plot $\log[N_2O_5]$ against $t$ .
If the graph is a straight line, the reaction is first order.

3. Determine the rate law (part iv)

From the straight-line plot, the reaction follows:

\text{Rate} = k [N_2O_5]

This is the rate law.

4. Calculate the rate constant $k$ (part v)

Slope of the $\log[N_2O_5]$ vs $t$ graph = $-\dfrac{k}{2.303}$
Pick two points far apart on the best-fit line (not data points necessarily):

\text{slope} = \frac{\log[N_2O_5]_2 - \log[N_2O_5]_1}{t_2 - t_1}

Then:

k = -2.303 \times \text{slope}

Using the endpoints of the best-fit line ( $\log[N_2O_5]$ falls from $-1.788$ at $t = 0$ to $-2.456$ at $t = 3200$ s):

\text{slope} = \frac{-2.456 - (-1.788)}{3200} = \frac{-0.668}{3200} = -2.0875 \times 10^{-4} \ \text{s}^{-1}

Result: $k = 2.303 \times 2.0875 \times 10^{-4} = 4.81 \times 10^{-4} \ \text{s}^{-1}$

5. Calculate half-life from $k$ (part vi)

For a first-order reaction:

t_{1/2} = \frac{0.693}{k}

Substitute the $k$ from step 4:

t_{1/2} = \frac{0.693}{4.81 \times 10^{-4}} \approx 1441 \ \text{s}

6. Find half-life directly from $[N_2O_5]$ vs $t$ graph (part ii)

On the $[N_2O_5]$ vs $t$ plot, find the time when concentration falls to half of its initial value ( $1.63 \to 0.815$ ). …

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Common Mistakes Students Make on "Average Rate of Reaction" (N₂O₅ Decomposition)

Mistake 1: Confusing Average Rate with Instantaneous Rate

The error: Students often calculate the average rate over a large time interval (like 0–3200 s) and treat it as the rate constant or instantaneous rate.

Why it's wrong: The average rate changes with time because concentration decreases. The rate constant $k$ is a constant at a given temperature — it does not equal the average rate.

How to avoid:

Average rate = $-\frac{\Delta[N_2O_5]}{\Delta t}$ over a specific interval.
For rate law, use integrated rate equation or plot $\log[N_2O_5]$ vs $t$ to find $k$ .

Mistake 2: Plotting $[N_2O_5]$ vs $t$ and Calling it a Straight Line

The error: Students assume the graph is linear and try to find slope directly.

Why it's wrong: For a first-order reaction, $[N_2O_5]$ vs $t$ is exponential decay — a curve, not a straight line.

How to avoid:

Plot $[N_2O_5]$ vs $t$ — you'll get a smooth decreasing curve.
Only $\log[N_2O_5]$ vs $t$ gives a straight line for first-order kinetics.

Mistake 3: Using Wrong Formula for Half-Life

The error: Students use $t_{1/2} = \frac{[A]_0}{2k}$ (zero-order formula) or $t_{1/2} = \frac{1}{k[A]_0}$ (second-order formula).

Why it's wrong: This reaction is first-order (as confirmed by the log plot being linear).

How to avoid:

For first-order:

t_{1/2} = \frac{0.693}{k}

Half-life is independent of initial concentration for first-order reactions.

Mistake 4: Reading Half-Life Incorrectly from the Graph

The error: Students pick any two points where concentration halves but don't check if the time interval is constant.

Why it's wrong: For first-order, $t_{1/2}$ should be constant throughout. If you pick $[N_2O_5]$ = 1.63 → 0.815 (half), the time should equal $t_{1/2}$ from any other pair (e.g., 1.36 → 0.68).

How to avoid:

From the table:
- At $t=0$ , $[N_2O_5] = 1.63 \times 10^{-2}$
- Half of that = $0.815 \times 10^{-2}$ — this value is not in the table, so interpolate or use the $k$ value.
Alternatively, find $k$ first, then calculate $t_{1/2}$ .

Mistake 5: Forgetting to Convert Units or Scale

The error: Students treat $10^2 \times [N_2O_5]$ as the actual concentration.

Why it's wrong: The table gives $10^2 \times [N_2O_5]$ , so actual $[N_2O_5]$ = (table value) $\times 10^{-2}$ mol L⁻¹.

How to avoid:

Always write:

[N_2O_5] = \frac{\text{table value}}{100}

When plotting or calculating $\log[N_2O_5]$ , use the actual concentration.

Mistake 6: Writing the Rate Law Incorrectly

The error: Students write $\text{Rate} = k[N_2O_5]^2$ or forget the stoichiometric coefficient. …

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

KEAM 2026Set eng-2026-04214 marksMCQ
$[A]$
›Reveal solutionSolution

Between 20 s and 40 s, $[A]$ falls $0.46\to0.26$ , so rate $=0.20/20=0.01$ mol L $^{-1}$ s $^{-1}$ .

From the table, $[A]_{20}=0.46$ and $[A]_{40}=0.26$ mol lit $^{-1}$ . The average rate of consumption of A over this interval is …
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KEAM 2022Set eng-2022-P1-A14 marksMCQ
$3A \rightarrow$
›Reveal solutionSolution
The rate of decrease of [A] is 0.3/20 = 0.015 mol L^{-1} min^{-1}.

Concept and Intuition

The rate of decrease in [A] is simply -\Delta[A]/\Delta t, the change in concentration of A over the time interval; the stoichiometric coefficient 3 is only used if one asks for the overall reaction rate, not the rate of disappearance of A itself.

Step-by-Step Solution

\Delta[A] = 0.4 - 0.1 = 0.3 mol L^{-1}.

Rate of decrease = 0.3 / 20 min. …
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Time /s	0	10	20	30	40	50	60
$[A]$ /mol lit $^{-1}$	0.65	0.55	0.46	0.38	0.26	0.20	0.13

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