Consider a first order gas phase decomposition reaction given below: $A(g) \rightarrow B(g) + C(g)$ The initial pressure of the system before decomposition of A was $p_i$. After lapse of time $t$, total pressure of the system increased by $x$ units and became $p_t$. The rate constant $k$ for the reaction is given as _________. (i) $k = \frac{2.303}{t} \log \frac{p_i}{p_i - x}$ (ii) $k = \frac{2.303}{t} \log \frac{p_i}{2p_i - p_t}$ (iii) $k = \frac{2.303}{t} \log \frac{p_i}{2p_i + p_t}$ (iv) $k = \frac{2.303}{t} \log \frac{p_i}{p_i + x}$

Consider a first order gas phase decomposition reaction given below: A(g) arrow B(g) + C(g) The initial pressu

NCERT Exemplar · Q5

A(g) \rightarrow B(g) + C(g)

(i)

k = \frac{2.303}{t} \log \frac{p_i}{p_i - x}

(ii)

k = \frac{2.303}{t} \log \frac{p_i}{2p_i - p_t}

(iii)

k = \frac{2.303}{t} \log \frac{p_i}{2p_i + p_t}

(iv)

k = \frac{2.303}{t} \log \frac{p_i}{p_i + x}

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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.

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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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For a first-order gas-phase reaction, the pressure increases because each mole of reactant produces two moles of products. The key is to relate the partial pressure of A remaining to the total pressure, then substitute into the first-order rate law. The correct expression is $k = \frac{2.303}{t} \log \frac{p_i}{2p_i - p_t}$ .

The average rate of a reaction tells us how fast reactants are consumed or products are formed. For a first-order reaction, the rate depends linearly on the concentration (or, for gases, the partial pressure) of the reactant. The integrated rate law is:

k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}

For gases at constant temperature and volume, partial pressure is proportional to concentration (from $PV = nRT$ ). So we can use pressures directly — but we must be careful: the total pressure changes because the number of moles changes.

Here, one mole of $A$ decomposes into one mole of $B$ and one mole of $C$ . So for every mole of $A$ that reacts, the total number of moles increases by 1. This means the total pressure increases as the reaction proceeds.

Let’s work through it step by step.

Set up the initial condition.

Initially, only $A$ is present at pressure $p_i$ . So the initial partial pressure of $A$ is $p_i$ , and the initial total pressure is also $p_i$ .
Define the change.

Let the decrease in pressure of $A$ at time $t$ be $y$ units. Since the stoichiometry is 1:1:1, when $y$ pressure of $A$ decomposes, $y$ pressure of $B$ and $y$ pressure of $C$ are formed. So at time $t$ :
- Partial pressure of $A$ remaining: $p_A = p_i - y$
- Partial pressure of $B$ : $p_B = y$
- Partial pressure of $C$ : $p_C = y$
Relate total pressure to the change.

The total pressure at time $t$ is:

p_t = p_A + p_B + p_C = (p_i - y) + y + y = p_i + y

So $y = p_t - p_i$ .

The problem also says the total pressure increased by $x$ units, so $x = p_t - p_i$ . That means $y = x$ . We’ll keep using $y$ for clarity.

Express the remaining pressure of A. From step 2:

p_A = p_i - y = p_i - (p_t - p_i) = 2p_i - p_t

This is the key relation: the partial pressure of A at time $t$ is $2p_i - p_t$ . …

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Method: First-Order Integrated Rate Law Using Partial Pressures

Concept

For a first-order gas-phase reaction, the rate constant is given by:

k = \frac{2.303}{t} \log \frac{[\text{A}]_0}{[\text{A}]_t}

In terms of pressure, concentration is proportional to partial pressure of A.

Step-by-step

Step 1: Write the reaction and initial conditions

A(g) \rightarrow B(g) + C(g)

Initial pressure of A = $p_i$
B and C are initially absent.

Step 2: Express pressure changes

Let $p$ = pressure of A that has decomposed at time $t$ .

Species	Initial pressure	Pressure at time $t$
A	$p_i$	$p_i - p$
B	0	$p$
C	0	$p$

Step 3: Relate total pressure $p_t$ to $p$

p_t = (p_i - p) + p + p = p_i + p

Given that total pressure increased by $x$ units:

p_t = p_i + x \quad\Rightarrow\quad p = x = p_t - p_i

Step 4: Find partial pressure of A at time $t$ , expressed via the measured quantities $p_i$ and $p_t$

p_A = p_i - p = p_i - (p_t - p_i) = 2p_i - p_t

Step 5: Apply first-order rate law

$k = \frac{2.303}{t} \log \frac{p_i}{p_A} = \frac{2.303}{t} \log \frac{p_i}{2p_i - p_t}$ …

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Common Mistakes & How to Avoid Them

Mistake 1: Confusing "increase by $x$ units" with "final pressure $p_t$ "

The error:

Students often treat $x$ and $p_t$ as unrelated or incorrectly assume $p_t = p_i + x$ without checking the stoichiometry.

Why it happens:

The problem says "total pressure increased by $x$ units and became $p_t$ ". This means:

p_t = p_i + x

But many students forget this simple relation and try to derive $k$ using $x$ and $p_i$ directly without linking to $p_t$ .

How to avoid:

Always write the given relation explicitly:

x = p_t - p_i

Then use $p_t$ in the formula, not $x$ alone. The correct expression must involve $p_t$ (or $p_i$ and $x$ together).

Mistake 2: Forgetting stoichiometric pressure changes

The error:

Students assume that if initial pressure of $A$ is $p_i$ , then after time $t$ , pressure of $A$ is simply $p_i - x$ .

Why it happens:

They think the increase in total pressure equals the decrease in pressure of $A$ . But the reaction is:

A(g) \rightarrow B(g) + C(g)

1 mole of $A$ produces 2 moles of gas.
So if $p_A$ decreases by $y$ , total pressure increases by $y$ (from $B$ ) + $y$ (from $C$ ) = $2y$ .

Correct relation:

Let $p_A$ at time $t$ be $p_i - y$ . Then:

$p_B = y$ , $p_C = y$
Total pressure: $p_t = (p_i - y) + y + y = p_i + y$

So $y = p_t - p_i = x$ .

Thus pressure of $A$ at time $t$ is:

p_A = p_i - x = 2p_i - p_t

How to avoid:

Always set up a pressure table:

Species	Initial	At time $t$
$A$	$p_i$	$p_i - y$
$B$	$0$	$y$
$C$	$0$	$y$
Total	$p_i$	$p_i + y$

Then $p_t = p_i + y \Rightarrow y = p_t - p_i$ .

Mistake 3: Using wrong expression for $k$ in the formula

The error:

Students plug $p_A = p_i - x$ directly into:

k = \frac{2.303}{t} \log \frac{p_i}{p_A}

and get option (i) or (iv).

Why it happens:

They don't convert $x$ to $p_t$ or forget that $p_A = 2p_i - p_t$ .

Correct derivation:

k = \frac{2.303}{t} \log \frac{p_i}{p_A} = \frac{2.303}{t} \log \frac{p_i}{p_i - y}

Since $y = p_t - p_i$ :

p_i - y = p_i - (p_t - p_i) = 2p_i - p_t

Thus:

k = \frac{2.303}{t} \log \frac{p_i}{2p_i - p_t}

This matches option (ii).

How to avoid:

Always express $p_A$ in terms of $p_i$ and $p_t$ (not $x$ alone) before substituting into the first-order formula.

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