Concept: First-Order Rate Constant from Total Pressure

For a first-order gas-phase reaction at constant volume, the rate constant is found using the integrated rate law:

$k = \frac{2.303}{t} \log \frac{P_0}{P_{N_2O_5}}$

where $P_0$ is initial pressure and $P_{N_2O_5}$ is the partial pressure of $N_2O_5$ remaining at time $t$.

Step 1 – Initial pressure

$P_0 = 0.5$ atm (only $N_2O_5$ present).

Step 2 – Pressure change

From stoichiometry: $2N_2O_5 \rightarrow 2N_2O_4 + O_2$

If $p$ is the decrease in pressure of $N_2O_5$, then $N_2O_4$ increases by $p$ and $O_2$ by $p/2$.

Total pressure at time $t$: $P_t = (0.5 - p) + p + p/2 = 0.5 + p/2$

Given $P_t = 0.512$ atm at $t = 100$ s, so $0.5 + p/2 = 0.512 \Rightarrow p = 0.024$ atm.

So $P_{N_2O_5} = 0.5 - p = 0.5 - 0.024 = 0.476$ atm (equivalently, $P_{N_2O_5} = 3P_0 - 2P_t = 1.5 - 1.024 = 0.476$ atm — the form NCERT's own Solution uses).

Step 3 – Apply the first-order formula (NCERT's route)

$k = \frac{2.303}{100} \log \frac{0.5}{0.476}$ …

The key to this problem is understanding that in a gas-phase reaction at constant volume, total pressure is proportional to the total number of moles. So as the reaction proceeds, the pressure changes — and that change tells us how much reactant has decomposed.

For a first-order reaction, the rate constant $k$ is given by:

where $[A]_0$ and $[A]_t$ are the concentrations (or partial pressures) of the reactant at time 0 and time $t$.

Here, we don't have partial pressures directly — only total pressure. But we can use the stoichiometry to relate them.

Let’s work through it step by step.

1. Write the reaction and initial conditions. The reaction is:

At $t = 0$, only $N_2O_5$ is present. Initial total pressure $P_0 = 0.5 \, \text{atm}$.

Let the initial partial pressure of $N_2O_5$ be $p_i = 0.5 \, \text{atm}$.

2. Define the progress variable. Let $p$ be the decrease in partial pressure of $N_2O_5$ at time $t$. Then:

Initial: $0.5$ atm of $N_2O_5$, $0$ of others.

At time $t$:

• $N_2O_5$: $0.5 - p$
• $N_2O_4$: $p$ (since 2 moles of $N_2O_4$ form from 2 moles of $N_2O_5$, so the pressure increase of $N_2O_4$ equals the decrease of $N_2O_5$)
• $O_2$: $p/2$ (since 1 mole of $O_2$ forms from 2 moles of $N_2O_5$)

3. Express total pressure at time $t$. Total pressure $P_t$ = sum of partial pressures:

Given $P_t = 0.512 \, \text{atm}$ at $t = 100 \, \text{s}$, we have:

4. Find the partial pressure of $N_2O_5$ at $t = 100 \, \text{s}$.

(This matches NCERT's own intermediate, $p_{N_2O_5} = 1.5 - 2 \times 0.512 = 0.476$ atm.)

5. Apply the first-order rate law, following NCERT's printed evaluation. For a first-order reaction:

Here $[A]_0 = 0.5 \, \text{atm}$, $[A]_t = 0.476 \, \text{atm}$, $t = 100 \, \text{s}$:

$k = \frac{2.303}{100} \log \frac{0.5}{0.476}$ …

Method: First-Order Integrated Rate Law (Using Total Pressure)

This is a first-order gas-phase reaction where we track total pressure changes. Since volume and temperature are constant, pressure is proportional to the number of moles.

Step 1: Write the reaction and initial conditions

At $t = 0$: only $N_2O_5$ is present.

Initial pressure of $N_2O_5$, $P_0 = 0.5$ atm.

Step 2: Set up the pressure relationships

Let $p$ be the pressure of $N_2O_5$ that has decomposed at time $t$.

Total pressure at time $t$:

Step 3: Find $p$ from the given data

At $t = 100$ s, $P_t = 0.512$ atm and $P_0 = 0.5$ atm.

Step 4: Find pressure of $N_2O_5$ remaining

Step 5: Apply the first-order rate law

For a first-order reaction:

Substitute values:

NCERT's printed Solution evaluates the logarithm as $\log(0.5/0.476) = 0.0216$, so: …

Here are the most common mistakes students make when solving this exact problem, along with how to avoid each one.

1. Confusing Total Pressure with Partial Pressure of Reactant

The Mistake:

Students plug the total pressure values (0.5 atm and 0.512 atm) directly into the first-order rate equation for partial pressure:

This is wrong because $P_t$ in the formula refers to the partial pressure of $N_2O_5$ at time $t$, not the total pressure of the mixture.

How to Avoid:

Always ask: “What does the data actually measure?” Here, the manometer reads total pressure of the gas mixture. You must first convert total pressure into the partial pressure of the reactant using stoichiometry.

2. Forgetting the Stoichiometric Mole-Pressure Relationship

The Mistake:

Even when students realise they need $P_{N_2O_5}$, they often guess the relation incorrectly (e.g., assuming $P_{total} = P_{N_2O_5}$ or using a wrong factor).

How to Avoid:

Use the ICE table method (Initial, Change, Equilibrium) in terms of pressure:

Total pressure at time t:

So $x = P_{total} - 0.5$.

Then the partial pressure of $N_2O_5$ at time t is:

Key result:

At $t = 100$ s, $P_{total} = 0.512$ atm

$\Rightarrow P_{N_2O_5} = 1.5 - 2(0.512) = 0.476$ atm

3. Using the Wrong Order Formula

The Mistake:

Applying the zero-order or second-order integrated rate equation because the problem “looks like” a pressure problem.

How to Avoid:

The problem explicitly states first order thermal decomposition. Always trust the given order. Use:

where $a$ = initial pressure of $N_2O_5$ (0.5 atm) and $a-x$ = pressure at time t (0.476 atm).

4. Arithmetic Errors in the Log Calculation

The Mistake:

Miscalculating $\frac{0.5}{0.476}$ or using $\log$ instead of $\ln$ (or vice versa) without the conversion factor 2.303.

How to Avoid:

• Compute carefully: $\frac{0.5}{0.476} \approx 1.0504$
• NCERT's printed Solution evaluates $\log(0.5/0.476) = 0.0216$
• Then $k = \frac{2.303}{100} \times 0.0216$ …