Q.The decomposition of $N_2O_5$ in $CCl_4$ at 318 K has been studied by monitoring the concentration of $N_2O_5$ in the solution. Initially the concentration of $N_2O_5$ is $2.33\ \text{mol L}^{-1}$ and after 184 minutes, it is reduced to $2.08\ \text{mol L}^{-1}$. The reaction takes place according to the equation
$2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$
Calculate the average rate of this reaction in terms of hours, minutes and seconds. What is the rate of production of $NO_2$ during this period?

Concept: Average Rate of Reaction — the change in concentration of a reactant or product per unit time, divided by its stoichiometric coefficient.

Step 1: Average rate of reaction (using $N_2O_5$)

The average rate of reaction is

Step 2: Convert to hours and seconds

• Per hour: $6.79 \times 10^{-4} \times 60 = 4.074 \times 10^{-2} \approx 4.07 \times 10^{-2}\ \text{mol L}^{-1} \text{h}^{-1}$ (NCERT's printed value; the unrounded chain $0.125/184 \times 60$ gives $4.08 \times 10^{-2}$)
• Per second: $6.79 \times 10^{-4} \div 60 = 1.13 \times 10^{-5}\ \text{mol L}^{-1} \text{s}^{-1}$

Step 3: Rate of production of $NO_2$

From the stoichiometry: $2N_2O_5 \rightarrow 4NO_2$, so …

The key idea here is that the average rate of reaction is defined as the change in concentration of any reactant or product, divided by its stoichiometric coefficient and by the time interval. For a reactant, the change is negative (it decreases), so we take the negative of that change to get a positive rate. For a product, the change is positive directly.

We are given the decomposition of $N_2O_5$:

The concentration of $N_2O_5$ drops from $2.33\ \text{mol L}^{-1}$ to $2.08\ \text{mol L}^{-1}$ over 184 minutes. That’s a decrease of $0.25\ \text{mol L}^{-1}$.

Let’s work through this step by step.

1. Write the definition of average rate of reaction For a general reaction $aA \rightarrow bB$, the average rate is:

Here, $a = 2$ for $N_2O_5$, so:

2. Calculate $\Delta[N_2O_5]$

The negative sign shows it’s being consumed.

3. Plug into the rate expression

So the numerical part is $0.125\ \text{mol L}^{-1}$ divided by the time interval.

4. Express the rate in different time units

   The time interval is 184 minutes. We need the rate in hours, minutes, and seconds.

   • Per minute:

• Per hour: Since 1 hour = 60 minutes, multiply the per-minute rate by 60:

• Per second: Since 1 minute = 60 seconds, divide the per-minute rate by 60:

5. Find the rate of production of $NO_2$ From the stoichiometry: $2\ \text{mol}\ N_2O_5$ produces $4\ \text{mol}\ NO_2$. So the rate of appearance of $NO_2$ is twice the rate of disappearance of $N_2O_5$ (since $4/2 = 2$). But careful: the average rate of reaction we just calculated is already $-\frac{1}{2}\frac{\Delta[N_2O_5]}{\Delta t}$. The rate of production of $NO_2$ is:

Because the average rate = $\frac{1}{4}\frac{\Delta[NO_2]}{\Delta t}$.

So: …

Method: Average Rate of Reaction from Concentration Change

This method uses the definition of average rate for a chemical reaction — the change in concentration of a reactant or product divided by the time interval, adjusted for the stoichiometric coefficient.

Step 1: Write the formula for average rate

For a reactant $R$ with stoichiometric coefficient $a$ in $aR \rightarrow \text{products}$:

The negative sign is used because the concentration of a reactant decreases with time, making the rate positive.

Step 2: Identify the given data

• Initial concentration of $N_2O_5$: $[N_2O_5]_i = 2.33\ \text{mol L}^{-1}$
• Final concentration: $[N_2O_5]_f = 2.08\ \text{mol L}^{-1}$
• Time interval: $\Delta t = 184\ \text{minutes}$
• Coefficient of $N_2O_5$ in $2N_2O_5 \rightarrow 4NO_2 + O_2$ is $a = 2$

Step 3: Calculate $\Delta [N_2O_5]$

Step 4: Compute average rate of reaction

Step 5: Convert to different time units

In hours ($1\ \text{h} = 60\ \text{min}$):

In seconds ($1\ \text{min} = 60\ \text{s}$):

Step 6: Find rate of production of $NO_2$

From the balanced equation:

The average rate of reaction already equals $\frac{1}{4}\frac{d[NO_2]}{dt}$, so:

$\text{Rate of production of } NO_2 = 4 \times \text{Rate of reaction}$ …

Common Mistakes & How to Avoid Them

1. Forgetting the Stoichiometric Coefficient in the Rate Definition

The Mistake:

Students often write the average rate as simply:

This is incorrect for the given reaction.

Why it's wrong:

The rate of reaction is defined as:

The coefficient 2 comes from the balanced equation: $2N_2O_5 \rightarrow 4NO_2 + O_2$

How to Avoid:

Always write the general definition first:

where $a$ and $b$ are stoichiometric coefficients. Then plug in numbers.

2. Using Wrong Sign for Reactant Concentration Change

The Mistake:

Writing $\Delta[N_2O_5] = 2.08 - 2.33 = -0.25$ and then dropping the negative sign without understanding why.

Why it's wrong:

The negative sign indicates concentration is decreasing. The rate must be positive, so we use:

How to Avoid:

• Always compute $\Delta[\text{Reactant}] = [\text{Final}] - [\text{Initial}]$
• Then apply the minus sign from the definition — two negatives make a positive rate.

3. Confusing Rate of Reaction with Rate of Production

The Mistake:

Stating that the rate of production of $NO_2$ equals the rate of reaction.

Why it's wrong:

From stoichiometry:

So:

How to Avoid:

• Write the relationship explicitly: $\displaystyle \frac{\Delta[NO_2]}{\Delta t} = \frac{4}{2} \times \left(-\frac{\Delta[N_2O_5]}{\Delta t}\right)$
• Or simply: Rate of production = (coefficient ratio) × Rate of reaction

4. Unit Conversion Errors

The Mistake:

Calculating rate in mol L⁻¹ min⁻¹ and then incorrectly converting to hours or seconds.

Common errors:

• Multiplying by 60 instead of dividing (or vice versa)
• Forgetting that time is in the denominator

Correct conversions:

• $\text{mol L}^{-1} \text{min}^{-1} \times 60 = \text{mol L}^{-1} \text{h}^{-1}$
• $\text{mol L}^{-1} \text{min}^{-1} \div 60 = \text{mol L}^{-1} \text{s}^{-1}$

How to Avoid:

• Treat units like numbers: cancel and multiply
• Write the conversion factor clearly: …