Q.The initial concentration of $N_2O_5$ in the following first order reaction $N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g)$ was $1.24\times10^{-2}\ \text{mol L}^{-1}$ at 318 K. The concentration of $N_2O_5$ after 60 minutes was $0.20\times10^{-2}\ \text{mol L}^{-1}$. Calculate the rate constant of the reaction at 318 K.

Concept: Average Rate Of Reaction — For a first-order reaction, the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.

Step 1: Identify the given values.

$[A]_0 = 1.24 \times 10^{-2}\ \text{mol L}^{-1}$, $[A]_t = 0.20 \times 10^{-2}\ \text{mol L}^{-1}$, $t = 60\ \text{min}$.

Step 2: Apply the first-order integrated rate law.

Step 3: Simplify the ratio and compute.

The key to solving this lies in understanding what the average rate of reaction actually tells us — and why, for a first-order reaction, we don’t use the average rate directly. Instead, we use the integrated rate law, which relates concentration to time in a way that accounts for the fact that the rate continuously changes as the reactant is used up.

For a first-order reaction like $N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$, the rate at any instant is proportional to the concentration of $N_2O_5$ remaining. That proportionality constant is $k$, the rate constant we need. The beauty of the integrated form is that it gives a straight line when $\log[\text{reactant}]$ is plotted against time — and from any single pair of concentration and time, we can calculate $k$ directly.

Let’s walk through it step by step.

1. Identify the order and the correct formula. The problem states this is a first-order reaction. For a first-order process, the integrated rate law is:

where $[A]_0$ is the initial concentration, $[A]_t$ is the concentration after time $t$, and $k$ is the rate constant. This formula comes from integrating $-\frac{d[A]}{dt} = k[A]$.

2. Write down the given data clearly.

   • Initial concentration, $[N_2O_5]_0 = 1.24 \times 10^{-2} \ \text{mol L}^{-1}$
   • Concentration after 60 minutes, $[N_2O_5]_t = 0.20 \times 10^{-2} \ \text{mol L}^{-1}$
   • Time, $t = 60 \ \text{min}$

   Notice that both concentrations are in the same units ($\text{mol L}^{-1}$) and have the same power of 10, which will simplify the ratio.

3. Set up the ratio inside the logarithm.

The $10^{-2}$ cancels out neatly. Now compute:

4. Take the logarithm (base 10).

You can recall that $\log(6.2) \approx 0.7924$ (since $\log(6) \approx 0.7782$ and $\log(6.3) \approx 0.7993$, so 6.2 is about halfway). More precisely, using a calculator or log table: $\log(6.2) = 0.7924$.

5. Plug into the formula.

First, compute $\frac{2.303}{60}$:

Then multiply by $0.7924$:

6. Express in proper scientific notation.

Rounding to three significant figures (since the given concentrations have three significant figures: $1.24$ and $0.20$), we get:

Method: Integrated Rate Law for a First-Order Reaction

For a first-order reaction, the rate constant $k$ is found using the integrated rate equation:

Steps

1. Write the first-order integrated rate law For a reaction $A \rightarrow \text{products}$:

2. Identify the given values

   • Initial concentration, $[N_2O_5]_0 = 1.24 \times 10^{-2}\ \text{mol L}^{-1}$
   • Concentration after time $t$, $[N_2O_5]_t = 0.20 \times 10^{-2}\ \text{mol L}^{-1}$
   • Time, $t = 60\ \text{minutes}$

3. Substitute into the formula

4. Simplify the ratio

5. Calculate the logarithm

6. Compute $k$

Final answer:

Why this works: The integrated rate law directly relates concentration change to time for a first-order reaction, giving the rate constant without needing initial rate data.

Common Mistakes Students Make on This Problem

Mistake 1: Using the Wrong Formula for First-Order Reactions

The error: Students often confuse the integrated rate laws. For a first-order reaction, the correct formula is:

Some mistakenly use the zero-order or second-order formula, or write the log term upside down.

How to avoid: Memorise the three distinct forms of the first-order integrated rate law:

• $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ (most common for calculations)
• $\ln \frac{[A]_0}{[A]_t} = kt$
• $[A]_t = [A]_0 e^{-kt}$

Always check: first-order → log of concentration ratio is directly proportional to time.

Mistake 2: Fumbling the Time Unit Instead of Committing to One

The error: Time is given in minutes. Some students half-convert — plugging $t = 60$ but labelling the answer $\text{s}^{-1}$, or converting to $3600$ s but then quoting a min$^{-1}$ value — and end up with a number/unit mismatch.

How to avoid: Either unit choice is valid as long as it is used consistently. NCERT's own printed Solution works entirely in minutes and reports $k = 0.0304\ \text{min}^{-1}$ — so working in minutes is the expected route here, not an error. If you do want the SI form, convert at the end:

Pro tip: Write the unit of $k$ explicitly in your final answer — it forces you to check that the number and unit belong together.

Mistake 3: Incorrect Substitution of Concentrations

The error: Students sometimes substitute $[A]_0$ and $[A]_t$ in the wrong places, e.g.:

This gives a negative value of $k$, which is impossible for a rate constant.

How to avoid: Remember: initial concentration goes on top because $[A]_0 > [A]_t$ for a reactant. The ratio $\frac{[A]_0}{[A]_t} > 1$, so $\log$ is positive.

Mistake 4: Arithmetic Errors with Powers of 10

The error: The concentrations are $1.24 \times 10^{-2}$ and $0.20 \times 10^{-2}$. Students often mishandle the $10^{-2}$ factor when computing the ratio:

But some incorrectly write $6.2 \times 10^{0}$ or mess up the subtraction of exponents.

How to avoid: Cancel the $10^{-2}$ factor explicitly on paper before calculating:

Mistake 5: Using $\log$ When the Formula Requires $\ln$ (or Vice Versa)

The error: The formula $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ uses base-10 log. Some students use natural log ($\ln$) without the 2.303 conversion factor.

How to avoid: Remember the relationship:

• If you use $\log_{10}$, include 2.303.
• If you use $\ln$, omit 2.303: $k = \frac{1}{t} \ln \frac{[A]_0}{[A]_t}$

Mistake 6: Rounding Too Early

The error: Students round intermediate values (e.g., $\log 6.2 = 0.79$ instead of $0.7924$), shifting the last digit of the final answer away from the correct $0.0304\ \text{min}^{-1}$.

How to avoid: Keep at least 3–4 significant figures in intermediate steps and round only the final answer. The given concentrations ($1.24$, $0.20 \times 10^{-2}$) support quoting the answer to 3 significant figures: $k = 3.04 \times 10^{-2}\ \text{min}^{-1}$.

Correct Solution (Quick Reference)

(As a labelled conversion to SI units: $k = 0.0304/60 = 5.07 \times 10^{-4}\ \text{s}^{-1}$.)