Concept: Average Rate of Reaction — the change in concentration of a reactant (or product) per unit time over a finite interval.

For the reaction $C_4H_9Cl + H_2O \rightarrow C_4H_9OH + HCl$, the average rate is:

The negative sign is used because the concentration of butyl chloride decreases with time.

Step 1: Choose an interval.

For example, between $t = 50$ s and $t = 100$ s:

• $\Delta t = 100 - 50 = 50$ s
• $\Delta [C_4H_9Cl] = 0.0820 - 0.0905 = -0.0085$ mol L$^{-1}$

Step 2: Apply the formula.

Step 3: Repeat for other intervals as needed.

Here are the average rates for all consecutive intervals:

Note: NCERT's printed Table 3.1 shows a 400–500 s row (rate $1.04 \times 10^{-4}$) computed from a $t = 500$ s concentration (0.0335) that does not appear in the question's own data listing; the 400–700 s row above is computed from the data exactly as printed in the question.

The average rate of a reaction tells us how fast the concentration of a reactant or product changes over a specific time interval. For the reaction

$C_4H_9Cl + H_2O \rightarrow C_4H_9OH + HCl$,

butyl chloride ($C_4H_9Cl$) is consumed. Since water is in large excess, its concentration hardly changes, so we track only $[C_4H_9Cl]$.

The average rate over a time interval $\Delta t = t_2 - t_1$ is defined as:

The negative sign ensures the rate is positive (concentration of reactant decreases). We’ll compute this for each consecutive time interval given in the table.

1. Interval 0 s to 50 s

   $\Delta t = 50 - 0 = 50$ s

   $\Delta [C_4H_9Cl] = 0.0905 - 0.100 = -0.0095$ mol L$^{-1}$

   Rate $= -\frac{-0.0095}{50} = \frac{0.0095}{50} = 1.90 \times 10^{-4}$ mol L$^{-1}$ s$^{-1}$

2. Interval 50 s to 100 s

   $\Delta t = 100 - 50 = 50$ s

   $\Delta [C_4H_9Cl] = 0.0820 - 0.0905 = -0.0085$ mol L$^{-1}$

   Rate $= -\frac{-0.0085}{50} = \frac{0.0085}{50} = 1.70 \times 10^{-4}$ mol L$^{-1}$ s$^{-1}$

3. Interval 100 s to 150 s

   $\Delta t = 150 - 100 = 50$ s

   $\Delta [C_4H_9Cl] = 0.0741 - 0.0820 = -0.0079$ mol L$^{-1}$

   Rate $= -\frac{-0.0079}{50} = \frac{0.0079}{50} = 1.58 \times 10^{-4}$ mol L$^{-1}$ s$^{-1}$

4. Interval 150 s to 200 s

   $\Delta t = 200 - 150 = 50$ s

   $\Delta [C_4H_9Cl] = 0.0671 - 0.0741 = -0.0070$ mol L$^{-1}$

   Rate $= -\frac{-0.0070}{50} = \frac{0.0070}{50} = 1.40 \times 10^{-4}$ mol L$^{-1}$ s$^{-1}$

5. Interval 200 s to 300 s

   $\Delta t = 300 - 200 = 100$ s

   $\Delta [C_4H_9Cl] = 0.0549 - 0.0671 = -0.0122$ mol L$^{-1}$

   Rate $= -\frac{-0.0122}{100} = \frac{0.0122}{100} = 1.22 \times 10^{-4}$ mol L$^{-1}$ s$^{-1}$

6. Interval 300 s to 400 s

   $\Delta t = 400 - 300 = 100$ s

   $\Delta [C_4H_9Cl] = 0.0439 - 0.0549 = -0.0110$ mol L$^{-1}$

   Rate $= -\frac{-0.0110}{100} = \frac{0.0110}{100} = 1.10 \times 10^{-4}$ mol L$^{-1}$ s$^{-1}$

7. Interval 400 s to 700 s

   $\Delta t = 700 - 400 = 300$ s

   $\Delta [C_4H_9Cl] = 0.0210 - 0.0439 = -0.0229$ mol L$^{-1}$

   Rate $= -\frac{-0.0229}{300} = \frac{0.0229}{300} = 7.63 \times 10^{-5}$ mol L$^{-1}$ s$^{-1}$

8. Interval 700 s to 800 s $\Delta t = 800 - 700 = 100$ s $\Delta [C_4H_9Cl] = 0.017 - 0.0210 = -0.0040$ mol L$^{-1}$ Rate $= -\frac{-0.0040}{100} = \frac{0.0040}{100} = 4.0 \times 10^{-5}$ mol L$^{-1}$ s$^{-1}$

Method: Average Rate from Concentration–Time Data

The average rate of reaction over a time interval is defined as the negative of the change in concentration of a reactant divided by the change in time.

Steps

1. Identify the reactant whose concentration is given.

   Here, it is $C_4H_9Cl$. Since it is a reactant, its concentration decreases with time.

2. Write the formula for average rate:

The negative sign makes the rate positive (since $\Delta[\text{reactant}]$ is negative).

3. Choose a time interval $(t_1, t_2)$ and read the corresponding concentrations $[C_4H_9Cl]_1$ and $[C_4H_9Cl]_2$ from the table.

4. Calculate:

Example Calculation

Take the interval $t = 0$ s to $t = 50$ s:

• $[C_4H_9Cl]_1 = 0.100\ \text{mol L}^{-1}$ at $t_1 = 0$ s
• $[C_4H_9Cl]_2 = 0.0905\ \text{mol L}^{-1}$ at $t_2 = 50$ s

Result for 0–50 s: $\boxed{1.9 \times 10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}}$

General Note

Repeat the same steps for any other interval (e.g., 50–100 s, 100–150 s, etc.) using the corresponding concentration values. The average rate will generally decrease as time increases because the reactant concentration becomes smaller.

Here are the most common mistakes students make when calculating the average rate of reaction from concentration-time data, along with how to avoid each.

1. Confusing “average rate” with “instantaneous rate”

Mistake:

Students try to draw a tangent or use calculus concepts when the question explicitly asks for the average rate over an interval.

How to avoid:

Remember:

• Average rate = $\frac{-\Delta[\text{reactant}]}{\Delta t}$ over a finite time interval.
• Use the formula directly:

2. Forgetting the negative sign (or misinterpreting it)

Mistake:

Students write the rate as positive without the negative sign, or they put the negative sign in the wrong place.

How to avoid:

• The concentration of reactant decreases with time, so $\Delta[\text{reactant}]$ is negative.
• The rate must be positive, so we use:

• Example: For $t=0$ to $50$ s: $\Delta[\text{C}_4\text{H}_9\text{Cl}] = 0.0905 - 0.100 = -0.0095$ Rate $= -\frac{-0.0095}{50} = +1.9 \times 10^{-4} \ \text{mol L}^{-1} \text{s}^{-1}$

3. Using the wrong time interval (mixing up $t_1$ and $t_2$)

Mistake:

Students calculate $\Delta t$ as $t_1 - t_2$ instead of $t_2 - t_1$, leading to a negative denominator.

How to avoid:

• Always use: $\Delta t = t_{\text{later}} - t_{\text{earlier}}$
• The order in the numerator must match: $[\text{conc}]_{\text{later}} - [\text{conc}]_{\text{earlier}}$

4. Not using the correct units

Mistake:

Students forget to include units or write them incorrectly (e.g., $\text{mol L}^{-1}$ without time, or $\text{s}^{-1}$ without concentration).

How to avoid:

• The unit of average rate is always:

• Check that both concentration and time units are consistent.

5. Computing one overall 0–800 s average instead of interval-by-interval rates

Mistake:

Some students compute a single average rate over the entire 0–800 s span and present that one number as “the” average rate of the reaction.

Why it's wrong:

The question asks for the average rate during different intervals of time — the whole point is to see the rate fall as the reactant is consumed. NCERT's own Solution (Table 3.1) computes the rate for every consecutive interval of the data; that is the expected approach here.

How to avoid:

• Compute $-\frac{\Delta[\text{C}_4\text{H}_9\text{Cl}]}{\Delta t}$ for each consecutive pair of readings, exactly as in Table 3.1.
• Present the results as a table so the decreasing trend is visible.

6. Forgetting that rate changes with time

Mistake:

Students assume the average rate is constant and report only one value.

How to avoid:

• The average rate decreases as the reaction proceeds (because reactant concentration falls).
• Compute every consecutive interval (as NCERT's Table 3.1 does) so the decreasing trend is explicit.

7. Arithmetic errors in subtraction or division

Mistake:

Simple calculation mistakes, especially when dealing with numbers like $0.0905 - 0.100 = -0.0095$ (some write $-0.0095$ as $-0.095$).

How to avoid:

• Write each step clearly.
• Double-check decimal places.
• Use a calculator carefully.

✓ Quick checklist to avoid all mistakes

Final tip:

Always show the formula in your answer before plugging numbers — examiners love seeing that you know the concept, not just the calculation.