Concept: Average Rate Of Reaction – the change in concentration per unit time, averaged over a finite interval. For a first-order reaction, the half-life is constant and independent of initial concentration.

(i) Plot $[N_2O_5]$ against $t$

Plot the given data with time on the x-axis and $[N_2O_5]$ on the y-axis. The curve falls exponentially, indicating first-order kinetics.

(ii) Half-life from the graph

Half-life is the time taken for the concentration to fall to half its initial value.

Initial $[N_2O_5] = 1.63 \times 10^{-2} \ \text{mol L}^{-1}$. Half of this is $0.815 \times 10^{-2} \ \text{mol L}^{-1}$.

From the table, this value lies between $t = 1200$ s (0.93) and $t = 1600$ s (0.78).

Reading the smooth decay curve gives $t_{1/2} \approx 1450 \ \text{s}$. (A straight-line interpolation between the two table points would give $\approx 1500$ s — an overestimate, because the exponential decay curve lies below the straight chord.)

(iii) Plot $\log[N_2O_5]$ against $t$

Take $\log_{10}$ of each concentration. A straight line confirms first-order kinetics. Slope $= -\frac{k}{2.303}$.

(iv) Rate law

Since the $\log$ vs $t$ plot is linear, the reaction is first order:

(v) Calculate the rate constant

For a first-order reaction: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$. …

(i) Plot $[N_2O_5]$ vs $t$. Plotting the concentration against time gives a smooth downward curve (steep at first, then flattening). It is not a straight line, so the reaction is not zero order.

(ii) Half-life from the graph. The initial concentration is $[N_2O_5]_0 = 1.63 \times 10^{-2}\ \text{mol L}^{-1}$, so half of it is $0.815 \times 10^{-2}\ \text{mol L}^{-1}$. Reading the curve, this value is reached at about $t_{1/2} \approx 1450\ \text{s}$.

(iii) Plot $\log[N_2O_5]$ vs $t$. Computing $\log[N_2O_5]$ (with $[N_2O_5]$ in $\text{mol L}^{-1}$):

These points fall on a straight line, which confirms first-order kinetics.

(iv) Rate law. A linear $\log[N_2O_5]$ vs $t$ plot means the reaction is first order in $N_2O_5$:

(v) Rate constant. For a first-order reaction, $\log[N_2O_5] = \log[N_2O_5]_0 - \dfrac{k}{2.303}\,t$, so the slope is $-\dfrac{k}{2.303}$. Using the endpoints $t = 0$ and $t = 3200\ \text{s}$: …

Method: Integrated Rate Law Analysis for First-Order Reactions

This method uses the integrated rate equation for a first-order reaction to determine the rate law, rate constant, and half-life from concentration–time data.

Steps

1. Plot $[N_2O_5]$ vs $t$ (part i)

• Plot the given data with time $t$ (s) on the x-axis and $[N_2O_5]$ (mol L⁻¹) on the y-axis.
• The curve shows a continuous decrease in concentration, typical of a first-order decay.

2. Test for first-order kinetics — plot $\log[N_2O_5]$ vs $t$ (part iii)

• For a first-order reaction:

• Calculate $\log_{10}[N_2O_5]$ for each time point.
• Plot $\log[N_2O_5]$ against $t$.
• If the graph is a straight line, the reaction is first order.

3. Determine the rate law (part iv)

• From the straight-line plot, the reaction follows:

• This is the rate law.

4. Calculate the rate constant $k$ (part v)

• Slope of the $\log[N_2O_5]$ vs $t$ graph = $-\dfrac{k}{2.303}$
• Pick two points far apart on the best-fit line (not data points necessarily):

• Then:

• Using the endpoints of the best-fit line ($\log[N_2O_5]$ falls from $-1.788$ at $t = 0$ to $-2.456$ at $t = 3200$ s):

• Result: $k = 2.303 \times 2.0875 \times 10^{-4} = 4.81 \times 10^{-4} \ \text{s}^{-1}$

5. Calculate half-life from $k$ (part vi)

• For a first-order reaction:

• Substitute the $k$ from step 4:

6. Find half-life directly from $[N_2O_5]$ vs $t$ graph (part ii)

• On the $[N_2O_5]$ vs $t$ plot, find the time when concentration falls to half of its initial value ($1.63 \to 0.815$). …

Common Mistakes Students Make on "Average Rate of Reaction" (N₂O₅ Decomposition)

Mistake 1: Confusing Average Rate with Instantaneous Rate

The error: Students often calculate the average rate over a large time interval (like 0–3200 s) and treat it as the rate constant or instantaneous rate.

Why it's wrong: The average rate changes with time because concentration decreases. The rate constant $k$ is a constant at a given temperature — it does not equal the average rate.

How to avoid:

• Average rate = $-\frac{\Delta[N_2O_5]}{\Delta t}$ over a specific interval.
• For rate law, use integrated rate equation or plot $\log[N_2O_5]$ vs $t$ to find $k$.

Mistake 2: Plotting $[N_2O_5]$ vs $t$ and Calling it a Straight Line

The error: Students assume the graph is linear and try to find slope directly.

Why it's wrong: For a first-order reaction, $[N_2O_5]$ vs $t$ is exponential decay — a curve, not a straight line.

How to avoid:

• Plot $[N_2O_5]$ vs $t$ — you'll get a smooth decreasing curve.
• Only $\log[N_2O_5]$ vs $t$ gives a straight line for first-order kinetics.

Mistake 3: Using Wrong Formula for Half-Life

The error: Students use $t_{1/2} = \frac{[A]_0}{2k}$ (zero-order formula) or $t_{1/2} = \frac{1}{k[A]_0}$ (second-order formula).

Why it's wrong: This reaction is first-order (as confirmed by the log plot being linear).

How to avoid:

• For first-order:

• Half-life is independent of initial concentration for first-order reactions.

Mistake 4: Reading Half-Life Incorrectly from the Graph

The error: Students pick any two points where concentration halves but don't check if the time interval is constant.

Why it's wrong: For first-order, $t_{1/2}$ should be constant throughout. If you pick $[N_2O_5]$ = 1.63 → 0.815 (half), the time should equal $t_{1/2}$ from any other pair (e.g., 1.36 → 0.68).

How to avoid:

• From the table:
  • At $t=0$, $[N_2O_5] = 1.63 \times 10^{-2}$
  • Half of that = $0.815 \times 10^{-2}$ — this value is not in the table, so interpolate or use the $k$ value.
• Alternatively, find $k$ first, then calculate $t_{1/2}$.

Mistake 5: Forgetting to Convert Units or Scale

The error: Students treat $10^2 \times [N_2O_5]$ as the actual concentration.

Why it's wrong: The table gives $10^2 \times [N_2O_5]$, so actual $[N_2O_5]$ = (table value) $\times 10^{-2}$ mol L⁻¹.

How to avoid:

• Always write:

• When plotting or calculating $\log[N_2O_5]$, use the actual concentration.

Mistake 6: Writing the Rate Law Incorrectly

The error: Students write $\text{Rate} = k[N_2O_5]^2$ or forget the stoichiometric coefficient. …