In a reaction between A and B, the initial rate of reaction ($r_0$) was measured for different initial concentrations of A and B as given below: | A/$\text{mol L}^{-1}$ | 0.20 | 0.20 | 0.40 | |---|---|---|---| | B/$\text{mol L}^{-1}$ | 0.30 | 0.10 | 0.05 | | $r_0$/$\text{mol L}^{-1}\text{s}^{-1}$ | $5.07\times10^{-5}$ | $5.07\times10^{-5}$ | $1.43\times10^{-4}$ | What is the order of the reaction with respect to A and B?

In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrat

Exercises · 3.10

In a reaction between A and B, the initial rate of reaction ( $r_0$ ) was measured for different initial concentrations of A and B as given below:

A/ $\text{mol L}^{-1}$	0.20	0.20	0.40
B/ $\text{mol L}^{-1}$	0.30	0.10	0.05
$r_0$ / $\text{mol L}^{-1}\text{s}^{-1}$	$5.07\times10^{-5}$	$5.07\times10^{-5}$	$1.43\times10^{-4}$

What is the order of the reaction with respect to A and B?

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Concept understanding — Average Rate Of Reaction

Reaction Rate Stoichiometry – From Intuition to Precision

Imagine you are watching a simple reaction:

$2 \text{NO}_2 \rightarrow 2 \text{NO} + \text{O}_2$

As NO₂ disappears, NO appears twice as fast as O₂ appears. Why? Because the balanced equation says: for every 2 molecules of NO₂ that break apart, you get 2 molecules of NO and 1 molecule of O₂. The numbers in front of the species — the stoichiometric coefficients — tell you the relative speeds at which reactants vanish and products appear.

That is the core idea: reaction rate stoichiometry is the relationship between the rates of change of different species in a chemical reaction, dictated by their coefficients in the balanced equation.

The Intuitive Picture

Think of a factory assembly line. The balanced equation is like a recipe:

2 units of raw material A → 2 units of product B + 1 unit of byproduct C

If the line runs steadily, every time 2 units of A are consumed, 2 units of B are produced and 1 unit of C is produced. So the rate at which A disappears must be twice the rate at which C appears. The rate at which B appears equals the rate at which A disappears (both have coefficient 2).

Note

The stoichiometric coefficients are not speeds themselves — they are scaling factors that connect the speeds of different species.

The Precise Statement

For a general reaction:

aA + bB \rightarrow cC + dD

The rate of reaction (often called the rate of the process, $r$ ) is defined as:

r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

Here:

$\frac{d[X]}{dt}$ is the instantaneous rate of change of concentration of species X (in mol L⁻¹ s⁻¹).
The minus sign is used for reactants (their concentration decreases with time).
The plus sign is used for products (their concentration increases with time).
Dividing by the coefficient normalises the rate — so $r$ is the same number for every species.

r = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}

This single value $r$ is the intrinsic rate of the reaction, independent of which species you measure.

How to Use It (Step by Step)

Suppose you are given:

$2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}$

And you measure that $\frac{d[\text{H}_2]}{dt} = -0.040 \text{ M/s}$ (negative because H₂ is being consumed).

Step 1: Write the rate relation:

r = -\frac{1}{2}\frac{d[\text{H}_2]}{dt} = -\frac{1}{1}\frac{d[\text{O}_2]}{dt} = \frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt}

Step 2: Plug in the known value:

r = -\frac{1}{2}(-0.040) = +0.020 \text{ M/s}

Step 3: Find the rate for O₂:

-\frac{d[\text{O}_2]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{O}_2]}{dt} = -0.020 \text{ M/s}

Step 4: Find the rate for H₂O:

\frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{H}_2\text{O}]}{dt} = 2r = 0.040 \text{ M/s}

Tip

A quick check: the coefficients tell you the relative rates. Here, H₂ disappears twice as fast as O₂, and H₂O appears at the same rate as H₂ disappears (both coefficient 2). Always verify your numbers match the coefficient ratios.

--- …

Why this formula?

Average Rate of Reaction — Why the Formula Holds

Let’s build this from the ground up. The goal is to understand why the average rate formula looks the way it does — not just memorise it.

1. What does "rate of reaction" mean physically?

A chemical reaction changes the concentration of reactants (decreasing) and products (increasing) over time.

Rate = how fast this change happens.
If you measure the change over a finite time interval, you get the average rate.

2. The core idea: change per unit time

For any quantity $X$ that changes from $X_1$ to $X_2$ over time $t_1$ to $t_2$ :

\text{Average rate of change of } X = \frac{\Delta X}{\Delta t} = \frac{X_2 - X_1}{t_2 - t_1}

This is just the slope of the straight line connecting the two points on a concentration vs. time graph.

3. Applying this to a reaction

Consider a simple reaction:

A \rightarrow B

Reactant A is consumed: $[A]$ decreases.
Product B is formed: $[B]$ increases.

For reactant A (disappearing):

\text{Average rate} = -\frac{\Delta [A]}{\Delta t}

Why the minus sign?

Because $\Delta [A] = [A]_2 - [A]_1$ is negative (concentration drops). The rate itself must be positive (speed is never negative). So we multiply by $-1$ .

For product B (appearing):

\text{Average rate} = +\frac{\Delta [B]}{\Delta t}

Here $\Delta [B]$ is positive, so no minus sign needed.

4. The general formula for any reaction

For a balanced reaction:

aA + bB \rightarrow cC + dD

The average rate is defined per mole of reaction — so it’s the same number regardless of which species you track.

We divide each $\frac{\Delta [\text{species}]}{\Delta t}$ by its stoichiometric coefficient:

\text{Average rate} = -\frac{1}{a} \frac{\Delta [A]}{\Delta t} = -\frac{1}{b} \frac{\Delta [B]}{\Delta t} = +\frac{1}{c} \frac{\Delta [C]}{\Delta t} = +\frac{1}{d} \frac{\Delta [D]}{\Delta t}

Why divide by the coefficient?

If 2 moles of A disappear for every 1 mole of C formed, then $\frac{\Delta [A]}{\Delta t}$ is twice as large as $\frac{\Delta [C]}{\Delta t}$ . Dividing by the coefficient normalises them to the same "per mole of reaction" rate.

5. Key exam point: the formula in one line

For any species $X$ with stoichiometric coefficient $\nu_X$ (negative for reactants, positive for products):

$\text{Average rate} = \frac{1}{\nu_X} \frac{\Delta [X]}{\Delta t}$ …

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The reaction is order 1.5 with respect to A and zero order with respect to B, so the rate law is $r_0 = k[A]^{1.5}[B]^{0} = k[A]^{1.5}$ .

Let the rate law be $r_0 = k[A]^m[B]^n$ . We find $m$ and $n$ by the method of initial rates, comparing experiments in which only one concentration changes.

Order with respect to B. Compare experiments 1 and 2, where $[A]$ is held constant at $0.20\ \text{mol L}^{-1}$ while $[B]$ changes from $0.30$ to $0.10\ \text{mol L}^{-1}$ :

\frac{r_{0,1}}{r_{0,2}} = \frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}} = 1 = \left(\frac{0.30}{0.10}\right)^n = 3^n \implies n = 0

The rate is unchanged when $[B]$ changes, so the reaction is zero order in B.

Order with respect to A. Since $n = 0$ , $r_0 = k[A]^m$ . Compare experiments 1 and 3 (where $[A]$ doubles from $0.20$ to $0.40$ ): …

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Method: Initial Rate Method (Comparing Experiments)

This method determines reaction order by comparing how the initial rate changes when the concentration of only one reactant is varied while keeping the other constant.

Step 1 – Write the general rate law

The rate law is:

r_0 = k [A]^m [B]^n

We need to find $m$ (order w.r.t. A) and $n$ (order w.r.t. B).

Step 2 – Find order with respect to B

Compare experiments 1 and 2 (where $[A]$ is constant at 0.20 mol L⁻¹):

Experiment	$[A]$	$[B]$	$r_0$
1	0.20	0.30	$5.07 \times 10^{-5}$
2	0.20	0.10	$5.07 \times 10^{-5}$

Since $[A]$ is constant, the ratio of rates is:

\frac{r_{0,1}}{r_{0,2}} = \left( \frac{[B]_1}{[B]_2} \right)^n

Substitute values:

\frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}} = \left( \frac{0.30}{0.10} \right)^n

1 = (3)^n

Therefore: $n = 0$

The reaction is zero order with respect to B.

Step 3 – Find order with respect to A

Now compare experiments 2 and 3 (where $[B]$ changes, but we already know $n=0$ , so B does not affect rate).

Experiment	$[A]$	$[B]$	$r_0$
2	0.20	0.10	$5.07 \times 10^{-5}$
3	0.40	0.05	$1.43 \times 10^{-4}$

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Common Mistakes Students Make on This Concept (Average Rate of Reaction & Order Determination)

Mistake 1: Confusing "Average Rate" with "Initial Rate"

The error: Students often try to calculate an average rate over a time interval, but the table gives initial rates ( $r_0$ ) — the rate at the very start of the reaction. These are not the same as average rates over time.

How to avoid:

Read the problem carefully: "initial rate of reaction ( $r_0$ )" means the instantaneous rate at $t = 0$ .
The data is already in rate form — no need to compute $\frac{\Delta [\text{product}]}{\Delta t}$ from concentration vs. time data.

Mistake 2: Assuming the Order is an Integer Without Checking

The error: Jumping to conclusions like "order = 1" or "order = 2" without systematically comparing experiments.

How to avoid:

Use the method of initial rates: Compare two experiments where only one reactant's concentration changes.
Write the rate law:

r_0 = k [A]^m [B]^n

Take ratios to eliminate $k$ :

For order w.r.t. B:

Compare Expt 1 and Expt 2 (both have $[A] = 0.20$ ):

\frac{r_{0,1}}{r_{0,2}} = \frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}} = 1 = \left(\frac{0.30}{0.10}\right)^n = (3)^n

So $3^n = 1 \Rightarrow n = 0$ — order w.r.t. B is zero.

For order w.r.t. A:

No pair of experiments holds $[B]$ constant while changing $[A]$ alone. That is not a problem: we have just established $n = 0$ , so $[B]$ does not affect the rate at all. Compare Expt 1 and Expt 3 directly (the change in $[B]$ between them is irrelevant because B is zero order):

Expt 1: $[A] = 0.20$ , $[B] = 0.30$ , $r_0 = 5.07 \times 10^{-5}$
Expt 3: $[A] = 0.40$ , $[B] = 0.05$ , $r_0 = 1.43 \times 10^{-4}$

With $[B]$ playing no role:

\frac{r_{0,3}}{r_{0,1}} = \frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}} \approx 2.82 = \left(\frac{0.40}{0.20}\right)^m = (2)^m

So $2^m = 2.82 \Rightarrow m \approx 1.5$ — order w.r.t. A is 1.5 (not an integer!).

Mistake 3: Forgetting to Check if B Affects the Rate at All

The error: Assuming both reactants affect the rate, then getting confused when the ratio for B equals 1.

How to avoid:

When the rate doesn't change despite changing $[B]$ , order w.r.t. B = 0 immediately. …

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