The following results have been obtained during the kinetic studies of the reaction: $2A + B \rightarrow C + D$ | Experiment | [A]/$\text{mol L}^{-1}$ | [B]/$\text{mol L}^{-1}$ | Initial rate of formation of D/$\text{mol L}^{-1}\text{min}^{-1}$ | |---|---|---|---| | I | 0.1 | 0.1 | $6.0\times10^{-3}$ | | II | 0.3 | 0.2 | $7.2\times10^{-2}$ | | III | 0.3 | 0.4 | $2.88\times10^{-1}$ | | IV | 0.4 | 0.1 | $2.40\times10^{-2}$ | Determine the rate law and the rate constant for the reaction.

> [!TLDR] > The rate law is $\text{Rate} = k[A]^1[B]^2$, and the rate constant is $k = 6.0\ \text{L}^2\text{mol}^{-2}\text{min}^{-1}$. The key to solving this lies in understanding what the **average rate of reaction** actually tells us. When we say "initial rate of formation of D", that number is directly proportional to the overall rate of the reaction — because for every molecule of D formed, the stoichiometry tells us exactly how much A and B are consumed. So we can treat that measured rate as a proxy for the reaction rate itself. The rate law is an *experimental* equation, not something you can guess from the balanced equation. It has the form: $$\text{Rate} = k[A]^x[B]^y$$ where $x$ and $y$ are the **orders** with respect to A and B, and $k$ is the rate constant. Our job is to find $x$, $y$, and $k$ from the data table. 1. **Find the order with respect to A ($x$).** Look for two experiments where $[B]$ is constant, so any change in rate is due only to A. Experiments **I** and **IV** both have $[B] = 0.1\ \text{mol L}^{-1}$. - Experiment I: $[A] = 0.1$, rate $= 6.0 \times 10^{-3}$ - Experiment IV: $[A] = 0.4$, rate $= 2.40 \times 10^{-2}$ When $[A]$ increases by a factor of $\frac{0.4}{0.1} = 4$, the rate increases by a factor of $\frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = 4$. Since $4^x = 4$, we get $x = 1$. The reaction is **first order** in A. 2. **Find the order with respect to B ($y$).** Now look for experiments where $[A]$ is constant. Experiments **II** and **III** both have $[A] = 0.3$. - Experiment II: $[B] = 0.2$, rate $= 7.2 \times 10^{-2}$ - Experiment III: $[B] = 0.4$, rate $= 2.88 \times 10^{-1}$ Here $[B]$ doubles (factor of 2), and the rate increases by a factor of $\frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = 4$. Since $2^y = 4$, we get $y = 2$. The reaction is **second order** in B. > [!WARNING] > A common mistake is to assume the order matches the stoichiometric coefficient. Here th

The following results have been obtained during th – Class 12 Chemistry

Exercises · 3.11

The following results have been obtained during the kinetic studies of the reaction:

$2A + B \rightarrow C + D$

Experiment	[A]/ $\text{mol L}^{-1}$	[B]/ $\text{mol L}^{-1}$	Initial rate of formation of D/ $\text{mol L}^{-1}\text{min}^{-1}$
I	0.1	0.1	$6.0\times10^{-3}$
II	0.3	0.2	$7.2\times10^{-2}$
III	0.3	0.4	$2.88\times10^{-1}$
IV	0.4	0.1	$2.40\times10^{-2}$

Determine the rate law and the rate constant for the reaction.

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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 rate law is $\text{Rate} = k[A]^1[B]^2$ , and the rate constant is $k = 6.0\ \text{L}^2\text{mol}^{-2}\text{min}^{-1}$ .

The key to solving this lies in understanding what the average rate of reaction actually tells us. When we say "initial rate of formation of D", that number is directly proportional to the overall rate of the reaction — because for every molecule of D formed, the stoichiometry tells us exactly how much A and B are consumed. So we can treat that measured rate as a proxy for the reaction rate itself.

The rate law is an experimental equation, not something you can guess from the balanced equation. It has the form:

\text{Rate} = k[A]^x[B]^y

where $x$ and $y$ are the orders with respect to A and B, and $k$ is the rate constant. Our job is to find $x$ , $y$ , and $k$ from the data table.

Find the order with respect to A ( $x$ ).

Look for two experiments where $[B]$ is constant, so any change in rate is due only to A. Experiments I and IV both have $[B] = 0.1\ \text{mol L}^{-1}$ .
- Experiment I: $[A] = 0.1$ , rate $= 6.0 \times 10^{-3}$
- Experiment IV: $[A] = 0.4$ , rate $= 2.40 \times 10^{-2}$
When $[A]$ increases by a factor of $\frac{0.4}{0.1} = 4$ , the rate increases by a factor of $\frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = 4$ .

Since $4^x = 4$ , we get $x = 1$ . The reaction is first order in A.
Find the order with respect to B ( $y$ ).

Now look for experiments where $[A]$ is constant. Experiments II and III both have $[A] = 0.3$ .
- Experiment II: $[B] = 0.2$ , rate $= 7.2 \times 10^{-2}$
- Experiment III: $[B] = 0.4$ , rate $= 2.88 \times 10^{-1}$
Here $[B]$ doubles (factor of 2), and the rate increases by a factor of $\frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = 4$ .

Since $2^y = 4$ , we get $y = 2$ . The reaction is second order in B.

Watch out

A common mistake is to assume the order matches the stoichiometric coefficient. Here the coefficient of B is 1, but the order is 2 — they are not the same thing. Always use experimental data, not the balanced equation.

Write the rate law. Putting it together: …

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Method: Initial Rates Method (Comparing Experiments)

This method determines the order of reaction with respect to each reactant by comparing how the initial rate changes when the concentration of one reactant is changed while the other is kept constant.

Step 1: Write the general rate law

For the reaction $2A + B \rightarrow C + D$ , the rate law is:

\text{Rate} = k[A]^x[B]^y

where $x$ = order with respect to A, $y$ = order with respect to B, and $k$ = rate constant.

The rate given is the initial rate of formation of D, which equals the rate of reaction (since 1 mole of D is produced per reaction).

Step 2: Find order with respect to B ( $y$ )

Compare Experiment II and Experiment III — here $[A]$ is constant at $0.3\ \text{mol L}^{-1}$ .

Experiment	[B]	Rate
II	0.2	$7.2 \times 10^{-2}$
III	0.4	$2.88 \times 10^{-1}$

When $[B]$ doubles ( $0.2 \rightarrow 0.4$ ), the rate changes from $7.2 \times 10^{-2}$ to $2.88 \times 10^{-1}$ .

\frac{\text{Rate}_{\text{III}}}{\text{Rate}_{\text{II}}} = \frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = 4

Since $[B]$ doubled and rate quadrupled:

2^y = 4 \quad \Rightarrow \quad y = 2

Order with respect to B is 2.

Step 3: Find order with respect to A ( $x$ )

Compare Experiment I and Experiment IV — here $[B]$ is constant at $0.1\ \text{mol L}^{-1}$ .

Experiment	[A]	Rate
I	0.1	$6.0 \times 10^{-3}$
IV	0.4	$2.40 \times 10^{-2}$

When $[A]$ quadruples ( $0.1 \rightarrow 0.4$ ), the rate changes from $6.0 \times 10^{-3}$ to $2.40 \times 10^{-2}$ .

\frac{\text{Rate}_{\text{IV}}}{\text{Rate}_{\text{I}}} = \frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = 4

Since $[A]$ quadrupled and rate quadrupled: …

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Common Mistakes & How to Avoid Them

1. Confusing "Rate of Formation of D" with "Rate of Reaction"

The Mistake:

Students often treat the given rate of formation of D as the overall rate of reaction directly, without accounting for stoichiometric coefficients.

Why it's wrong:

For the reaction $2A + B \rightarrow C + D$ , the rate of reaction is defined as:

\text{Rate} = -\frac{1}{2}\frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[C]}{dt} = \frac{d[D]}{dt}

So the rate of reaction equals the rate of formation of D (since coefficient of D is 1).

How to avoid:

Always check the stoichiometric coefficient of the species whose rate is given.
If the coefficient is 1, rate of reaction = rate of formation. If it's 2, divide by 2.
Here, no conversion is needed — the given data directly represents the rate of reaction.

2. Assuming Order = Stoichiometric Coefficient

The Mistake:

Students write: $2A + B \rightarrow$ so order = 2 w.r.t A and 1 w.r.t B, giving Rate $= k[A]^2[B]$ .

Why it's wrong:

Order is determined experimentally, not from the balanced equation. This reaction could have fractional or zero order.

How to avoid:

Always use the method of initial rates to find orders.
Compare experiments where only one concentration changes.

3. Incorrect Comparison of Experiments

The Mistake:

Comparing experiments where both concentrations change, then trying to deduce order.

Example: Comparing I and II — both [A] and [B] change — leads to wrong conclusions.

How to avoid:

To find order w.r.t A: pick experiments where [B] is constant (I and IV).
To find order w.r.t B: pick experiments where [A] is constant (II and III).

4. Arithmetic Errors in Ratio Calculations

The Mistake:

Mishandling powers of 10 or decimal divisions when comparing rates.

Example:

For I and IV:

Rate ratio $= \frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = 4$
Concentration ratio of A $= \frac{0.4}{0.1} = 4$ So $4 = (4)^x \Rightarrow x = 1$ → order w.r.t A = 1

For II and III:

Rate ratio $= \frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = 4$
Concentration ratio of B $= \frac{0.4}{0.2} = 2$ So $4 = (2)^y \Rightarrow y = 2$ → order w.r.t B = 2

How to avoid:

Write ratios clearly: $\frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[B]_2}{[B]_1}\right)^y$
Cancel powers of 10 carefully. …

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