Concept: Average Rate Of Reaction — we use the method of initial rates to find the order with respect to each reactant.

Step 1: Write the general rate law.

Rate $= k[A]^x[B]^y$, where $x$ and $y$ are the orders.

Step 2: Find $y$ (order in B).

Compare experiments II and III (same [A] = 0.3):

$\frac{\text{Rate III}}{\text{Rate II}} = \frac{2.88\times10^{-1}}{7.2\times10^{-2}} = 4 = \left(\frac{0.4}{0.2}\right)^y = 2^y \Rightarrow y = 2$.

Step 3: Find $x$ (order in A).

Compare experiments I and IV (same [B] = 0.1):

$\frac{\text{Rate IV}}{\text{Rate I}} = \frac{2.40\times10^{-2}}{6.0\times10^{-3}} = 4 = \left(\frac{0.4}{0.1}\right)^x = 4^x \Rightarrow x = 1$. …

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:

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.

3. Write the rate law. Putting it together: …

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:

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 $0.3\ \text{mol L}^{-1}$.

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

Since $[B]$ doubled and rate quadrupled:

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 $0.1\ \text{mol L}^{-1}$.

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

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

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:

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. …