Q.For the reaction:
$2A + B \rightarrow A_2B$
the rate $= k[A][B]^2$ with $k = 2.0\times10^{-6}\ \text{mol}^{-2}\,\text{L}^{2}\,\text{s}^{-1}$. Calculate the initial rate of the reaction when $[A] = 0.1\ \text{mol L}^{-1}$, $[B] = 0.2\ \text{mol L}^{-1}$. Calculate the rate of reaction after $[A]$ is reduced to $0.06\ \text{mol L}^{-1}$.

Concept: Average Rate Of Reaction — the rate law is given directly, so we simply substitute concentrations into $r = k[A][B]^2$.

Step 1: Initial rate

$r_0 = (2.0 \times 10^{-6})(0.1)(0.2)^2$

$= (2.0 \times 10^{-6})(0.1)(0.04)$

$= 8.0 \times 10^{-9}\ \text{mol L}^{-1} \text{s}^{-1}$

Step 2: Rate after $[A]$ drops to $0.06\ \text{mol L}^{-1}$

From stoichiometry $2A + B \rightarrow A_2B$, when $[A]$ falls by $0.04\ \text{mol L}^{-1}$, $[B]$ falls by half of that: $0.02\ \text{mol L}^{-1}$.

So $[B] = 0.2 - 0.02 = 0.18\ \text{mol L}^{-1}$.

Step 3: Substitute

$r = (2.0 \times 10^{-6})(0.06)(0.18)^2$

$= (2.0 \times 10^{-6})(0.06)(0.0324)$

$= 3.888 \times 10^{-9}\ \text{mol L}^{-1} \text{s}^{-1}$

The rate law $\text{Rate} = k[A][B]^2$ is first order in $A$ and second order in $B$. For the second part, note that as $A$ is consumed, $B$ is consumed too, in the ratio set by the balanced equation $2A + B \rightarrow A_2B$.

1. Initial rate. Substitute $[A] = 0.1$, $[B] = 0.2$:

2. Amount of $A$ reacted. $\Delta[A] = 0.1 - 0.06 = 0.04\ \text{mol L}^{-1}$.

3. Amount of $B$ reacted. From $2A + B \rightarrow A_2B$, one mole of $B$ is consumed for every two moles of $A$:

4. New rate with $[A] = 0.06$, $[B] = 0.18$:

Method: Direct Substitution into Rate Law

This method uses the experimentally determined rate equation to calculate reaction rates at given concentrations.

Steps:

Step 1: Identify the rate law

Given:

where $k = 2.0 \times 10^{-6}\ \text{mol}^{-2}\,\text{L}^{2}\,\text{s}^{-1}$.

Step 2: Calculate the initial rate

Substitute the initial concentrations $[A]_0 = 0.1\ \text{mol L}^{-1}$ and $[B]_0 = 0.2\ \text{mol L}^{-1}$:

First, $(0.2)^2 = 0.04$

Then, $0.1 \times 0.04 = 0.004$

Finally, $2.0 \times 10^{-6} \times 0.004 = 8.0 \times 10^{-9}$

Step 3: Find $[B]$ when $[A]$ drops to $0.06\ \text{mol L}^{-1}$

From the stoichiometry: $2A + B \rightarrow A_2B$

• Change in $[A]$ = $0.10 - 0.06 = 0.04\ \text{mol L}^{-1}$ consumed
• For every 2 moles of A consumed, 1 mole of B is consumed
• So, $[B]$ consumed = $\frac{0.04}{2} = 0.02\ \text{mol L}^{-1}$
• Remaining $[B] = 0.20 - 0.02 = 0.18\ \text{mol L}^{-1}$

Step 4: Calculate the rate after change

Substitute $[A] = 0.06$ and $[B] = 0.18$:

$(0.18)^2 = 0.0324$

$0.06 \times 0.0324 = 0.001944$

$2.0 \times 10^{-6} \times 0.001944 = 3.888 \times 10^{-9}$

Key Exam Tip:

• Always check stoichiometric ratios when finding changed concentrations — the rate law exponents are not the stoichiometric coefficients.

Here are the common mistakes students make with this exact problem, and how to avoid each.

Mistake 1: Forgetting the stoichiometric link between $[A]$ and $[B]$

The error:

When $[A]$ drops from $0.1$ to $0.06\ \text{mol L}^{-1}$, students often plug $[B] = 0.2$ into the rate law again — as if $[B]$ didn’t change.

Why it’s wrong:

The reaction consumes $A$ and $B$ together. From the equation

$2A + B \rightarrow A_2B$,

for every 2 moles of A used, 1 mole of B is used. So when $[A]$ falls, $[B]$ must also fall.

How to avoid:

Always calculate the change in $[A]$, then use stoichiometry to find the change in $[B]$.

• Change in $[A] = 0.10 - 0.06 = 0.04\ \text{mol L}^{-1}$
• Since $2\ \text{A}$ consumed per $1\ \text{B}$, $\Delta[B] = \frac{1}{2} \times 0.04 = 0.02\ \text{mol L}^{-1}$
• So $[B]_{\text{new}} = 0.20 - 0.02 = 0.18\ \text{mol L}^{-1}$

Mistake 2: Confusing the rate law with the stoichiometric coefficient

The error:

Some students write the rate as

$\text{Rate} = -\frac{1}{2}\frac{d[A]}{dt} = k[A][B]^2$

and then try to use the $-\frac{1}{2}$ factor when plugging numbers.

Why it’s wrong:

The given rate law $\text{Rate} = k[A][B]^2$ already defines the rate in terms of appearance of product or disappearance of reactant in the standard way. You don’t adjust $k$ or the concentrations with coefficients unless the problem explicitly asks for $-\frac{d[A]}{dt}$.

How to avoid:

Read the problem carefully. If it says “the rate $= k[A][B]^2$”, use that expression exactly as given. Do not insert stoichiometric factors unless asked.

Mistake 3: Using the wrong units or forgetting units entirely

The error:

Students compute the numeric value but write the rate with wrong units (e.g., $\text{s}^{-1}$ or $\text{mol L}^{-1}$).

Why it’s wrong:

The rate of reaction always has units of concentration per time. Here, $k$ has units $\text{mol}^{-2}\,\text{L}^{2}\,\text{s}^{-1}$, and $[A][B]^2$ has units $\text{mol}^3\,\text{L}^{-3}$. Multiplying gives:

How to avoid:

Always do a unit check after every calculation. If the final unit isn’t $\text{mol L}^{-1} \text{s}^{-1}$ (or similar), you’ve made an error.

Mistake 4: Forgetting to square $[B]$ in the rate law

The error:

Plugging $[B] = 0.2$ instead of $[B]^2 = 0.04$ in the initial rate calculation.

Why it’s wrong:

The rate law explicitly says $[B]^2$. Missing the square changes the answer by a factor of 5 in this case.

How to avoid:

Write the rate law in full before substituting numbers:

Then substitute step-by-step.

Mistake 5: Not checking if the reaction is elementary (over-interpreting the rate law)

The error:

Assuming the rate law matches the stoichiometric coefficients (i.e., thinking order = coefficient for every reactant).

Why it’s wrong:

Here it happens to match (order 1 in A, order 2 in B), but that’s not guaranteed in general. The rate law is experimental, not derived from the balanced equation.

How to avoid:

Treat the given rate law as fixed data. Never change the exponents based on the balanced equation unless told the reaction is elementary.

✓ Quick summary — correct calculations

Initial rate:

Rate after $[A]$ drops to 0.06:

Final tip: Always write the stoichiometric link between reactants before doing the second part — it’s the most commonly missed step.