Q.For a reaction, $A + B \rightarrow$ Product; the rate law is given by, $r = k[A]^{1/2}[B]^2$. What is the order of the reaction?

The key idea is that the order of a reaction is the sum of the exponents of the concentration terms in the experimentally determined rate law.

Reasoning:

1. The rate law is $r = k[A]^{1/2}[B]^2$.
2. The exponent of $[A]$ is $\frac{1}{2}$, and the exponent of $[B]$ is $2$.
3. The overall order is the sum of these exponents: $\frac{1}{2} + 2 = 2.5$.

The order of a reaction tells you how the rate depends on the concentration of each reactant. It’s not something you guess from the balanced equation — it comes directly from the experimentally determined rate law. Here, the rate law is given: $r = k[A]^{1/2}[B]^2$.

The exponent on $[A]$ is $1/2$, meaning if you double $[A]$, the rate increases by a factor of $2^{1/2} \approx 1.41$. The exponent on $[B]$ is $2$, so doubling $[B]$ quadruples the rate. The overall order is simply the sum of these individual exponents.

1. Identify the exponents in the rate law:

   For $[A]$, exponent = $\frac{1}{2}$.

   For $[B]$, exponent = $2$.

2. Add them:

   $\frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2} = 2.5$.

3. That’s the overall order of the reaction.

Method: Sum of Exponents (Order from Rate Law)

The order of a reaction is the sum of the powers (exponents) of the concentration terms in the experimentally determined rate law.

Steps

1. Write the given rate law

   $r = k[A]^{1/2}[B]^2$

2. Identify the exponents

   • Exponent of $[A]$ = $\frac{1}{2}$
   • Exponent of $[B]$ = $2$

3. Add the exponents

   $\text{Order} = \frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2}$

4. State the result

   The overall order of the reaction is $\frac{5}{2}$ (or 2.5).

Key Exam Tip

• The order is not the same as the stoichiometric coefficients (here, both are 1 from the balanced equation, but the order is 2.5).
• Order is experimentally determined from the rate law, not from the balanced chemical equation.

Here are the common mistakes students make when determining the order of a reaction from a rate law like $r = k[A]^{1/2}[B]^2$, and how to avoid each.

Mistake 1: Forgetting to add the exponents

What students do wrong:

They see $[A]^{1/2}$ and $[B]^2$, and answer “order = 2” (only looking at $[B]$) or “order = 1/2” (only looking at $[A]$).

Why it’s wrong:

The overall order is the sum of all exponents in the rate law.

How to avoid:

Always write the sum explicitly:

Key rule: Overall order = sum of powers of all concentration terms in the rate law.

Mistake 2: Confusing stoichiometric coefficients with exponents

What students do wrong:

They see $A + B \rightarrow \text{Product}$ and think the order is $1 + 1 = 2$ (from the balanced equation).

Why it’s wrong:

The rate law is determined experimentally, not from the balanced chemical equation. The exponents in the rate law are not the stoichiometric coefficients.

How to avoid:

• Ignore the balanced equation when finding order from a given rate law.
• Only look at the exponents in $r = k[A]^{1/2}[B]^2$.

Key rule: Order comes from the rate law, not the reaction equation.

Mistake 3: Misreading fractional exponents

What students do wrong:

They treat $[A]^{1/2}$ as $[A]^1$ or $[A]^2$, or think $1/2$ means “half-order” is not a valid order.

Why it’s wrong:

Fractional orders (like $1/2$, $3/2$) are perfectly valid in chemical kinetics.

How to avoid:

• Treat $1/2$ as a number: $0.5$.
• Add it normally: $0.5 + 2 = 2.5$.

Key rule: Orders can be integers, fractions, or even zero — just add the numbers as given.

Mistake 4: Forgetting to include the constant $k$ in the order

What students do wrong:

They think $k$ has an exponent that contributes to the order.

Why it’s wrong:

$k$ is the rate constant — it is a proportionality factor, not a concentration term. Its units change with order, but its exponent does not add to the order.

How to avoid:

• Only look at the exponents on $[A]$, $[B]$, etc.
• $k$ is just a multiplier; ignore it when summing exponents.

Key rule: Order depends only on concentration exponents, not on $k$.

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