Q.Identify the reaction order from each of the following rate constants.

The key idea is that the units of the rate constant $k$ directly reveal the overall order of the reaction. For a general reaction of order $n$, the rate has units of $\text{mol L}^{-1} \text{s}^{-1}$, and since $\text{rate} = k[\text{conc}]^n$, the units of $k$ are $\text{(mol L}^{-1})^{1-n} \text{s}^{-1}$. …

The rate constant $k$ is not just a number — its units carry a fingerprint of the reaction’s overall order. This works because the rate of a reaction always has units of concentration per time (typically $\text{mol L}^{-1}\text{s}^{-1}$). When you write the rate law $\text{Rate} = k[\text{A}]^n$, the units of $k$ must adjust so that the right-hand side matches the units of rate.

For a reaction of order $n$, the general unit of $k$ is:

In molar terms, concentration is in $\text{mol L}^{-1}$, so:

Let’s apply this to each case.

1. Case (i): $k = 2.3 \times 10^{-5}\ \text{L mol}^{-1}\text{s}^{-1}$

   • The units are $\text{L}^{1} \text{mol}^{-1} \text{s}^{-1}$.
   • Compare with the general form $\text{L}^{n-1} \text{mol}^{1-n} \text{s}^{-1}$.
   • For the exponent of $\text{mol}$: $1-n = -1 \implies n = 2$.
   • Check the exponent of $\text{L}$: $n-1 = 1$, which matches.
   • So the reaction is second order overall.

2. Case (ii): $k = 3 \times 10^{-4}\ \text{s}^{-1}$

   • The units are simply $\text{s}^{-1}$, with no concentration term.
   • From the general form, this means $(\text{concentration})^{1-n}$ must be dimensionless, so $1-n = 0 \implies n = 1$.
   • The reaction is first order overall. …

Method: Unit Analysis of Rate Constant

This method uses the units of the rate constant ($k$) to determine the reaction order. The logic is based on the general rate law:

where:

• Rate has units of $\text{mol L}^{-1} \text{s}^{-1}$ (or $\text{concentration} \cdot \text{time}^{-1}$)
• $[\text{A}]$ has units of $\text{mol L}^{-1}$
• $k$ has units that depend on $n$

Steps

1. Write the general unit equation Units of Rate = Units of $k$ × (Units of concentration)$^n$

2. Solve for units of $k$

3. Match given units to find $n$ Compare the given units of $k$ with the expression above.

(i) $k = 2.3 \times 10^{-5}\ \text{L mol}^{-1} \text{s}^{-1}$

• Given units: $\text{L mol}^{-1} \text{s}^{-1}$

  This can be rewritten as: $(\text{mol L}^{-1})^{-1} \cdot \text{s}^{-1}$

• From step 2: $(\text{mol L}^{-1})^{1-n} \cdot \text{s}^{-1}$

• Equate exponents:

Answer: Second order reaction

(ii) $k = 3 \times 10^{-4}\ \text{s}^{-1}$

• Given units: $\text{s}^{-1}$ (no concentration term)

• From step 2: $(\text{mol L}^{-1})^{1-n} \cdot \text{s}^{-1}$ …

Here are the most common mistakes students make when identifying reaction order from the units of the rate constant ($k$), along with how to avoid each.

Mistake 1: Forgetting the General Formula for Units of $k$

Students often try to memorize the units for each order individually (e.g., "first order is $\text{s}^{-1}$") but fail to connect them to a single, logical formula.

How to avoid:

Always derive the units using the general rate law:

Where:

• Rate has units of $\text{concentration} \cdot \text{time}^{-1}$ (usually $\text{mol L}^{-1} \text{s}^{-1}$).
• $[\text{Reactant}]$ has units of $\text{concentration}$ (usually $\text{mol L}^{-1}$).

Rearrange to solve for the units of $k$:

Simplify:

For your specific problems:

• (i) $k = 2.3 \times 10^{-5}\ \text{L mol}^{-1}\text{s}^{-1}$

  This is $(\text{mol L}^{-1})^{-1} \text{s}^{-1}$.

  Set $1-n = -1 \implies n = 2$. Second order.

• (ii) $k = 3 \times 10^{-4}\ \text{s}^{-1}$

  This is $(\text{mol L}^{-1})^{0} \text{s}^{-1}$.

  Set $1-n = 0 \implies n = 1$. First order.

Mistake 2: Confusing $\text{L mol}^{-1} \text{s}^{-1}$ with $\text{mol L}^{-1} \text{s}^{-1}$

A very common slip: reading the unit as "mol per liter per second" (which is actually the unit of rate, not $k$).

How to avoid:

Read the unit backwards or in words:

• $\text{L mol}^{-1} \text{s}^{-1}$ = "liters per mole per second" → This is second order.
• $\text{mol L}^{-1} \text{s}^{-1}$ = "moles per liter per second" → This is zero order (rate = $k$).

Quick check: If the unit has a negative exponent on concentration (like $\text{L mol}^{-1}$), the order is greater than 1. If it has a positive exponent (like $\text{mol L}^{-1}$), the order is less than 1 (zero order).

Mistake 3: Ignoring the Time Unit (e.g., $\text{s}^{-1}$ vs $\text{min}^{-1}$)

Students sometimes think that any $k$ with $\text{s}^{-1}$ must be first order. But $\text{s}^{-1}$ alone is not enough — you must check the concentration unit too.

How to avoid:

• First order: $k$ has units of only $\text{time}^{-1}$ (e.g., $\text{s}^{-1}$, $\text{min}^{-1}$). No concentration unit appears.
• Second order: $k$ has units of $\text{concentration}^{-1} \cdot \text{time}^{-1}$ (e.g., $\text{L mol}^{-1} \text{s}^{-1}$). …