Q.From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

Concept: Average Rate Of Reaction — the order is the sum of exponents in the rate law; the rate constant’s dimension depends on the overall order.

Step 1: Determine overall order by adding the exponents of concentration terms in the given rate expression.

Step 2: Find dimensions of $k$ using the relation:

$\text{Rate} = k[\text{conc}]^n$, where rate has units $\text{mol L}^{-1} \text{s}^{-1}$ and concentration is $\text{mol L}^{-1}$.

So $[k] = (\text{mol L}^{-1})^{1-n} \text{s}^{-1}$.

1. Order = 2. $[k] = (\text{mol L}^{-1})^{1-2} \text{s}^{-1} = \text{L mol}^{-1} \text{s}^{-1}$.
2. Order = 1 + 1 = 2. $[k] = \text{L mol}^{-1} \text{s}^{-1}$.
3. Order = 3/2. $[k] = (\text{mol L}^{-1})^{1-3/2} \text{s}^{-1} = \text{L}^{1/2} \text{mol}^{-1/2} \text{s}^{-1}$.
4. Order = 1. $[k] = \text{s}^{-1}$.
   ✓Final answer

   1. Order 2, $k$ in $\text{L mol}^{-1} \text{s}^{-1}$.
   2. Order 2, $k$ in $\text{L mol}^{-1} \text{s}^{-1}$.
   3. Order 3/2, $k$ in $\text{L}^{1/2} \text{mol}^{-1/2} \text{s}^{-1}$.
   4. Order 1, $k$ in $\text{s}^{-1}$.

The core idea: what “order” really means

The order of a reaction is not the same as the stoichiometric coefficients in the balanced equation. It is an experimentally determined number that tells you how the rate depends on the concentration of each reactant. In a rate law like

the overall order is $m + n$. The rate constant $k$ is the proportionality constant that makes the equation dimensionally consistent. Since rate always has dimensions of concentration per time (e.g., $\text{mol L}^{-1} \text{s}^{-1}$), the units of $k$ must adjust to match the total exponent.

Let’s apply this to each case.

(i) $3NO(g) \rightarrow N_2O(g)$; Rate $= k[NO]^2$

1. Find the order. The rate law has only one reactant, $NO$, raised to the power 2. So the overall order is simply $2$.

2. Find the dimensions of $k$.

   Rate has units of $\text{mol L}^{-1} \text{s}^{-1}$ (concentration per time).

   $[NO]^2$ has units of $(\text{mol L}^{-1})^2 = \text{mol}^2 \text{L}^{-2}$.

   So:

Equivalently, $\text{L mol}^{-1} \text{s}^{-1}$.

(ii) $H_2O_2(aq) + 3I^-(aq) + 2H^+ \rightarrow 2H_2O(l) + I_3^-$; Rate $= k[H_2O_2][I^-]$

1. Find the order. The exponents are 1 on $H_2O_2$ and 1 on $I^-$. The $H^+$ ion does not appear in the rate law (its concentration may be constant or it does not affect the rate). So overall order = $1 + 1 = 2$.

2. Find the dimensions of $k$.

   $[H_2O_2][I^-]$ has units of $(\text{mol L}^{-1}) \times (\text{mol L}^{-1}) = \text{mol}^2 \text{L}^{-2}$.

   Hence:

Same as case (i): $\text{L mol}^{-1} \text{s}^{-1}$.

(iii) $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$; Rate $= k[CH_3CHO]^{3/2}$

1. Find the order. The exponent is $3/2 = 1.5$. So overall order = $1.5$.

2. Find the dimensions of $k$.

   $[CH_3CHO]^{3/2}$ has units of $(\text{mol L}^{-1})^{3/2} = \text{mol}^{3/2} \text{L}^{-3/2}$.

   Therefore:

Which is usually written as $\text{L}^{1/2} \text{mol}^{-1/2} \text{s}^{-1}$.

(iv) $C_2H_5Cl(g) \rightarrow C_2H_4(g) + HCl(g)$; Rate $= k[C_2H_5Cl]$

1. Find the order. The exponent is 1 (implied). So overall order = $1$.

2. Find the dimensions of $k$.

   $[C_2H_5Cl]$ has units of $\text{mol L}^{-1}$.

   So:

For a first-order reaction, $k$ always has units of $\text{time}^{-1}$ (e.g., $\text{s}^{-1}$, $\text{min}^{-1}$).

Method: Order of Reaction from Rate Law Expression

This method uses the experimentally determined rate law to directly read the order with respect to each reactant and the overall order. The dimensions of the rate constant $k$ are then derived using dimensional analysis.

Steps

1. Identify the exponents on each concentration term in the given rate expression.

   • The exponent on a reactant is its individual order.
   • The overall order is the sum of all individual orders.

2. Write the general form of the rate law:

   $\text{Rate} = k[\text{A}]^m[\text{B}]^n$

   where $m$ and $n$ are the orders.

3. Find dimensions of $k$ using:

   $\text{Dimensions of Rate} = \text{concentration} \cdot \text{time}^{-1}$

   So, $[k] = \dfrac{[\text{Rate}]}{[\text{concentration}]^{\text{overall order}}}$

Applying to each reaction

(i) $3NO(g) \rightarrow N_2O(g)$; Rate $= k[NO]^2$

• Order w.r.t. NO = 2
• Overall order = 2
• Dimensions of $k$: $[k] = \dfrac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^2} = \text{mol}^{-1} \text{L} \text{s}^{-1}$ → $k$ has units of $\text{L mol}^{-1} \text{s}^{-1}$

(ii) $H_2O_2(aq) + 3I^-(aq) + 2H^+ \rightarrow 2H_2O(l) + I_3^-$; Rate $= k[H_2O_2][I^-]$

• Order w.r.t. $H_2O_2$ = 1
• Order w.r.t. $I^-$ = 1
• Overall order = $1 + 1 = 2$
• Dimensions of $k$: $[k] = \dfrac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^2} = \text{mol}^{-1} \text{L} \text{s}^{-1}$ → $k$ has units of $\text{L mol}^{-1} \text{s}^{-1}$

(iii) $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$; Rate $= k[CH_3CHO]^{3/2}$

• Order w.r.t. $CH_3CHO$ = $3/2$
• Overall order = $1.5$
• Dimensions of $k$: $[k] = \dfrac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^{3/2}} = \text{mol}^{-1/2} \text{L}^{1/2} \text{s}^{-1}$ → $k$ has units of $\text{mol}^{-1/2} \text{L}^{1/2} \text{s}^{-1}$

(iv) $C_2H_5Cl(g) \rightarrow C_2H_4(g) + HCl(g)$; Rate $= k[C_2H_5Cl]$

• Order w.r.t. $C_2H_5Cl$ = 1
• Overall order = 1
• Dimensions of $k$: $[k] = \dfrac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1}$ → $k$ has units of $\text{s}^{-1}$

Final Answer Summary

Common Mistakes Students Make on "Order of Reaction & Dimensions of Rate Constants"

Mistake 1: Confusing Order with Molecularity

The error: Students often think the stoichiometric coefficient (like 3 in $3NO$) determines the order. Here, the order comes only from the rate law exponent, not the balanced equation.

How to avoid:

• Order = sum of exponents in the rate law.
• For (i): Rate $= k[NO]^2$ → order = 2 (not 3).
• For (ii): Rate $= k[H_2O_2][I^-]$ → order = $1 + 1 = 2$ (ignore $H^+$ and $I_3^-$).

Mistake 2: Forgetting Units of $k$ Depend on Overall Order

The error: Using the same unit for all reactions (e.g., always $\text{s}^{-1}$ or always $\text{L mol}^{-1} \text{s}^{-1}$).

How to avoid:

• General formula:

where $n$ = overall order.

• For zero order: $k$ has units of $\text{mol L}^{-1} \text{s}^{-1}$
• For first order: $k$ has units of $\text{s}^{-1}$
• For second order: $k$ has units of $\text{L mol}^{-1} \text{s}^{-1}$
• For order $3/2$: $k$ has units of $\text{L}^{1/2} \text{mol}^{-1/2} \text{s}^{-1}$

Mistake 3: Mishandling Fractional Orders

The error: Treating $3/2$ order as either 1 or 2, or writing wrong units.

How to avoid:

• For (iii): Rate $= k[CH_3CHO]^{3/2}$ → $n = 1.5$
• Units:

(Often written as $\text{L}^{1/2} \text{mol}^{-1/2} \text{s}^{-1}$)

Mistake 4: Ignoring Concentration Units in Dimensional Analysis

The error: Using $\text{mol}$ without volume, or forgetting that concentration is $\text{mol L}^{-1}$.

How to avoid:

• Always write concentration as $[\text{mol L}^{-1}]$
• For (iv): first order → $k$ has units $\text{s}^{-1}$ (time$^{-1}$ only, no concentration term)

Mistake 5: Misreading the Rate Law for Multi-reactant Reactions

The error: Including all reactants in the order calculation even if they don't appear in the rate law.

How to avoid:

• In (ii), the rate law is $k[H_2O_2][I^-]$ — $H^+$ does not appear, so its exponent is 0.
• Order = $1 + 1 = 2$, not 3.

Quick Summary Table (Exam-Ready)

Final tip: Always write the general formula for $k$ first, then substitute $n$. This prevents unit errors even for fractional orders.