Reaction Order - Chemistry

1. Fundamental Concepts

Definition: Reaction order describes how the rate of a reaction depends on the concentration of the reactants. It is determined experimentally and is not related to the stoichiometric coefficients in the balanced equation.

Rate Law Equation:

$$Rate = k[A]^m [B]^n$$

$k$ : Rate constant (temperature dependent).

$[A], [B]$ : Molar concentrations of reactants.

$m, n$ : Orders with respect to reactants A and B (must be determined by experiment).

Overall Order: The sum of the individual orders ( $m + n$ ).

2. Key Concepts

Zero Order: Rate is independent of concentration ( $Rate = k$ ). Concentration decreases linearly with time.

First Order: Rate is directly proportional to concentration ( $Rate = k[A]$ ). Concentration decays exponentially. Half-life ( $t_{1/2}$ ) is constant.

Second Order: Rate is proportional to the square of the concentration ( $Rate = k[A]^2$ ). Rate is very sensitive to concentration changes.

Units of k:

0th Order: $M \cdot s^{-1}$

1st Order: $s^{-1}$

2nd Order: $M^{-1} \cdot s^{-1}$

Integrated Rate Laws (Linear Plots):

0th: $[A]_t = -kt + [A]_0$ ( $[A]$ vs $t$ is linear)

1st: $\ln[A]_t = -kt + \ln[A]_0$ ( $\ln[A]$ vs $t$ is linear)

2nd: $\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$ ( $\frac{1}{[A]}$ vs $t$ is linear)

3. Examples

Easy

Question: Given the rate law $Rate = k[NO]^2[O_2]$ , what is the overall order of the reaction?

Answer: 3rd Order.

Explanation: Sum the exponents: $2 (\text{from } NO) + 1 (\text{from } O_2) = 3$ .

Medium

Question: A first-order reaction has a rate constant $k = 0.02 s^{-1}$ . Calculate the half-life ( $t_{1/2}$ ) of the reaction.

Answer: 34.7 seconds.

Explanation: Use the first-order half-life formula:

$$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{0.02} = 34.65 s$$

Hard

Question: Determine the order of the reaction $A + B \rightarrow Products$ using the following experimental data.

Answer: The reaction is second order with respect to A, zero order with respect to B, and second order overall.

Explanation:

1. Compare Exp 1 and 2: [A] doubles ( $\times 2$ ), [B] stays same. Rate quadruples ( $\times 4$ ). Since $2^m = 4$ , $m = 2$ . (Order in A is 2).

2. Compare Exp 1 and 3: [B] doubles ( $\times 2$ ), [A] stays same. Rate stays same ( $\times 1$ ). Since $2^n = 1$ , $n = 0$ . (Order in B is 0).

3. Overall: $2 + 0 = 2$ .

4. Problem-Solving Techniques

Method of Initial Rates:

Find two experiments where only one reactant's concentration changes.

Use the ratio: $\frac{Rate_2}{Rate_1} = (\frac{[A]_2}{[A]_1})^m$ .

If Rate $\times 2$ when Conc $\times 2 \rightarrow$ 1st Order.

If Rate $\times 4$ when Conc $\times 2 \rightarrow$ 2nd Order.

If Rate $\times 1$ when Conc $\times 2 \rightarrow$ 0th Order.

Graphical Analysis:

Plot the data. If the y-axis creates a straight line, that defines the order.

Y-axis = $[A]$ $\rightarrow$ 0th Order.

Y-axis = $\ln[A]$ $\rightarrow$ 1st Order.

Y-axis = $1/[A]$ $\rightarrow$ 2nd Order.

Half-life Trend:

If $t_{1/2}$ is constant $\rightarrow$ 1st Order.

If $t_{1/2}$ doubles as concentration halves $\rightarrow$ 2nd Order.