6.3 Kinetics: Rate Law Expressions, Reaction Orders, and the Method of Initial Rates

Introduction to Rate Laws

A rate law, also referred to as a rate equation, is a mathematical expression used to communicate the rates of chemical reactions. It specifically describes the relationship between the rate of a reaction and the concentration of the reactants involved.

The Function and Structure of Rate Laws

The rate law for a generic chemical equation where substance A reacts with substance B to form products is expressed as:

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

Components of the Rate Law Expression

• Rate: Listed on the left-hand side of the equation.
• Rate Constant ($k$): A proportionality constant that must be determined experimentally through data.
• Concentration ([A], [B]): The molar concentration of the reactants.
• Reaction Order ($m, n$): The exponents to which the reactant concentrations are raised. These values characterize how a change in concentration affects the rate of the reaction.

Characteristics of Reaction Orders

• Reaction orders ($m$ and $n$) must always be determined experimentally; they cannot be reliably predicted from stoichiometric coefficients.
• These exponents are typically positive integers but can also be fractions, negative numbers, or zero.
• Zero-Order Reactions: When a reactant has a reaction order of zero, any concentration of that reactant raised to the zero power equals one ($[A]^0 = 1$). Consequently, zero-order reactants are omitted from the written rate law as they do not influence the specific rate of that reaction.

Determining the Overall Reaction Order

The overall order of a reaction is the sum of the individual reaction orders ($m + n$) for each reactant present in the rate law.

Examples of Overall Order Calculation

1. Example One: In the rate law $Rate = k[A]^1 [B]^2$, the overall reaction order is $1 + 2 = 3$.
2. Example Two: In the rate law $Rate = k[A]^1 [B]^{-3}$, the overall reaction order is $1 + (-3) = -2$.

Guided Practice: Methanol and Ethyl Acetate

In the reaction between methanol ($CH_3OH$) and ethyl acetate, the provided experimental rate law is:

$Rate = k[CH_3OH]$

Analysis

• Order with respect to Methanol: Since the exponent not listed is assumed to be one, methanol is first-order ($m = 1$).
• Order with respect to Ethyl Acetate: Ethyl acetate is absent from the expression, indicating its reaction order is zero ($n = 0$).
• Overall Order: The sum of the exponents ($1 + 0$) is 1.

The Method of Initial Rates

Experimental data can be used to determine reaction rates across multiple trials. By comparing these trials, researchers find the reaction order for each reactant, the rate constant ($k$), and the final rate law.

Case Study: Decomposition of Acetaldehyde

Acetaldehyde ($CH_3CHO$) decomposes when heated to yield methane and carbon monoxide.

Experimental Data Table:

• Trial 1: Initial concentration = $1.75 \times 10^{-3}$; Rate = $2.06 \times 10^{-11} mol/L \, s$
• Trial 2: Initial concentration = $3.5 \times 10^{-3}$; Rate = $8.24 \times 10^{-11} mol/L \, s$

Step-by-Step Calculation of Reaction Order ($m$):

1. Set up a ratio of the rate laws for Trial 1 and Trial 2:     $\frac{Rate_1}{Rate_2} = \frac{k[CH_3CHO]_1^m}{k[CH_3CHO]_2^m}$
2. Plug in the values:     $\frac{2.06 \times 10^{-11}}{8.24 \times 10^{-11}} = \frac{(1.75 \times 10^{-3})^m}{(3.5 \times 10^{-3})^m}$
3. Simplify the ratios:     $0.25 = (0.5)^m$
4. Solve for $m$ using natural logarithms:     $\ln(0.25) = m \times \ln(0.5)$$m = \frac{\ln(0.25)}{\ln(0.5)} = 2$

Final Rate Law for Acetaldehyde: $Rate = k[CH_3CHO]^2$

Step-by-Step Calculation of Rate Constant ($k$):

1. Rearrange the rate law: $k = \frac{Rate}{[CH_3CHO]^2}$
2. Substitute Trial 1 data:     $k = \frac{2.06 \times 10^{-11} mol/L \, s}{(1.75 \times 10^{-3} mol/L)^2}$
3. Calculate the result:     $k = 6.726530612 \times 10^{-6}$
4. Applying significant figures: $k = 6.73 \times 10^{-6} L/mol \, s$

Units of the Rate Constant $k$

The units for the rate constant vary depending on the overall reaction order ($x$). A general form for determining the units is:

$L^{x-1} \times mol^{1-x} \times s^{-1}$

Common Units by Order

• Zero-Order ($x=0$): $mol \, L^{-1} \, s^{-1}$
• First-Order ($x=1$): $s^{-1}$
• Second-Order ($x=2$): $L \, mol^{-1} \, s^{-1}$
• Third-Order ($x=3$): $L^2 \, mol^{-2} \, s^{-1}$

Guided Practice: Hypochlorite and Iodide Reaction

This experiment involves two reactants: Hypochlorite ($OCl^-$) and Iodide ($I^-$). The general rate law is:

$Rate = k[OCl^-]^m [I^-]^n$

Experimental Data Table:

• Trial 1: $[OCl^-] = 0.0040$, $[I^-] = 0.0020$, Rate = $0.00184 mol/L \, s$
• Trial 2: $[OCl^-] = 0.0020$, $[I^-] = 0.0040$, Rate = $0.00092 mol/L \, s$
• Trial 3: $[OCl^-] = 0.0020$, $[I^-] = 0.0020$, Rate = $0.00046 mol/L \, s$

Part 1: Determination of Order for Hypochlorite ($m$) Compare trials where iodide concentration is held constant (Trials 1 and 3):

1. $\frac{Rate_1}{Rate_3} = \frac{[OCl^-]_1^m}{[OCl^-]_3^m}$
2. $\frac{0.00184}{0.00046} = \frac{(0.0040)^m}{(0.0020)^m}$
3. $4 = 2^m \Rightarrow m = 2$

Part 2: Determination of Order for Iodide ($n$) Compare trials where hypochlorite concentration is held constant (Trials 2 and 3):

1. $\frac{Rate_2}{Rate_3} = \frac{[I^-]_2^n}{[I^-]_3^n}$
2. $\frac{0.00092}{0.00046} = \frac{(0.0040)^n}{(0.0020)^n}$
3. $2 = 2^n \Rightarrow n = 1$

Part 3: Complete Rate Law and Rate Constant

• Overall Rate Law: $Rate = k[OCl^-]^2 [I^-]^1$
• Overall Order: $2 + 1 = 3$ (Third-order)
• Solving for $k$ using Trial 1 data:   $k = \frac{0.00184}{(0.0040)^2 \times (0.0020)}$$k = 57,500 = 5.75 \times 10^4$
• Units for Third-Order: $L^2 / mol^2 \, s$ (or $mol^{-2} \, L^2 \, s^{-1}$)