Kinetics Notes: Rate Laws, Orders, and Integrated Rate Laws

Rate Laws and Reaction Orders

• Exponent concepts in rate laws: the rate of a reaction depends on the starting concentrations through a rate law of the form
  \text{rate} = k [A]^n
  where A is a reactant (or a product of reactants). When there are multiple reactants, the rate law is a product of terms, e.g. for two reactants A and B, rate = k [A]^m [B]^p, and the overall order is the sum of the exponents (m + p).

• Major order categories (for a single reactant A):

  • First order: \text{rate} = k [A]^1, often written as \text{rate} = k [A]. Overall order = 1.
  • Second order: \text{rate} = k [A]^2. Overall order = 2.
  • Zero order: \text{rate} = k (i.e., independent of [A]). Overall order = 0.

• Exponent possibilities beyond integers:

  • Exponents can be fractions (e.g., halves), zero, or even negative in some reactions. Negative exponents imply that increasing the reactant concentration can slow the reaction.
  • Example note from lecture: exponents can be zero, 1/2, -1, etc.; this is possible though not the most common in simple cases.

• Practical quiz/homework setup mentioned in lecture:

  • A practice quiz (14.2) with a problem of three reactants to determine three exponents (x, y, z) and then solve for k.
  • In Thursday’s quiz, the same method but with three exponents to determine for a tri-reactant system; solved in the same way by finding exponents and then k.

• Why orders matter:

  • The order with respect to each reactant determines how the rate changes with concentration and how the rate changes over time.
  • For multi-reactant systems, the total (overall) order is the sum of the individual exponents in the rate law.

• Connection to lab and data analysis:

  • The lecture previews “method of initial rates” used to infer the exponents by measuring initial rates at different initial concentrations.
  • “Integrated rate laws” (calculus-based) allow predictions over time once the order and k are known.

From Rate Laws to Integrated Rate Laws

• Integrated rate laws are the time-evolved versions of the rate laws for simple orders (zero, first, second).

• First order integrated rate law:

  • Regular rate law: \text{rate} = k [A]
  • Integrated form (taking time derivative and integrating):
    \ln [A]t = -k t + \ln [A]0
  • Linearization form: if you plot time t on the x-axis and \ln [A] on the y-axis, the data should fall on a straight line with slope -k and intercept \ln [A]_0.
  • Intercept relation: [A]0 = e^{b} where b is the intercept (since \ln [A]0 = b).

• Second order integrated rate law:

  • Regular rate law: \text{rate} = k [A]^2
  • Integrated form: \frac{1}{[A]t} = k t + \frac{1}{[A]0}
  • Linearization form: plot time t on x-axis and \frac{1}{[A]}t on y-axis; slope = k, intercept = \frac{1}{[A]0}.

• Zero order integrated rate law:

  • Regular rate law: \text{rate} = k
  • Integrated form: [A]t = [A]0 - k t
  • Linearization form: plot time t on x-axis and [A] on y-axis; slope = -k, intercept = [A]_0.

• Practical notes on using the integrated laws:

  • They enable predictions such as how much A remains after a given time, given k and [A]_0.
  • They are derived via calculus, but you only need to use the resulting forms for problems (you don’t have to perform the calculus on exams).

Linearization and Data Analysis with Graphs

• How to identify the order from plots (the common linearizations):

  • First order: plot t on x-axis vs \ln [A] on y-axis; a straight line indicates first order. Slope = -k; intercept = \ln [A]_0.
  • Second order: plot t on x-axis vs \frac{1}{[A]} on y-axis; a straight line indicates second order. Slope = k; intercept = \frac{1}{[A]_0}.
  • Zero order: plot t on x-axis vs [A] on y-axis; a straight line indicates zero order. Slope = -k; intercept = [A]_0.

• R-squared (R^2) value as a measure of linearity:

  • Closer to 1.0 indicates a better linear fit.
  • A practical rule of thumb from lecture: R^2 ≈ 0.995 or higher is considered very linear; 0.95–0.99 may be acceptable depending on experimental conditions.

• Practical workflow when you don’t know the order a priori:

  • Try the three linearizations (ln[A] vs t, 1/[A] vs t, and [A] vs t) and compare how linear each is using R^2.
  • The best linear fit identifies the order and yields k from the slope.
  • For first order, the intercept gives [A]0 via [A]0 = e^b; for second order, intercept gives 1/[A]0; for zero order, intercept gives [A]0 directly.

• Excel-based workflow (high level):

  • Create a data table with Time t and measured [A].
  • For first order: add a new column with \ln [A] (use Excel function LN).
  • Create scatter plots with Time on x-axis and the linearized quantity on y-axis.
  • Add a trend line and display the equation and the R^2 value.
  • Repeat for second order using 1/[A] as the y-axis.
  • Repeat for zero order using [A] as the y-axis.
  • Compare R^2 values to determine the best fit and thus the order.

• Practical caution:

  • In practice, data may not be perfectly linear due to experimental error; an R^2 very close to 1 (e.g., 0.995 or higher) is typically considered strong evidence of a linear relationship.
  • In exams, you might be given unlabeled plots and asked to identify the order by inspecting the most linear plot or by inspecting the axes in each plot.

Determing k, [A]_0, and units from a First-Order Example

• Example scenario from lecture (illustrative values):

  • First-order fit yields an intercept b = -2.3 for the ln[A] vs t plot.
  • Then the initial concentration is
    [A]_0 = e^{b} = e^{-2.3} \approx 1.00\times 10^{-1}\, \text{M}.
  • The slope of the line is -k, so if the slope is -0.0125 s^{-1} then
    k = 1.25\times 10^{-2}\, \text{s}^{-1}.
  • In the example, a later calculation gave k = 2.9\times 10^{-4} s^{-1} with a corresponding intercept around -2.3, and [A]_0 ≈ 0.100 M.

• Worked calculation for a common first-order problem:

  • If k = 2.9\times 10^{-4}\ \text{s}^{-1} and [A]0 \approx 1.00\times 10^{-1}\ \text{M}, then after time t = 75 s the concentration is [A]{75} = [A]_0 \exp(-kt) = (1.00\times 10^{-1}) \exp(-2.9\times 10^{-4} \cdot 75) \approx 0.098\ \text{M}.

• Units for k by order (quick guide):

  • First order: k\text{ units} = \text{s}^{-1}.
  • Zero order: k\text{ units} = \text{M s}^{-1}.
  • Second order: k\text{ units} = \text{M}^{-1} \text{s}^{-1}.
  • Rationale: rate has units of [M]/s; for rate law \text{rate} = k [A]^n, the units of k must supply the remaining units to yield left-hand side units.

• How to extract [A]_0 from the intercept when the data is first order:

  • If the first-order integrated rate law is written as
    \ln [A]t = -kt + \ln [A]0,
  • then the intercept b equals \ln [A]0, so [A]0 = e^{b}.

Practical Example: Using Integrated Laws to Answer Questions

• Given a first-order reaction with k and [A]_0 determined from a plot, you can answer:

  • How much A remains after a certain time t: use [A]t = [A]0 e^{-kt} (equivalently from the integrated form).
  • What is [A] after any time t: plug t into the same formula.
  • If you’re given an intercept b and told that b = \ln([A]0), you can compute [A]0 with [A]_0 = e^{b}.

• Practical lab/exam context:

  • You may be asked to determine the order by looking at the best linearization and then extract k from the slope.
  • You may be given data for three repeated experiments (different initial concentrations) and asked to deduce x, y, z (the exponents) and k for a tri-reactant system.
  • In a lab setting, Excel is commonly used to plot the data and obtain the trend line equation and R^2 to judge linearity.

Quick Reference: How to Plot and Interpret in Excel (Summary of Lecture Steps)

• Steps to prepare plots for zero, first, and second order:

  • Ensure Time t is in the leftmost column (x-axis for all plots).
  • For zero order: plot [A] vs t.
  • For first order: plot ln([A]) vs t (compute ln with a calculator or Excel LN function).
  • For second order: plot 1/[A] vs t (compute 1/[A]).
  • Add a trend line for each plot and display the equation and R^2 value.
  • The plot with the highest R^2 (closest to 1) indicates the correct order.
  • From the trend line, extract k from the slope (slope = -k for first and zero order; slope = k for second order).
  • The intercept gives [A]0 for zero order, and 1/[A]0 for second order, and ln([A]_0) for first order.

• Example workflow with three datasets:

  • Dataset 1: best fit is first order; slope = -k1; intercept gives [A]_0.
  • Dataset 2: best fit is zero order; slope = -k2; intercept gives [A]_0.
  • Dataset 3: best fit is second order; slope = k3; intercept gives 1/[A]_0.
  • Use the order that yields the closest R^2 to 1 to determine the correct k and initial concentration.

• Practical note about data spread and realism:

  • Real lab data may not be perfectly linear due to experimental error and multiple variables affecting the measurements.
  • In some contexts (e.g., biology labs), R^2 values slightly below 1 may still be acceptable; the educator’s threshold may vary.

Practice Problems and Exam Prep (What to Expect)

• You may be given data and asked to determine the order by plotting and evaluating R^2 values, then compute k and [A]_0 from the best-fit line.

• You may be asked to compute [A] at a given time using the appropriate integrated rate law once you know the order.

• You may encounter problems with three reactants where you must determine three exponents (x, y, z) in a rate law and then solve for k.

• You should be comfortable performing these steps with a calculator or software (Excel) and interpreting the results, including unit analysis for k.

• Final takeaway: memorize the integrated rate laws and the corresponding linearizations, know how to extract k and [A]_0 from the plot, and be able to determine the order from the best linear fit and R^2 value.

Commonly Used Formulas (Quick Reference)

• General rate law for a single reactant A:
  \text{rate} = k [A]^n

• First order integrated rate law:
  \ln [A]t = -k t + \ln [A]0
  \text{Plot: } t \text{ vs } \ln [A] \text{ (slope } -k, \text{ intercept } \ln [A]_0) }

• Second order integrated rate law:
  \frac{1}{[A]t} = k t + \frac{1}{[A]0}

• Zero order integrated rate law:
  [A]t = [A]0 - k t

• Units for k by order:

  • First order: k\text{ has units of } \text{s}^{-1}
  • Zero order: k\text{ has units of } \text{M s}^{-1}
  • Second order: k\text{ has units of } \text{M}^{-1} \text{s}^{-1}

• Example numerical note:

  • If [A]0 = 0.100\ \text{M} and the plot intercept gives \ln [A]0 = -2.3, then
    [A]_0 = e^{-2.3} \approx 1.0\times 10^{-1}\ \text{M}.
  • If k = 2.9\times 10^{-4}\ \text{s}^{-1} and t = 75 s,
    [A]{75} = [A]0 e^{-k t} \approx 0.100\times e^{-0.00029\times 75} \approx 0.098\ \text{M}.