Lecture 9.3 Notes: Reaction Orders and Their Determination

Purpose of reaction order

• Reaction orders tell us how each reactant affects the overall rate of a reaction.

• They describe the relationship between reactant concentrations and the rate, not what appears in the balanced equation.

• Orders are experimentally determined, not deduced from the stoichiometry.

• Typical values : zero, one, or two (often integers). They can be fractional or negative, though fractions are rare in exams.

• Negative order: increasing a reactant slows the rate (possible in complex mechanisms).

• Fractional order: possible in some mechanisms, but not commonly encountered in standard problems.

• Overall order = sum of the individual orders with respect to each reactant.

• When more than one reactant is present, the rate law can be first order in one reactant and second order in another, giving an overall order that is the sum (e.g., rate = K [A]^1 [B]^1 "," \rightarrow overall order 2).

• If a reactant is effectively not involved in the rate-determining step, its order with respect to that reactant can be zero (even if it participates in a preceding equilibrium).

What a rate law looks like

• General form: $r = k [A]^m [B]^n \times \text{(other factors)}$ where m and n are the orders with respect to A and B.

• If a reaction involves only one reactant, the overall order is the exponent on that reactant (m for A, etc.).

• If there are multiple reactants, the exponents add to the overall order.

• The rate constant k depends on the overall order; units of k change with order (see quick reference below).

Typical orders and their qualitative meaning

• Zero order: doubling [A] does not change the rate. Rate is independent of [A].

• First order: doubling [A] doubles the rate (linear in [A]).

• Second order (single reactant): doubling [A] increases rate by a factor of 4 ($\propto [A]^2$).

• Higher orders: doubling can increase rate by $2^{\text{order}}$; a higher-order implies stronger sensitivity to concentration.

• Third order is rare; many textbook problems stay at zero, one, or two.

• Common multi-reactant cases: rate = k [A]^m [B]^n with m + n = overall order (e.g., m = 1, n = 1 \rightarrow overall order 2).

Zero order specifics

• Rate law (zero order): $r = k$

• Integrated rate law: $[A] = [A]_0 - kt$

• Graphical signature: concentration vs time is a straight line with slope
  

  -k; rate constant k equals the magnitude of the slope, and the rate is constant over time.

• When initiated with a zero-order process, the plot of [A] vs t is linear.

First order specifics

• Rate law (first order): $r = k[A]$

• Integrated rate law: $[A] = [A]_0 e^{-kt}$

• Graphical signature: plotting

• ln[A] vs t yields a straight line with slope
  

  -k (since ln[A] = ln[A]_0 - kt).

• The linear plot for a first-order reaction is a plot of
  

  $</p><p>\text{ln}([A]) \text{ vs } t</p><p>$

• Slope of the line is the negative rate constant:
  

  $\text{slope} = -k$

Second order specifics

• Two common forms:

• Single reactant second order: $r = k[A]^2$

  • Integrated rate law: $\frac{1}{[A]} = kt + \frac{1}{[A]_0}$

  • Graphical signature: plotting $1/[A]$ vs t gives a straight line with slope k.

• Two reactants each first order: $r = k[A][B]$

  • If both A and B are involved with order 1, the overall order is 2.

  • Integrated rate law for the two-reactant-case is more complicated unless one reactant is held constant.

• For a second-order reaction, the inverse plot (1/[A] vs t) is linear (for the single-reactant case).

Graphs and plots overview

• Concentration vs time plots:

• Zero order: linear decrease, [A] vs t is linear.

• First order: nonlinear decay when plotting [A] vs t (but ln[A] vs t is linear).

• Second order (single reactant): nonlinear decay when plotting [A] vs t; 1/[A] vs t is linear.

• Rate vs concentration plots: not strictly linear in all cases; the linearity depends on the order as shown by the integrated forms above.

Overall reaction orders and rate laws (conceptual)

• Rate law example: rate = K [N2O]^2 [X]^1 \rightarrow orders are: N2O is 2, X is 1; overall order = 3.

• When the transcript mentions “nitrous oxide order being second and this one first,” the total is third order for that rate law.

• If there are multiple reactants, the total order is the sum of their individual orders; the rate constant k corresponds to that overall order.

• Orders are with respect to each individual molecule or species, not the entire reaction equation.

How to determine reaction orders experimentally (minimum three experiments)

• Core idea: hold the concentration of one reactant constant while varying the other, then observe the effect on the rate.

• Procedure outline:

• Do at least three experiments with two reactants (A and B).

• In two experiments, hold B constant and double A; in the third, hold A constant and double B (or use a different pair of experiments).

• Use rate data to determine the order with respect to each reactant.

• Practical measurement using rate ratios:

• Rate laws: $r<em>1 = k [A]</em>1^m [B]<em>1^n, \quad r</em>2 = k [A]<em>2^m [B]</em>2^n.$

• If B is held constant and A is doubled: $\frac{r<em>2}{r</em>1} = \left(\frac{[A]<em>2}{[A]</em>1}\right)^m.$
  

  Therefore, $m = \frac{\log(r<em>2/r</em>1)}{\log([A]<em>2/[A]</em>1)}.$

• If A is held constant and B is doubled: $\frac{r<em>2}{r</em>1} = \left(\frac{[B]<em>2}{[B]</em>1}\right)^n,$
  

  so $n = \frac{\log(r<em>2/r</em>1)}{\log([B]<em>2/[B]</em>1)}.$

• Alternate calculation method when data are not perfect (use logs):

• If you know that doubling the concentration leads to a rate change by a factor, you can use logs to extract the order.

• Example: if r2 = 2.00 r1 and [A]2 = 2[A]1, then
  

  $m = \frac{\log(r<em>2/r</em>1)}{\log([A]<em>2/[A]</em>1)} = \frac{\log 2.00}{\log 2.00} = 1.$

• If r2 = 1.99 r1 and [A]2 = 2[A]1, then
  

  $m = \frac{\log(1.99)}{\log 2} \approx 1.0 <br></p><p>\text{(close to 1, rounding)}.$

• Summary of the data interpretation approach:

• Identify which experiment varies which concentration while holding others constant.

• Compute rate ratios for pairs of experiments.

• Solve for the exponent(s) m, n by the appropriate ratio equations.

• Cross-check by using a second pair to verify the order with respect to the other reactant.

Worked practice notes (as described in the lecture)

• Example 1: Rate law components

• Suppose a rate law has A with order 2 and O2 with order 1: rate = k [A]^2 [O2]^1.

• From experiments, A is second order (order 2) and O2 is first order (order 1). Overall order = 3.

• Example 2: A single reactant with a non-integer order

• If doubling [A] results in a rate increase by a factor of $2^{3/2} \approx 2.83$, then the order with respect to A is 3/2.

• Example 3: Missing reactant with zero order

• If a reactant (say B) is present but its concentration doubling does not change the rate, then the order with respect to B is 0; rate law could be r = k [A]^m (if B does not affect the rate).

• Example 4: Two experiments with a factor of 4 increase in rate when a concentration is doubled

• If doubling a concentration leads to quadrupling the rate, the exponent with respect to that concentration is 2 (second order).

• Example 5: Confirming a zero-order effect via rate vs concentration data

• If doubling a reactant's concentration while holding others constant yields no change in rate, that reactant's order is 0.

Quick reference: key formulas and concepts

• Rate law: $r = k \prod<em>i [\text{species}</em>i]^{\text{order}_i}$

• Overall order: $\text{Overall order} = \sum<em>i \text{order}</em>i$

• For two experiments with a single varying reactant X:

• If [X] doubles and rate scales by a factor F: $F = 2^{\text{order}<em>X} \Rightarrow \text{order}</em>X = \frac{\log F}{\log 2}$

• If two experiments yield rate ratio and concentration ratio:

• General form: $\frac{r<em>2}{r</em>1} = \left(\frac{[X]<em>2}{[X]</em>1}\right)^{\text{order}_X}$

• So: $\text{order}<em>X = \frac{\log(r</em>2/r<em>1)}{\log([X]</em>2/[X]_1)}$

• Integrated rate laws and plots (for common orders):

• Zero order: $r = k, \quad [A] = [A]_0 - kt$

  • Plot: [A] vs t is linear with slope -k; k = -slope.

• First order: $r = k[A], \quad [A] = [A]_0 e^{-kt}$

  • Plot: \ln[A] vs t is linear with slope -k.

• Second order (single reactant): $r = k[A]^2, \quad \frac{1}{[A]} = kt + \frac{1}{[A]_0}$

  • Plot: 1/[A] vs t is linear with slope k.

• Common units note (vary with order):

• Zero order: k has units of concentration/time (e.g., M/s).

• First order: k has units of 1/time (e.g., s^{-1}).

• Second order: k has units of 1/(concentration·time) (e.g., M^{-1}s^{-1}).

Connections to broader concepts

• Reaction orders reflect the mechanism and the rate-determining steps, not just the stoichiometry.

• Experimentally determined orders provide insight into which steps control the pace of a reaction.

• Numerical practice with log transformations helps handle imperfect data and noisy measurements.

Practical tips for exam problems

• Always check whether a given reactant is held constant or varied between experiments.

• Use the ratio method to extract the exponent for the reactant that changes between two experiments.

• If data are not perfectly doubling, use the logarithm method to estimate the order.

• Remember: the missing reactant in a rate law corresponds to a zero order with respect to that reactant.

Summary takeaway

• Reaction orders tell you how sensitive the rate is to each reactant.

• They are determined experimentally, typically from at least three experiments.

• The overall order is the sum of the individual orders.

• Zero, first, and second orders have straightforward integrated rate laws and linear plots in the appropriate transformed variables.