12.1: Reaction Rates and Rate Laws

12.1 Reaction Rates

  • Question addressed: How fast can a reaction convert reactants into products?
  • Core idea: The rate of a reaction measures how fast product appears (or reactant disappears) over time.
  • Key terms:
    • Rate: speed of the reaction; depends on concentration, temperature, state of matter, and catalysts.
    • Average rate: rate calculated over a time interval.
    • Instantaneous rate: rate at a specific moment, often estimated from the slope of a concentration–time curve.
    • Initial rate: instantaneous rate at time t = 0.
    • Units of rate:
      Rate units=molL⋅s\text{Rate units} = \frac{\text{mol}}{\text{L} \cdot \text{s}}
  • Example reaction studied: \ce{H2 + Cl2 -> 2 HCl}
    • Consider how fast HCl is produced; this is the rate of the reaction. What affects the rate? concentration, temperature, catalysts, etc.
  • Conceptual takeaways from the data visuals (conceptual descriptions from the transcript):
    • A plot of concentration vs. time shows reactants decreasing and products increasing over time.
    • Rates can be evaluated from the slope of these curves:
    • Average rate over a time interval (e.g., 0 to 25 s) is calculated from the change in concentration divided by the change in time.
    • Instantaneous rate at a given time is the slope at that time (tangent line).
    • Initial rate is the instantaneous rate at t = 0.
  • General forms (from the transcript):
    • For a generic reaction with stoichiometric coefficients, the rate of reaction and the rate of formation/consumption of species are related by the stoichiometric coefficients. The convention used is:
      Rate<em>rxn=1νRate</em>chem\text{Rate}<em>{rxn} = \frac{1}{\nu} \text{Rate}</em>{chem}
  • Example relation for the reaction H2 + Cl2 → 2 HCl:
    • When 1 molecule of H2 disappears, 2 molecules of HCl appear.
    • Reactant rates are reported as negative because reactants are consumed:
      Rate<em>rxn=−Rate(H</em>2)=−Rate(Cl2)=12Rate(HCl)\text{Rate}<em>{rxn} = -\text{Rate}(\mathrm{H</em>2}) = -\text{Rate}(\mathrm{Cl_2}) = \frac{1}{2} \text{Rate}(\mathrm{HCl})
  • Practice example (12.1–12.2 style):
    • For the reaction: \ce{2 NO2 + O2 -> 2 NO3}
    • If the production rate of NO3 is 0.30 mol L−1s−10.30\ \text{mol} \, \text{L}^{-1} \text{s}^{-1}, determine:
    • rate(NO2)
    • rate(O2)
    • reaction rate (overall)
    • Solution:
    • Let the rate of reaction be r. Then: d[NO3]dt=2r⇒r=12d[NO3]dt=12×0.30=0.15 mol L−1s−1\frac{d[\mathrm{NO3}]}{dt} = 2r \Rightarrow r = \frac{1}{2} \frac{d[\mathrm{NO3}]}{dt} = \frac{1}{2} \times 0.30 = 0.15\ \text{mol L}^{-1} \text{s}^{-1}
    • Then: −d[NO2]dt=2r⇒rate(NO2)=2r=0.30 mol L−1s−1-\frac{d[\mathrm{NO2}]}{dt} = 2r \Rightarrow \text{rate}(\mathrm{NO2}) = 2r = 0.30\ \text{mol L}^{-1} \text{s}^{-1}
    • And: −d[O2]dt=r⇒rate(O2)=0.15 mol L−1s−1-\frac{d[\mathrm{O2}]}{dt} = r \Rightarrow \text{rate}(\mathrm{O2}) = 0.15\ \text{mol L}^{-1} \text{s}^{-1}
  • Clicker-style check (12.1): If the reaction: \ce{4 Fe + 3 O2 -> 2 Fe2O3} produces Fe2O3 at a rate of 0.60 mol L−1s−10.60\ \text{mol L}^{-1} \text{s}^{-1}, what is the reaction rate?
    • The relation: d[Fe2O3]dt=2r⇒r=d[Fe2O3]dt/2=0.60/2=0.30 mol L−1s−1\frac{d[\mathrm{Fe2O3}]}{dt} = 2r \Rightarrow r = \frac{d[\mathrm{Fe2O3}]}{dt} / 2 = 0.60 / 2 = 0.30\ \text{mol L}^{-1} \text{s}^{-1}
    • Answer: 0.30 (option C in the transcript).
  • Practical implications:
    • The initial rate is often used for comparing how fast different reactions proceed under the same starting conditions.
    • Rate conventions help in standardizing how rates are reported across reactions with different stoichiometries.

12.2 Four factors affecting reaction rates

  • Four main factors listed:
    • Concentration: Higher reactant concentrations lead to more collisions, increasing rate.
    • Temperature: Higher temperature increases molecular energy, more collisions exceed activation energy, increasing rate.
    • State of matter (phase): Reactants must be well mixed; gases and solutions typically react faster due to better mixing.
    • Catalyst: Provides an alternative pathway with lower activation energy, increasing rate.

12.3 Rate Law

  • Rate law concept:
    • For the reaction \ce{H2 + Cl2 -> 2 HCl} , the rate law is:
      Rate=k[H2]m[Cl2]n\text{Rate} = k [\mathrm{H2}]^{m} [\mathrm{Cl2}]^{n}
    • k is the rate constant at a fixed temperature.
    • m and n are the reaction orders with respect to H2 and Cl2, respectively.
    • Experimentally determined from data:
    • [H2] and [Cl2] are the initial concentrations used to probe how the rate changes when one concentration is varied.
  • Key details from the transcript:
    • Experimental results (rates) with initial reactant concentrations:
    • Data set:
      • Rate = 1.30×10−5 mol L−1s−11.30 \times 10^{-5} \ \text{mol L}^{-1} \text{s}^{-1}, [H2] = 0.001 mol L−10.001\ \text{mol L}^{-1}, [Cl2] = 0.001 mol L−10.001\ \text{mol L}^{-1}
      • Rate = 3.90×10−5 mol L−1s−13.90 \times 10^{-5} \ \text{mol L}^{-1} \text{s}^{-1}, [H2] = 0.003 mol L−10.003\ \text{mol L}^{-1}, [Cl2] = 0.001 mol L−10.001\ \text{mol L}^{-1}
      • Rate = 3.25×10−4 mol L−1s−13.25 \times 10^{-4} \ \text{mol L}^{-1} \text{s}^{-1}, [H2] = 0.001 mol L−10.001\ \text{mol L}^{-1}, [Cl2] = 0.005 mol L−10.005\ \text{mol L}^{-1}
      • Rate = 5.20×10−5 mol L−1s−15.20 \times 10^{-5} \ \text{mol L}^{-1} \text{s}^{-1}, [H2] = 0.001 mol L−10.001\ \text{mol L}^{-1}, [Cl2] = 0.002 mol L−10.002\ \text{mol L}^{-1}
    • From these data, determine the orders m and n and the rate constant k:
    • Compare data where [H2] changes and [Cl2] held constant to determine m:
      • When [H2] increases from 0.001 to 0.003 (factor of 3) while [Cl2] stays at 0.001, rate increases from 1.30×10^-5 to 3.90×10^-5 (factor of 3). This indicates m = 1.
    • Compare data where [Cl2] changes (and [H2] is fixed at 0.001) to determine n:
      • When [Cl2] increases from 0.001 to 0.005 (factor of 5) while [H2] stays at 0.001, rate increases from 1.30×10^-5 to 3.25×10^-4 (factor of 25). This is 5^2, so n = 2.
    • Therefore, rate law is:
      Rate=k[H2][Cl2]2\text{Rate} = k [\mathrm{H2}] [\mathrm{Cl2}]^{2}
    • Determine k using any data point; using the first point:
      1.30×10−5=k(1.0×10−3)(1.0×10−3)2=k(1.0×10−3)(1.0×10−6)=k×1.0×10−91.30 \times 10^{-5} = k (1.0 \times 10^{-3}) (1.0 \times 10^{-3})^{2} = k (1.0 \times 10^{-3}) (1.0 \times 10^{-6}) = k \times 1.0 \times 10^{-9}
      ⇒k=1.30×10−51.0×10−9=1.30×104\Rightarrow k = \frac{1.30 \times 10^{-5}}{1.0 \times 10^{-9}} = 1.30 \times 10^{4}
    • Verified with the second data point:
      k=3.90×10−5(3.0×10−3)(1.0×10−3)2=3.90×10−53.0×10−9=1.30×104k = \frac{3.90 \times 10^{-5}}{(3.0 \times 10^{-3}) (1.0 \times 10^{-3})^{2}} = \frac{3.90 \times 10^{-5}}{3.0 \times 10^{-9}} = 1.30 \times 10^{4}
    • Result:
    • Rate law: Rate=k[H2][Cl2]2\text{Rate} = k [\mathrm{H2}] [\mathrm{Cl2}]^{2} with k=1.3×104 (L2 mol−2s−1)  .k = 1.3 \times 10^{4} \ \text{(L}^{2}\text{ mol}^{-2} \text{s}^{-1}) \;. (Units follow from overall order 3.)
    • Overall reaction order: m + n = 1 + 2 = 3 (third order overall).
    • Units of k for a third-order rate law (Rate = k [A][B]^2):
      Units(k)=mol L−1s−1(mol L−1)3=L2mol2  s\text{Units}(k) = \frac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^{3}} = \frac{\text{L}^2}{\text{mol}^2 \; \text{s}}
  • Additional notes on rate law concepts (from the transcript):
    • Overall reaction order equals the sum of reaction orders with respect to each reactant (m + n).
    • The rate constant k depends on temperature; higher temperature increases rates for a given k.
    • The units of k depend on the overall order of the reaction.
  • Practice question (12.3 style): Given a generic rate law Rate=k[A]\text{Rate} = k [A], what is the overall reaction order? What are the units of Rate, and the units of k?
    • Overall order: 1 (first order).
    • Units of Rate: mol L−1s−1\mathrm{mol} \, \mathrm{L}^{-1} \mathrm{s}^{-1}
    • Units of k: s−1\mathrm{s}^{-1}

Additional practice and check points (12.1–12.3 wrap-ups)

  • Rate conventions and signs:
    • Rate of reaction is often defined using the species with a coefficient of 1 in the balanced equation.
    • For a reaction with coefficients aA + bB -> cC + dD, the rate relations satisfy:
      Raterxn=−1ad[A]dt=−1bd[B]dt=1cd[C]dt=1dd[D]dt .\text{Rate}_{rxn} = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt} \,.
  • Rate units table (as per the transcript):
    • Overall order 0: Units(k)=molL⋅s\text{Units}(k) = \frac{\text{mol}}{\text{L} \cdot \text{s}}
    • Overall order 1: Units(k)=1s\text{Units}(k) = \frac{1}{\text{s}}
    • Overall order 2: Units(k)=Lmol⋅s\text{Units}(k) = \frac{\text{L}}{\text{mol} \cdot \text{s}}
    • Overall order 3: Units(k)=L2mol2⋅s\text{Units}(k) = \frac{\text{L}^2}{\text{mol}^2 \cdot \text{s}}
    • The general pattern: each additional order adds one factor of L in the numerator and one factor of mol in the denominator.
  • Clicker insights (conceptual):
    • Second-order rate law units question: For a second-order rate law, the units of k can be represented as A. L/(mol s), B. M^{-1} s^{-1}, C. M^{-1} / s, D. 1/(M s), E. All of the above. Answer: All of the above, since L, M, and their inverses relate by unit conversions in aqueous solutions.
    • A zero- or first-order comparison question example: If tripling [Cl2] has no effect on the rate of a reaction Cl2 → 2 Cl, the rate law is Rate = k (i.e., zero-order with respect to Cl2). The correct choice is A: Rate = k.
  • Recap highlights (12.4-ish content in transcript):
    • Rate depends on concentration (rate law), temperature, and catalysts.
    • Rate is defined as the initial rate for a species with a coefficient of 1 in the balanced equation.
    • Reaction orders and the rate constant k are determined from experimental data.
    • Rate laws connect observable rates to concentrations via the rate constant and reaction orders, and the units of k depend on the overall order.

Quick reference formulas from the transcript (for rapid lookup)

  • Rate law form:
    Rate=k[H2]m[Cl2]n\text{Rate} = k [\mathrm{H2}]^{m} [\mathrm{Cl2}]^{n}
  • Relation between reaction rate and species rates:
    Rate<em>rxn=1νRate</em>chem\text{Rate}<em>{rxn} = \frac{1}{\nu} \text{Rate}</em>{chem}
    Rate<em>rxn=−Rate(H</em>2)=−Rate(Cl2)=12Rate(HCl)\text{Rate}<em>{rxn} = -\text{Rate}(\mathrm{H</em>2}) = -\text{Rate}(\mathrm{Cl_2}) = \frac{1}{2} \text{Rate}(\mathrm{HCl})
  • Example: NO3 formation in \ce{2 NO2 + O2 -> 2 NO3}
    • If d[NO3]dt=0.30 mol L−1s−1\dfrac{d[\mathrm{NO3}]}{dt} = 0.30 \ \text{mol L}^{-1} \text{s}^{-1}, then
    • Reaction rate r=12d[NO3]dt=0.15 mol L−1s−1r = \dfrac{1}{2} \dfrac{d[\mathrm{NO3}]}{dt} = 0.15 \ \text{mol L}^{-1} \text{s}^{-1}
    • rate(NO2) = -\dfrac{d[\mathrm{NO2}]}{dt} = 2r = 0.30 \ \text{mol L}^{-1} \text{s}^{-1}
    • rate(O2) = -\dfrac{d[\mathrm{O2}]}{dt} = r = 0.15 \ \text{mol L}^{-1} \text{s}^{-1}
  • Example: third-order rate law units
    • If Rate = k [A]^1 [B]^2, then overall order = 3 and
    • Units(k)=L2mol2  s\text{Units}(k) = \frac{\mathrm{L}^2}{\mathrm{mol}^2 \; \mathrm{s}}