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:
  \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:
  \text{Rate}{rxn} = \frac{1}{\nu} \text{Rate}{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:
  \text{Rate}{rxn} = -\text{Rate}(\mathrm{H2}) = -\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\ \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: \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: -\frac{d[\mathrm{NO2}]}{dt} = 2r \Rightarrow \text{rate}(\mathrm{NO2}) = 2r = 0.30\ \text{mol L}^{-1} \text{s}^{-1}
- And: -\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\ \text{mol L}^{-1} \text{s}^{-1}, what is the reaction rate?
- The relation: \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:
  \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 \times 10^{-5} \ \text{mol L}^{-1} \text{s}^{-1}, [H2] = 0.001\ \text{mol L}^{-1}, [Cl2] = 0.001\ \text{mol L}^{-1}
  - Rate = 3.90 \times 10^{-5} \ \text{mol L}^{-1} \text{s}^{-1}, [H2] = 0.003\ \text{mol L}^{-1}, [Cl2] = 0.001\ \text{mol L}^{-1}
  - Rate = 3.25 \times 10^{-4} \ \text{mol L}^{-1} \text{s}^{-1}, [H2] = 0.001\ \text{mol L}^{-1}, [Cl2] = 0.005\ \text{mol L}^{-1}
  - Rate = 5.20 \times 10^{-5} \ \text{mol L}^{-1} \text{s}^{-1}, [H2] = 0.001\ \text{mol L}^{-1}, [Cl2] = 0.002\ \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:
  \text{Rate} = k [\mathrm{H2}] [\mathrm{Cl2}]^{2}
- Determine k using any data point; using the first point:
  1.30 \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}
  \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 = \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: \text{Rate} = k [\mathrm{H2}] [\mathrm{Cl2}]^{2} with 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):
  \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 \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: \mathrm{mol} \, \mathrm{L}^{-1} \mathrm{s}^{-1}
- Units of k: \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:
  \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: \text{Units}(k) = \frac{\text{mol}}{\text{L} \cdot \text{s}}
- Overall order 1: \text{Units}(k) = \frac{1}{\text{s}}
- Overall order 2: \text{Units}(k) = \frac{\text{L}}{\text{mol} \cdot \text{s}}
- Overall order 3: \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:
\text{Rate} = k [\mathrm{H2}]^{m} [\mathrm{Cl2}]^{n}
Relation between reaction rate and species rates:
\text{Rate}{rxn} = \frac{1}{\nu} \text{Rate}{chem}
\text{Rate}{rxn} = -\text{Rate}(\mathrm{H2}) = -\text{Rate}(\mathrm{Cl_2}) = \frac{1}{2} \text{Rate}(\mathrm{HCl})
Example: NO3 formation in \ce{2 NO2 + O2 -> 2 NO3}
- If \dfrac{d[\mathrm{NO3}]}{dt} = 0.30 \ \text{mol L}^{-1} \text{s}^{-1} , then
- Reaction rate r = \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
- \text{Units}(k) = \frac{\mathrm{L}^2}{\mathrm{mol}^2 \; \mathrm{s}}