Chemistry Lecture 9

Instantaneous vs. Average Rates

• Instantaneous rate mirrors what you see on a speedometer: the rate at a single instant in time. It is the slope of the concentration vs. time curve at that moment.
  • For a reactant A: rate = $-\frac{d[A]}{dt}$; for a product P: rate = $\frac{d[P]}{dt}$.
  • In practice, you estimate it from the slope of the tangent to the curve at a specific time.
• Average rate
  • Defined over a time interval: $\text{rate}<em>{avg} = -\frac{\Delta[A]}{\Delta t}$ (for a reactant) or $\text{rate}</em>{avg} = \frac{\Delta[P]}{\Delta t}$ (for a product).
• Example: H(2) + I(2) → 2 HI
  • H(_2) is a reactant; HI is a product.
  • Rate relationships follow stoichiometry: for a reaction aA + bB → products with coefficients a, b, the rates relate as
  • $-\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[\text{Products}]}{dt} = r$
  • where r is the rate of the reaction, and c is the stoichiometric coefficient of the product (or its corresponding relation).

Rate Law and Rate Constant

• The rate law expresses how the rate depends on reactant concentrations:
  • $r = k [A]^m [B]^n \cdots$
  • m, n, etc. are the reaction orders with respect to each reactant; they are not necessarily equal to the stoichiometric coefficients.
  • The overall order of the reaction is the sum of the individual orders: $\text{overall order} = m + n + \cdots$
• The rate constant k is temperature- and condition-dependent (e.g., presence of a catalyst).
• Determining orders experimentally (not from the balanced equation): vary one reactant at a time while keeping others constant and observe how the rate changes.
  • If doubling [A] while holding others fixed doubles the rate, the order in A is 1. If doubling [B] increases the rate by a factor of 8, the order in B is 3, etc.
  • In a common example: for a two-reactant system, the final rate law might look like
  • $\text{rate} = k [O_2]^1 [NO]^2$
  • Overall order = 1 + 2 = 3.
• Practical notes:
  • The orders m, n are determined from experiments, not from coefficients in the balanced equation.
  • The rate constant k is the same for a given set of conditions (temperature, catalysts); if k varies, you may have changed conditions.
  • The units of k depend on the overall order of the reaction.

Experimental Determination of the Rate Law (Single Reaction Focus)

• Method: focus on one reactant at a time, holding others constant, to determine its order.
• Example approach with NO(2) decomposition NO(2) → NO + NO; or NO(_2) + NO → products (as used in class):
  • Collect data at different initial concentrations and measure initial rates.
  • Using two experiments where one reactant’s concentration changes while the other is fixed:
  • If the rate doubles when [NO(2)] changes by a factor of 2 (and [NO] fixed), order in NO(2) is 1.
  • If the rate changes by a factor of 8 when [NO] changes by a factor of 2, order in NO is 3, etc.
  • From a complete data set, determine the rate law: e.g., for a reaction with NO(_2) and NO, you might find
  • $\text{rate} = k [NO_2]^1 [NO]^2$
  • Overall order = 3.
• Determining k from experiments:
  • With temperature fixed and no catalyst added, use the data to solve for k with the determined orders.
  • Example: using an experiment with measured rate and known concentrations:
  • $\text{rate} = k [NO_2]^1 [NO]^2$
  • Solve for k from a chosen data point, then verify with another data point.
  • Units of k depend on the overall order; for a third-order reaction the units are M^-2 s^-1, etc. (check consistency with data).
• Practical considerations:
  • If two experiments yield inconsistent k, something is off (data quality, measurement error, or condition changes).
  • Temperature and catalysts must be held constant when determining k; you can use multiple experiments to validate k, but not mix conditions.

Graphical Methods for a Single Reactant (Arriving at the Order via Plots)

• When focusing on one reactant A with rate law r = k [A]^m, three standard plots help identify m:
  • Plot 1: [A] vs time. Linear if m = 0 (zero order).
  • Plot 2: (\ln[A]) vs time. Linear if m = 1 (first order).
  • Plot 3: 1/[A] vs time. Linear if m = 2 (second order).
• Rationale:
  • These plots arise from integrating the rate law for a single reactant and rewriting the relation into linear form.
• Common example (NO(_2) decomposition):
  • Data plotted as the three graphs, the one that is linear indicates the reaction order with respect to NO(_2).
  • If 1/[NO(2)] vs time is linear, the reaction is second order in NO(2).
  • Slope of the linear plot equals the rate constant times a factor dependent on the integrated form:
  • For second order in A: $\frac{1}{[A]<em>t} = k t + \frac{1}{[A]</em>0}$
• Determining k from the linear plot: the slope equals k (for the appropriate integrated form).
• Example conclusions: if the natural log plot is not linear but the 1/[A] plot is linear, the reaction is not first order in A; use the appropriate integrated equation to extract k.

Integrated Rate Laws and Half-Lives

• For a single reactant A with rate law r = k [A]^m, the integrated forms give linear relations:
  • First order (m = 1):
  • $\ln [A]<em>t = -k t + \ln [A]</em>0$
  • Graph: slope = -k; intercept = \ln [A]_0.
  • Half-life: $t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}$
  • Second order (m = 2):
  • $\frac{1}{[A]<em>t} = k t + \frac{1}{[A]</em>0}$
  • Graph: slope = k; intercept = \frac{1}{[A]_0}.
  • Half-life: $t<em>{1/2} = \frac{1}{k [A]</em>0}$
  • Zero order (m = 0):
  • $[A]<em>t = [A]</em>0 - k t$
  • Graph: slope = -k; intercept = [A]_0.
  • Half-life: $t<em>{1/2} = \frac{[A]</em>0}{2k}$
• Nuclear decay as a first-order process:
  • $\ln [A]<em>t = -k t + \ln [A]</em>0$
  • A linear plot of (\ln[A]_t) vs t yields slope = -k.
• General notes:
  • The half-life behavior differs by order: only first-order half-life is independent of initial concentration; second-order half-life depends on [A]_0.
  • For a two-reactant system, multi-variable plots are more complex; the one-reactant method is a common stepping-stone.

Dynamic Equilibrium in Kinetics

• Reversible reactions: aA + bB ⇌ cC + dD
  • At dynamic equilibrium, the forward and reverse rates are equal:
  • $r<em>{forward} = r</em>{reverse}$
  • This balance defines equilibrium constant expressions and the concentrations at which the system stops showing net change in concentrations.
• Connection to physical processes:
  • Dynamic equilibrium also appears in non-chemical contexts:
  • Vapor pressure: rate of evaporation equals rate of condensation.
  • Crystallization/dissolution: rate of dissolution equals rate of crystallization.
• Significance in kinetics:
  • Kinetics helps explain how systems approach equilibrium and how factors like temperature or catalysts shift the rates of forward vs. reverse reactions.

SETI Method and Study Tips for the Exam

• SETI: State, Explain, Example, Illustrate
  • State the concept clearly.
  • Explain the underlying relationships or equations.
  • Provide an example to illustrate the concept.
  • Illustrate with a mental model or a real-world analogy.
• Class tips discussed for exam readiness:
  • Expect math questions (derivations, rate laws, integrated forms) and concept questions.
  • There are practice problems and a practice exam in the files; some quizzes are timed with post-exam comments for reviewing mistakes.
  • In discussions, instructors asked students two reflective questions:
  • Why did you choose IU Indianapolis?
  • If there is one thing you would change (aside from grades or parking), what would it be and why?
• Practical exam prep outline:
  • Review chapters 12–14 content; use the exam prep files.
  • Practice solving instantaneous rate problems, rate laws, and graph-based determinations of reaction order.
  • Be comfortable with interpreting the slope of tangents, computing instantaneous rates from data, and applying integrated rate laws.