Kinetics Notes: Rate, Rate Law, and Instantaneous vs Average Rates

Instantaneous vs Average Rates

• When we compare two time points (e.g., 0 s and 50 s), we are talking about the average rate of reaction over that interval.

• For a reaction written in one direction, the rate is determined by what happens on the left-hand side (the reactants).

• Example system:

  • Reaction: NO₂ → products (direction fixed, products not important for rate calculation here).

  • We want the average rate of consumption of NO₂ over the first 50 s.

  • Concentrations used: [NO₂] at t = 0 s is 0.010 M; [NO₂] at t = 50 s is 0.0079 M.

• Calculation of the average rate over 0 to 50 s:

  • ext{rate}{avg} = rac{[NO2]{50s} - [NO2]_{0s}}{50 - 0} = rac{0.0079 - 0.010}{50} = -4.2 imes 10^{-5}\n \, ext{M s}^{-1}.

  • The negative sign indicates consumption of NO₂; the positive rate magnitude is used for the rate of reaction: | ext{rate}| = 4.2 imes 10^{-5}\n \, ext{M s}^{-1}.$n- In exams or problem sets, grid-based data may be provided so reading off concentrations is exact; in this lecture a textbook value is used.

• The “instantaneous rate” is the slope of the concentration–time curve at a single time point, not an average over an interval.

• If you want the rate at a specific time (e.g., t = 100 s), you approximate the tangent line to the curve at that time and compute the slope (rise/run) of that tangent.

• In this example, to illustrate instantaneous rate, a tangent is drawn at t = 100 s with a chosen run and rise from the textbook figure; the slope computed is used as the instantaneous rate at t = 100 s.

• Slopes generally decrease with time for a reaction proceeding in the forward direction because the reactant concentration is decreasing, so fewer collisions occur.

• Observations about time dependence:

  • The average rate over successive 50 s intervals (0–50 s, 50–100 s, 100–150 s, …) tends to decrease as the reaction proceeds.

  • This decreasing rate aligns with the intuitive idea that more reactants early on lead to faster rates, and fewer reactants later slow things down.

• Rate goes to zero as time → ∞ for many reactions that approach equilibrium (the curve flattens, slope → 0).

Consequences of the slope and instantaneous rate

• The instantaneous rate at a time t is the slope of the concentration vs time curve at t:

  • ext{rate}(t) = - rac{d[NO2]}{dt}igg|{t}.$</p></li></ul></li><li><p>The average rate over a time interval is based on finite differences, while the instantaneous rate uses a tangent slope at a single point.</p></li><li><p>In practice, instantaneous rates are used to discuss how fast the reaction is at a precise moment (e.g., to decide when to stop the reaction in a flask).</p></li><li><p>The rate constant k and the reaction order n in the rate law determine how the rate changes with time and concentration.</p></li></ul><h3 id="44ecdce5-ed81-4849-934c-383b6d0f7a91" data-toc-id="44ecdce5-ed81-4849-934c-383b6d0f7a91" collapsed="false" seolevelmigrated="true">Rate law, signs, and interpretation</h3><ul><li><p>The general differential rate law for a single reactant A decomposing is:</p><ul><li><p>$- rac{d[A]}{dt} = k [A]^n,$</p></li><li><p>where k is the rate constant and n is the reaction order with respect to A.</p></li></ul></li><li><p>The rate of reaction (a positive quantity) is often written as:</p><ul><li><p>$ ext{Rate} = k [A]^n,$</p></li><li><p>with the negative sign appearing when expressing the rate of consumption of a reactant (to emphasize decrease in concentration).</p></li></ul></li><li><p>If the reaction is written to go left to right, we typically report rates as positive values; the sign is handled by the minus sign in the differential form.</p></li><li><p>Stoichiometry does not fix the rate law/order: the reaction NO₂ → products with 2 NO₂ could be first, second, or zero order with respect to NO₂ depending on the mechanism. This is illustrated by comparing experiments rather than by stoichiometry alone.</p></li></ul><h3 id="98da2951-b5e7-4916-a178-bfcffb7f4831" data-toc-id="98da2951-b5e7-4916-a178-bfcffb7f4831" collapsed="false" seolevelmigrated="true">Slope, instantaneous rate, and calculus notes</h3><ul><li><p>In this course (Chem 122), calculus is not required in the rate-law derivations, but the concept of the derivative is used to discuss instantaneous rates.</p></li><li><p>The slope or tangent line method provides a calculus-free way to approximate instantaneous rates from experimental data.</p></li><li><p>If you want a calculus-based treatment, you can explore derivatives and the limit definition of slope, but it’s not required for this course.</p></li></ul><h3 id="57f7331d-be77-429f-9eef-2cbfdc5360d5" data-toc-id="57f7331d-be77-429f-9eef-2cbfdc5360d5" collapsed="false" seolevelmigrated="true">Instantaneous rate at a specific time (no calculus)</h3><ul><li><p>Example from the lecture: tangent to the curve at t = 100 s was formed; a triangle with base Δt = 110 s and height Δ[NO₂] = 0.026 M was used to estimate the slope:</p><ul><li><p>$ ext{slope} = rac{ ext{rise}}{ ext{run}} = rac{0.026 ext{ M}}{110 ext{ s}} imes (-1) = -2.36 imes 10^{-5} ext{ M s}^{-1}.$</p></li><li><p>Therefore, the instantaneous rate at t = 100 s is approximately$- rac{d[NO2]}{dt}igg|{t=100s}
    ext{ ≈ } 2.36 imes 10^{-5} ext{ M s}^{-1}.$</p></li></ul></li><li><p>The instructor notes: the rate law is typically written as$- rac{d[NO2]}{dt} = k [NO2]^n,$and the instantaneous rate is the left-hand side value with the appropriate [NO₂].</p></li></ul><h3 id="0f4b409f-8edf-48d9-a173-4a3a5dcdbc2b" data-toc-id="0f4b409f-8edf-48d9-a173-4a3a5dcdbc2b" collapsed="false" seolevelmigrated="true">Stoichiometry and comparative rates for NO₂ decomposition</h3><ul><li><p>For the reaction 2 NO₂ → 2 NO + O₂, the rates are related by stoichiometry:</p><ul><li><p>The rate of NO₂ consumption equals the rate of NO production:</p></li><li><p>$- rac{d[NO_2]}{dt} = rac{d[NO]}{dt}.$</p></li><li><p>The rate of O₂ production is half the rate of NO₂ consumption (because 2 NO₂ produce 1 O₂):</p></li><li><p>$ rac{d[O2]}{dt} = rac{1}{2} rac{d[NO]}{dt} = rac{1}{2}ig(- rac{d[NO2]}{dt}ig).$</p></li></ul></li><li><p>Example instantaneous rates (from the lecture): if$- rac{d[NO_2]}{dt} = 8.6 imes 10^{-6} ext{ M s}^{-1},$then</p><ul><li><p>NO production rate:$ rac{d[NO]}{dt} = 8.6 imes 10^{-6} ext{ M s}^{-1},$</p></li><li><p>O₂ production rate:$ rac{d[O_2]}{dt} = 4.3 imes 10^{-6} ext{ M s}^{-1}.$</p></li></ul></li></ul><h3 id="2606ad3a-1a9b-48d7-88a7-ebc60c332dcc" data-toc-id="2606ad3a-1a9b-48d7-88a7-ebc60c332dcc" collapsed="false" seolevelmigrated="true">Rate law concepts and the method of initial rates</h3><ul><li><p>Rate law definition (differential form):</p><ul><li><p>$- rac{d[A]}{dt} = k [A]^n,$</p></li><li><p>rate constant k (temperature-dependent), order n (indicates how the rate depends on concentration).</p></li></ul></li><li><p>Key ideas:</p><ul><li><p>The rate constant k depends only on temperature (same value at the same T).</p></li><li><p>The order n is determined by mechanism and is not tied to stoichiometry.</p></li><li><p>The rate law is usually written in terms of reactants (for reactions that go left to right).</p></li><li><p>If the reaction is more complex (multiple reactants), the rate law generalizes to:</p></li><li><p>$ ext{Rate} = k [A]^a [B]^b \,,

  • with the overall order being \(a + b\).

• The practical goal: with k and the orders, you can predict the rate for any concentrations and forecast how long a reaction will take to reach a given extent.

Method of initial rates (experimental approach)

• Idea: determine the rate law by measuring initial rates at different initial concentrations and comparing how the rate changes.

• Example framework (single reactant A):

  • Suppose we have rate data with [A] = 0.9 M giving rate ≈ 5.4 (in some units), and with [A] = 0.45 M giving rate ≈ 2.7 (half as large).

  • Since halving [A] halves the rate, the reaction is first order in A: ext{Rate} \

Rate law, signs, and interpretation

• The general differential rate law for a single reactant A decomposing is:

  • $-Rac{d[A]}{dt} = k [A]^n,$

  • where k is the rate constant and n is the reaction order with respect to A.

• The rate of reaction (a positive quantity) is often written as:

  • $ext{Rate} = k [A]^n,$

  • with the negative sign appearing when expressing the rate of consumption of a reactant (to emphasize decrease in concentration).

• If the reaction is written to go left to right, we typically report rates as positive values; the sign is handled by the minus sign in the differential form.

• Stoichiometry does not fix the rate law/order: the reaction NO₂ → products with 2 NO₂ could be first, second, or zero order with respect to NO₂ depending on the mechanism. This is illustrated by comparing experiments rather than by stoichiometry alone.

Rate law concepts and the method of initial rates

• Rate law definition (differential form):

  • $-Rac{d[A]}{dt} = k [A]^n,$

  • rate constant k (temperature-dependent), order n (indicates how the rate depends on concentration).

• Key ideas:

  • The rate constant k depends only on temperature (same value at the same T).

  • The order n is determined by mechanism and is not tied to stoichiometry.

  • The rate law is usually written in terms of reactants (for reactions that go left to right).

  • If the reaction is more complex (multiple reactants), the rate law generalizes to:

  • $ext{Rate} = k [A]^a [B]^b \,,$

  • with the overall order being $a + b.$

• The practical goal: with k and the orders, you can predict the rate for any concentrations and forecast how long a reaction will take to reach a given extent.

Method of initial rates (experimental approach)

• Idea: determine the rate law by measuring initial rates at different initial concentrations and comparing how the rate changes.

• Example framework (single reactant A):

  • Suppose we have rate data with [A] = 0.9 M giving rate ≈ 5.4 (in some units), and with [A] = 0.45 M giving rate ≈ 2.7 (half as large).

  • Since halving [A] halves the rate, the reaction is first order in A: $ext{Rate} \propto [A]^1.$