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Chemistry · Ch 7 — Chemical Kinetics

Rate Law and Rate Constant

7.3

Rate Law and Rate Constant

The rate law is an empirical fact, not a stoichiometric guess. For the general reaction xA+yB→productsxA+yB\rightarrow\text{products}, the observed relationship between rate and concentration is written as

Rate=k[A]m[B]n\text{Rate}=k[A]^m[B]^n

Here k, the rate constant, is a fixed proportionality factor for a given reaction at a given temperature; m and n are the order of the reaction with respect to A and B, and their SUM (m+n)(m+n) is the overall order. The one rule the chapter hammers on: m and n can ONLY be found by running the actual experiment and watching how the rate responds to changing concentrations -- they can never simply be copied from the balanced equation's own coefficients x and y.

Proof by example -- cyclopropane isomerisation. Table 7.2's data shows that halving [cyclopropane][\text{cyclopropane}] exactly halves the rate, so Rate∝[cyclopropane]1\text{Rate}\propto[\text{cyclopropane}]^1: the reaction is first order. Dividing Rate by [cyclopropane][\text{cyclopropane}] at each of the three data rows (Table 7.3) gives the SAME constant, 3.46×10−2 min−13.46\times10^{-2}\ \text{min}^{-1}, confirming the rate law Rate=k[cyclopropane]\text{Rate}=k[\text{cyclopropane}] and pinning down kk.

Proof by example -- oxidation of NO, a genuine two-reactant case. For 2NO(g)+O2(g)→2NO2(g)2NO(g)+O_2(g)\rightarrow2NO_2(g), three experiments vary one concentration at a time. Comparing experiments where [NO][NO] is fixed but [O2][O_2] changes shows Rate∝[O2]1\text{Rate}\propto[O_2]^1 (first order in O2O_2); comparing experiments where [O2][O_2] is fixed but [NO][NO] changes shows Rate∝[NO]2\text{Rate}\propto[NO]^2 (SECOND order in NO) -- even though NO's own coefficient in the balanced equation is 2, matching here purely by coincidence, not by rule. The full rate law is Rate=k[NO]2[O2]1\text{Rate}=k[NO]^2[O_2]^1, overall order =2+1=3=2+1=3. …

Table 7.3Rate constant for the isomerisation of cyclopropane
Rate (mol L−1^{-1} min−1^{-1})[cyclopropane] (mol L−1^{-1})k=Rate/[cyclopropane]k=\text{Rate}/[\text{cyclopropane}]
6.92×10−26.92\times10^{-2}23.46×10−23.46\times10^{-2}
3.46×10−23.46\times10^{-2}13.46×10−23.46\times10^{-2}
Table rate-vs-rate-constantDifferences between rate and rate constant of a reaction
#Rate of a reactionRate constant of a reaction
1Represents the speed at which reactants convert to products at any instantA proportionality constant
2Measured as the decrease in concentration of reactants or increase in concentration of productsEqual to the rate of reaction when the concentration of each reactant is unity
Misc example-2Example 2 -- reading off orders directly from a given rate law

Worked out. (a) Br−(aq)+BrO3−(aq)+6H+(aq)→3Br2(l)+3H2O(l)Br^-(aq)+BrO_3^-(aq)+6H^+(aq) \rightarrow 3Br_2(l)+3H_2O(l), experimental rate law Rate=k[Br−][BrO3−][H+]2\text{Rate}=k[Br^-][BrO_3^-][H^+]^2: first order in Br−Br^-, first order in BrO3−BrO_3^-, second order in H+H^+, overall order =1+1+2=4=1+1+2=4. (b) CH3CHO(g)→ΔCH4(g)+CO(g)CH_3CHO(g)\xrightarrow{\Delta}CH_4(g)+CO(g), experimental rate law Rate=k[CH3CHO]3/2\text{Rate}=k[CH_3CHO]^{3/2}: order with respect to acetaldehyde =3/2=3/2, and since it is the only species in the rate law, overall order is also $3 …

Misc example-3Example 3 -- finding overall order from a single rate/concentration/rate-constant reading

Worked out. For x+2y→productx+2y\rightarrow\text{product}, the rate is 4×10−3 mol L−1s−14\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1} when [x]=[y]=0.2[x]=[y]=0.2 M, and the rate constant at 400 K is 2×10−2 s−12\times10^{-2}\ \text{s}^{-1}. Find the overall order. Book's solution: with Rate=k[x]n[y]m\text{Rate}=k[x]^n[y]^m, substitute 4×10−3=(2×10−2)(0.2)n+m4\times10^{-3}=(2\times10^{-2})(0.2)^{n+m}, so (0.2)n+m=4×10−32×10−2=0.2(0.2)^{n+m}=\dfrac{4\times10^{-3}}{2\times10^{-2}}=0.2, giving (0.2)n+m=(0.2)1(0.2)^{n+m}=(0.2)^1 -- comparing powers, n+m=1n+m=1, so the overall order of the reaction is 1. …

Misc evaluate-yourself-2Evaluate Yourself 2 -- order from a rate-multiplier statement and from a 3-experiment data table

Worked out. Book's practice box (no printed solution). (1) For X+Y→productX+Y\rightarrow\text{product}: quadrupling [x][x] alone increases the rate by a factor of 8; quadrupling BOTH [x][x] and [y][y] increases the rate by a factor of 16. Find the order with respect to x and y, and the overall order. (2) Find the individual and overall order of 2NO(g)+Cl2(g)→2NOCl(g)2NO(g)+Cl_2(g)\rightarrow2NOCl(g) from three experiments: (1) [NO]=0.1,[Cl2]=0.1[NO]=0.1,[Cl_2]=0.1, rate =7.8×10−5=7.8\times10^{-5}; (2) [NO]=0.2,[Cl2]=0.1[NO]=0.2,[Cl_2]=0.1, rate =3.12×10−4=3.12\times10^{-4}; (3) [NO]=0.2,[Cl2]=0.3[NO]=0.2,[Cl_2]=0.3, rate =9.36×10−4 mol L−1s−1=9.36\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}. Working it through: (1) 4n=8⇒n=3/24^n=8\Rightarrow n=3/2 (order in x); combined with y, 4n+m=16⇒n+m=2⇒m=1/24^{n+m}=16\Rightarrow n+m=2\Rightarrow m=1/2 (order in y); overall order =2=2. (2) Comparing experiments 1 and 2 ([NO][NO] doubled, [Cl2][Cl_2] fixed): rate ratio ≈4.0⇒\approx4.0\Rightarrow order in NO =2=2. Comparing experiments 2 and 3 ([Cl2][Cl_2] tripled, [NO][NO] fixed): rate ratio =3.0⇒=3.0\Rightarrow order in Cl2=1Cl_2=1. Overall orde …

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