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Analyze this in terms of the overall reaction order. A first order reaction depends on the concentration of one of the reactants raised to the first power. We ...
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Red paint from the pond in Van Gogh's Wheat Stack Under a Cloudy Sky has degraded to grey/white over time.
Key topics: Reaction Rates Factors That Affect Reaction Rates Reaction Mechanisms How fast does a reaction occur? e.g. , many famous painting are changing color…
(from ibchem.com) Also see page 434. Collision Theory of Chemical Reactions A very simple picture of a chemical reaction between two molecules is that
The transition state (activated complex) is a transient species of high energy that falls apart to generate the products of the reaction. Transition state theory (TST), also known as Activated Complex Theory, is a more sophisticated theory that has two main assumptions
Reaction progress with time. The rate changes over time. Reaction progress can be monitored using a spectrometer. We can follow a color change, conductance change, pressure change, or whatever is convenient. Consider the reaction A → B. average rate = [A] t = [B] t
Value of k? k = rate / [Br 2 ]
5
1
3
1
5
1
3
1 (same within experimental error) Consider the reaction A → 2B. rate = [A] t = 1 2 [B] t
e.g. , Write the balanced equation corresponding to the following rate expression rate = 1 2 [N 2 O 5 ] t = 1 2 [N 2 ] t = 1 5 [O 2 ] t Solution:
Solution: We are told that [C] t = 0. 086 , [A] t = 0. 172 , [B] t = 0. 258 If we write the balanced equation a A + b B → c C, we know that
If we divide each expression by 0.086 we get 2 a = 3 b = 1 c Therefore a = 2, b = 3, and c = 1. Dependence of Reaction Rate on Reactant Concentration In the bromine reaction we considered earlier,
we determined the rate = k [Br 2 ]. In terms of the reactant concentrations, we write
x
y with x = 1 and y = 0. The exponents (x and y) are empirically determined. Without experimental data we can’t simply predict their values.
rate = k[A] x [B] y . Ø x = 0: the reaction is 0 th order with respect to A. Ø x = 1: the reaction is 1 st order with respect to A. Ø x = 2: the reaction is 2 nd order with respect to A. The overall reaction order is the sum of the exponents, so the bromine reaction is 1 st order overall (1 + 0 = 1).
Then we can determine k from any single exp.
3
1
1
1 The rate has units of M s
e.g. , For the reaction A + B → 2C, the rate = k [A] 2 with k = 1. × 10
First order kinetics; we obtain a straight line when the data is plotted on a logarithmic scale. From concentration vs. time data ( e.g. Lab 17 in Chem 1112) we would do the following:
Half-life: The half-life of a reaction (t1/2) is the time required for the reactant concentration to drop to half of its original value. We express this as
1 / 2
1 / 2
Then the integrated rate law gives
1 / 2
First order kinetics; the concentration is halved every half-life. (from chemwiki.ucdavis.edu) ln 1 2 = ln 2 ⇡ 0 .693 so that t 1 / 2 =
Second order kinetics: a plot of 1/[A] vs. time has slope k and intercept 1/[A] 0 To compute the half-life, we use
e.g. , The reaction 2A → B is second order in A with k = 32 M
Solution: 1 [A]t = kt +
) kt =
[A]t
) t =
k[A]t
k[A] 0 t = 1 (32 M ^1 s ^1 )(0. 0018 M ) 1 (32 M ^1 s ^1 )(0. 0075 M ) = 17.36 s 4 .17 s =13.2 s. For the half-life, t 1 / 2 = 1 (32 M ^1 s ^1 )(0. 0075 M ) = 4.17 s t 1 / 2 = 1 (32 M ^1 s ^1 )(0. 0025 M ) = 12.5 s Zeroth Order Reactions A zeroth order reaction is independent of reactant concentration: For aA → products, rate = k [A] 0 = k. Once again, we redefine the rate constant to include the stoichiometric coefficient and write
Now we change Δ to d (finite change to infinitesimal change):
Z (^) [A] t [A] 0 d[A] = k Z (^) t 0 dt ) [A]t [A] 0 = kt or [A]t = kt + [A] 0
We can obtain the activation energy and the frequency factor from a plot of ln k versus 1/T. Dependence of Reaction Rate on Temperature Reactions almost always occur faster at higher temperature. Quantified by the Arrhenius Equation for the rate constant k = Ae
A = frequency factor Ea = activation energy R = gas constant = 8.314 J K
Now, consider two different temperatures:
Subtract (1) – (2) to obtain ln k 1 k 2 = Ea R ✓ 1 T 2 1 T 1 ◆ e.g. , Determine Ea, and also k at 655 K, from the data: T (K) k (s
k
3 J mol 1 8 .314 J K ^1 mol 1
) ln 2. 0 ⇥ 10 4 ln k = 0. 291 ) ln k = 8. 226 ) k = e 8. 226 = 2. 7 ⇥ 10 4
