Chapter 12 – Kinetics Summary
Rate:
A→B
rate = −
∆[A]
∆𝑡
rate =
aA + bB  cC + dD
***Remember in "rate" = −
∆[B]
∆𝑡
∆[A]
∆𝑡
, the “rate” is the rate of the overall reaction.
***But the rate for the reaction of A would be = −
***The rate for the formation of C would be =
∆[C]
∆𝑡
∆[A]
∆𝑡
1 ∆[A]
. It is not − a
∆𝑡
. No coefficient.
1 ∆[A]
. It is not − c
∆𝑡
. No coefficient.
Rate Law:
Dependence on Concentrations:
A(g) + 2B(g) → 2C(g)
Rate = 𝑘[𝐀]𝑥 [𝐁]𝑦
1. Do experiments. Vary one reactant concentration (while holding the other reactant
concentration constant) and determine the rate of these reactions.
2. Divide one rate equation by another rate equation and solve for “x”.
3. Divide one rate equation by another rate equation and solve for “y”.
The Integrated Rate Law: The Dependence of Concentration on Time
Order of Reaction and Half-Life: (t1/2)
Zeroth order: Rate = k[A]0 = k
t ½ = [Ainitial]/2k
First Order: Rate law: rate = k[A]1 = k[A]
Second Order: Rate = k[A]2
1/[A] = kt + 1/[A]initial
t½ = 1/(k[A]0)
**Be able to solve equations involving natural log and exponents. For example
ln ([A]t/3.250 M) = (-6.250 x 10-4/s)(3600 s)
ln ([A]t/3.250 M) = -2.250
Take the exponent of both sides  ([A]t/3.250 M) = e-2.250
[A]t = (3.250 M)(e--2.250) = 0.3425 M
The Effect of Temperature on Reaction Rate
R = 8.314 J/mol-K, T is in Kelvins
**Be able to solve equations involving natural log and exponents. For example
-6.250 x 10-4 = e-Ea/ (8.314)(300 K)
Take the natural log of both sides 
ln(-6.250 x 10-4/s) = -Ea/ (8.314 J/mol-K)(300 K)
Ea = - (8.314 J/mol-K)(300 K) x ln(-6.250 x 10-4)
Ea = 18402 J/mol = 18.4 kJ/mol
ln k = ln A + (-Ea / RT)
**Be able to solve equations involving natural log and exponents. For example
ln ([A]t/3.250 M) = (-6.250 x 10-4/s)(3600 s)
ln ([A]t/3.250 M) = -2.250
Take the exponent of both sides  ([A]t/3.250 M) = e-2.250
[A]t = (3.250 M)(e--2.250) = 0.3425 M
Reaction Mechanisms
https://courses.lumenlearning.com/boundless-chemistry/chapter/reactionmechanisms/#:~:text=The%20reaction%20mechanism%20is%20the,is%20the%20rate%2Ddeter
mining%20step.
1. Slow step followed by a fast step. Easy. The overall rate = slow rate.
2. Fast step followed by a slow step.
a. Must rearrange the fast step to find out what the intermediate is equal to.
b. Substitute the intermediate in the slow rate with what you find in Step (a).
c. Set the overall (observed) rate = slow rate (where the intermediate has been
substituted).
d. Find what k is equal to in terms of k1, k-1, and k2.
3. Steady state approximation.
a. Must rearrange the fast step to find out what the intermediate is equal to. Use rate
forward = rate backward. (Steady state approximation.)
b. Substitute the intermediate in the slow rate with what you find in Step (a).
c. Set the overall (observed) rate = slow rate (where the intermediate has been
substituted).
d. Find what k is equal to in terms of k1, k-1, and k2.