Solving Kinetics Problems Writing Rate Expressions from Balanced Equations Finding Reaction Order from [A]- time Data Finding Reaction Order from Initial Rate Finding k from [A] & Rate Data Calculating ½ Life from k Drawing Reaction Profile from Data Creating Reaction Mechanisms from Rate Law & Finding the Slow Step
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Solving Kinetics ProblemsWriting Rate Expressions from Balanced
EquationsFinding Reaction Order from [A]-time
DataFinding Reaction Order from Initial Rate
Finding k from [A] & Rate DataCalculating ½ Life from k
Drawing Reaction Profile from DataCreating Reaction Mechanisms from Rate Law & Finding the Slow Step
Writing Rate Expressions from Balanced Equations
The rate expression is not the rate law It tells us what we are looking for in a rate
law experiment Example: H
2 (g) + I
2 (g) → 2 HI (g)
Rate expression could be rate of disappearance of hydrogen
Rate = ΔH2 = 1 ΔHI
Δt 2 Δt
It tells you in the lab what to measure
Your turn
The equation is Cu (s) + 2 Ag+ (aq) → Cu2+ (aq) + 2 Ag(s)
Write at least two expressions by which we could measure the rate
Answers:
Rate = ΔCu = 1 ΔAg+ = ΔCu2+ = 1ΔAg
Δt 2 Δt Δt 2 Δt
Finding Reaction Order from [A]-time Data
Needed: a chart of concentration of a reactant we want to study vs. time
Product: a fast graph of Ln[A] vs time or 1/[A] vs time
If the Ln[A] v time graph is linear, it's first order If the Ln[A] v time graph is a curve, it's 2nd order If [A] v time goes down in a linear fashion, it's
zero order (rare)
Your turn
Reactant A changes concentration with time. Here is the data:
What is the order of the reaction in A? Hint: find the natural log of each [A] and
graph on your graphing calculator
Solution
Plotting LN[A] v. time we get
The straight line plot suggests a first order equation Rate = k[A]
Your Turn
Butadiene changes concentration with time. Here is the data:
Time (s) 0 1000 1800 2800 3600 4400
[Butadiene]Mol/L
0.01000 0.00625 0.00476 0.00370 0,00313 0.00270
What is the order of the reaction in butadiene?
Again, find the natural log of each [A] and graph on your graphing calculator.
Solution
Plotting LN[butadiene] vs time we get
This is not a straight line as shown by the straight line between the first and last points, so the reaction must be 2nd order.
Finding Reaction Order from Initial Rate
This is a favorite of test writers! You are given concentrations of all
reactants and the rate of reaction for each set of conditions
You must identify the controls and variables and determine how the rate is affected
You are looking for doubling of rate when concentration doubles (1st), or quadrupling of rate when concentration doubles (2nd)
Your TurnSO2 + O2 → SO3
Given the following data, determine the order of reaction in SO
2 and O
2
Look for doubling of concentrations with other concentration held constant!
Solution
Given the following data, determine the order of reaction in SO
2 and O
2
In 2 and 1, oxygen concentration doubles while SO2 is held constant
Solution
Given the following data, determine the order of reaction in SO
2 and O
2
In 2 and 1, rate of formation of the trioxide goes from 0.60 to 1.20, also a doubling
Solution
Given the following data, determine the order of reaction in SO
2 and O
2
That means the rate is directly proportional to the concentration of the oxygen gas, so the reaction is first
order in O2
Solution
Given the following data, determine the order of reaction in SO
2 and O
2
When we look at Experiments 1 and 3, we see the oxygen concentration is held constant, and the SO2 concentration is
doubling.
Solution
Given the following data, determine the order of reaction in SO
2 and O
2
But at the same time, the rate of trioxide formation goes from 1.2 to 4.8, which is a quadrupling. That is 22 times the
initial rate, so the rate is going up faster than the concentration.
Solution
Given the following data, determine the order of reaction in SO
2 and O
2
This means that Rate = k[SO2]2
The reaction is 2nd order in SO2
Predicting Concentrations, Rates
Now that we have reaction order, let's see if we can fill in the table.
The reaction is 1st order in O2 and 2nd order in SO2
Solutions
To find the oxygen in Exp 4, we see that the rate is 17% lower than in Exp 1
The reaction is 1st order in O2 and 2nd order in SO2
Solutions
And the [SO2] is 33% lower than in Exp 1
That is predictable since Rx is 2nd order
The reaction is 1st order in O2 and 2nd order in SO2
Solutions
The change in sulfur dioxide accounts for all the rate change, so oxygen is 0.20 M
The reaction is 1st order in O2 and 2nd order in SO2
0.20 M
Solutions
Now predict the rate of forming trioxide in experiment 5
The reaction is 1st order in O2 and 2nd order in SO2
0.20 M
Solutions
Both concentrations change. Let's do oxygen first
The reaction is 1st order in O2 and 2nd order in SO2
0.20 M
Solutions
The oxygen is 50% higher than in Exp 2, so rate should be 50% higher, or 9.0 x 10-3
The reaction is 1st order in O2 and 2nd order in SO2
0.20 M
Solutions
But SO2 is 17% higher than in Exp 2, so rate from that is 33% higher yet, 1.2 x 10-2
The reaction is 1st order in O2 and 2nd order in SO2
0.20 M
1.2 x 10-2 M/s
Finding k from [A] & Rate Data
Let's go with a first order reaction we have already looked at:
Rate = -0.240[A] -2.49, from the equation of lineAnd k = negative of slope, or 0.240 here
Finding k from [A] & Rate Data
But suppose we have the data, but no equation, and know it's first order
Look at Exp 1 and 2 and assure yourself that the reaction is first order in A
Finding k from [A] & Rate Data
Look at Exp 1 and 3 and see that the reaction is 2nd order in B
Finding k from [A] & Rate Data
Look at Exp 3 & 4 and see that the reaction is 2nd order in C
Finding k from [A] & Rate Data
Thus the rate law isRate = k[A][B]2[C]2
Finding k from [A] & Rate Data
Rate = k[A][B]2[C]2
So solve for k and plug in the numbers from any of the fully known data lines
K = Rate [A][B]2[C]2
Finding k from [A] & Rate Data
Rate = k[A][B]2[C]2
So solve for k and plug in the numbers from any of the fully known data lines
K = 2.85 x 1012
This is harder than anything on a test
Your Turn
Find the value of the rate constant:
2 NO (g) + Cl2 (g) → 2NOCl (g)
Data:
[NO]o mol/L [Cl
2]
o mol/L Initial Rate
Mol/L-min
0.10 0.10 0.18
0.10 0.20 0.36
0.20 0.20 1.45
First you must find the order of the reaction in both reactants, and write the rate law, then plug in to solve for k.
Solution
Find the value of the rate constant:
The rate law is Rate = k [NO]2[Cl2]
So k = (Rate)/[NO]2[Cl2] = 0.18 x 10-3 L2mol-2min-1
[NO]o mol/L [Cl
2]
o mol/L Initial Rate
Mol/L-min
0.10 0.10 0.18
0.10 0.20 0.36
0.20 0.20 1.45
First you must find the order of the reaction in both reactants, and write the rate law, then plug in to solve for k.
Calculating ½ Life from kWe'll stick with first order reactions here
Remember that t ½ = 0.693
kSuppose the rate constant for a 1st order
reaction is 0.18 x 10-3s-1 What is the half-life? If the initial concentration is 2.0 M, what will it be after the reaction runs for 770 seconds?
Solution
Suppose the rate constant for a 1st order reaction is 0.18 x 10-3s-1 What is the half-life? If the initial concentration is 2.0 M, what will it be after the reaction runs for 770 seconds?
T ½ = 0.693 = 0.693 = 385 s
K 0.0018 s-1
Now 770/385 = 2.0 so that's 2 half-lives
In the first half-life, the concentration goes to 1.0 M
In the second half-life, the concentration goes to 0.50 M
Your Turn
A certain first-order reaction is 45.0% complete in 65 s. What are the rate constant and half-life for this process?
This is #37 on page 605
Your Turn
A certain first-order reaction is 45.0% complete in 65 s. What are the rate constant and half-life for this process?
If [A]o = 100.0, then after 65 s, [A] = 55.0. In 1st order Rxn,
LN([A]/[A]o = -kt, LN(55.0/100.0 = -k(65 s)
You do the arithmetic, but
K = 9.2 x 10-3 s-1 and t ½ = 0.693/k = 75 s
This is #37 on page 605
Drawing Reaction Profile from Data
This should be a little familiar to you if you recall the diagrams of endothermic and exothermic reaction
We just add the activation energy and diagram it like a hill the reactants have to get over by colliding at the right energy and orientation.
Drawing Reaction Profile from Data
Draw a reaction energy profile for an endothermic reaction with ΔH = +34 kJ/mol and a forward activation energy of 66 kJ/mol. Calculate the activation energy in the reverse direction.
Drawing Reaction Profile from Data
Draw a reaction energy profile for an endothermic reaction with ΔH = +34 kJ/mol and a forward activation energy of 66 kJ/mol. Calculate the activation energy in the reverse direction.
Your turn
Draw the reaction energy profile of an exothermic reaction with a forward activation energy of 116 kJ/mol and a ΔH of -225 kJ/mol. Calculate the activation energy in the reverse direction.
Your turn
Draw the reaction energy profile of an exothermic reaction with a forward activation energy of 116 kJ/mol and a ΔH of -225 kJ/mol. Calculate the activation energy in the reverse direction.
Ea for reverse direction is 341 kJ/mol
From UC Davis
Creating Reaction Mechanisms from Rate Law & Finding the Slow Step
The rate law gives us a mathematical picture of the initial and time-related concentrations or pressures at a given temperature.
However, to control reactions we need to understand how they run. So we derive reaction mechanisms from the rate law, whenever possible.
Example
For the reaction H2 (g) + 2 ICl → I
2 + 2 HCl the rate
law is found to be Rate = k [H2][ICl]
What is the most rational two-step mechanism that fits all the information given?
First, a termolecular collision is almost never seen. So a two-step mechanism is very reasonable.
The rate law implies that both hydrogen and ICl are involved in the slow step.
Example
For the reaction H2 (g) + 2 ICl → I
2 + 2 HCl the rate
law is found to be Rate = k [H2][ICl]
The rate law implies that both hydrogen and ICl are involved in the slow step.