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.

So we can reasonably say

• H2 + ICl → HI + HCl (slow)

• HI + ICl → I2 + HCl (fast)

Overall: H2 (g) + 2 ICl → I

2 + 2 HCl

Another mechanism

Here's one I found on YouTube:

Organic mechanismClick on the link above