Chemical

Kinetics By: Ms. Buroker

Chemical Kinetics

Spontaneity is important in determining if a reaction occurs- but it doesn’t tell us much about the speed of the reaction. Kinetics is the area of chemistry that deals with reaction rates.

Factors Affecting Reaction Rates

1.) Concentration of reactants

2.) Temperature (Increasing temperature increasesthe reaction rates- in what direction is a differentquestion.)

3.) Surface Area

4.) Catalysis (Substances that participate but are not

changed by the reaction. It is not included in thebalanced equation. Inhibitors slow down

reactions.)

2N2O5→ 4NO2 + O2

Let’s say that at t=0, the concentration of N2O5 is 0.0100mol/L and the concentrations of NO2 and O2 are 0.00mol/L. Over time, the concentration of N2O5 will decrease and the concentrations of NO2 and O2 will increase. Kinetics deals with the speed at which this takes place!!

Reaction Rate = change in concentration of a reactant or product

per unit time

Reaction Rates

Rate = D[A] Dt

Reaction Rate = concentration of A @ t2- concentration of A @t1

t2- t1

Reaction Rate = change in concentration of a reactant or product

per unit time

Change positive means increase & Negative means decrease

These brackets mean “concentration of”

Rate = D[A] Dt

Reaction Rates

Once the rate of one component is determined, others can be determined (i.e. the rate of appearance and disappearance) using stoichiometric ratios. (More on this later)

Plotting the concentration of a chemical in a reaction as a function of time is called a Kinetic Curve or Concentration vs. Time Curve.

Rates are not constant by change over time; however it is possible to calculate the slope at a given time (Instantaneous rate) by drawing a tangent at that point and calculating the slope (derivative) of the line.

Always use a positive value for rate

Kinetic Curves or Concentration vs. Time Curves

Rate =-D[A] Dt

2N2O5→ 4NO2 + O2

2N2O5→ 4NO2 + O2

Finding the instantaneous rates at several times during the reaction.

2N2O5 → 4NO2 + O2As 2mol of N2O5 is consumed

4 moles of NO2 is being formed

1 mole of O2 is being formed

→ +

Rate LawsChemical reactions are reversible!

2N2O5 → 4NO2 + O2

*When enough NO2 and O2 have formed,

they will react to form N2O5 so the D[N2O5] depends on the rate of the forward and reverse reaction.

We aren’t going to deal with anything that complicated in this class … so we’ll study on the rate of the FORWARD reaction!! In other words, we’ll study the rate shortly after the reactants are mixed, before the products have significant time to build up.

Rate Laws

Rate = k[A]n

Proportionality Constant

Concentration of reactant A

Order of the reactant

Remember that the products are not being considered- only the rate of the reactants disappearing! “n,” the order of the reaction must be determined experimentally!!

Rate LawsDifferential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction.

Integrated rate laws express (reveal) the relationship between concentration of reactants and time

*The differential rate law is usually just called “the rate law.”

Writing a (differential) Rate Law

2 NO(g) + Cl2(g) 2 NOCl(g)Experiment [NO]

Mol/L[Cl2]Mol/L

Rate Mol/L·s

1 0.250 0.250 1.43 x 10-6

2 0.500 0.250 5.72 x 10-6

3 0.250 0.500 2.86 x 10-6

4 0.500 0.500 11.4 x 10-6

Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction:

Step 1 – Determine the values for the exponents in the rate law:

Rate= k[NO]x[Cl2]y

Writing a (differential) Rate Law

Step 2 – We want to know how the reaction is affected according to each reactant, so we look at the data and choose a situation where the reactants do not change. Then we plug and chug using the Rate law from above to determine the order of the reaction … x and y!!

Experiment [NO]Mol/L

[Cl2]Mol/L

Rate Mol/L·s

1 0.250 0.250 1.43 x 10-6

2 0.500 0.250 5.72 x 10-6

3 0.250 0.500 2.86 x 10-6

4 0.500 0.500 11.4 x 10-6

Writing a (differential) Rate LawIn Experiments 1 and 3, the concentration of NO is held constant, so we can see how the rate is affected by changing the concentration of Cl2.

Writing a (differential) Rate Law

We solve by dividing the two rates where the data is the same: Rate 3/Rate 1

2.86 x 10-6 = k(0.250)x(0.500)y

1.43 x 10-6= k (0.250)x(0.250)y

SO…

AND …2 = 2y therefore y=1

2.86 x 10-6 = (0.500)y

1.43 x 10-6 = (0.250)y

So what this means is that the reaction is first order with respect to Cl2 so whatever you do to change [Cl2], the same thing will happen to the rate.

Experiment [NO]Mol/L

[Cl2]Mol/L

Rate Mol/L·s

1 0.250 0.250 1.43 x 10-6

2 0.500 0.250 5.72 x 10-6

3 0.250 0.500 2.86 x 10-6

4 0.500 0.500 11.4 x 10-6

Writing a (differential) Rate LawNow, we’re going to look at experiments 3 and 4, where the concentration of Cl2 is held constant, so we can see how the rate is affected by changing the concentration of NO.

Writing a (differential) Rate Law

We solve by dividing the two rates where the data is the same: Rate 3/Rate 4

2.86 x 10-6 = k(0.250)x(0.500)y

11.4 x 10-6= k (0.500)x(0.500)y

SO…

AND ….25 = .5x therefore y=2

2.86 x 10-6 = (0.250)x

11.4 x 10-6 = (0.500)x

So what this means is that the reaction is second order with respect to NO so whatever you do to change [NO], the rate will double.

Writing a Rate LawPart 2 – Determine the value for k, the rate constant, by using any set of experimental data:

Experiment

[NO] (mol/L)

[Cl2]

(mol/L)

Rate Mol/L·s

1 0.250 0.250 1.43 x 10-6

R = k[NO]2[Cl2]

261.43 10 0.250 0.250

mol mol molx k

L s L L

6 3 25

3 3 2

1.43 109.15 10

0.250

x mol L Lk x

L s mol mol s

Writing a Rate LawPart 3 – Determine the overall order for the reaction.

R = k[NO]2[Cl2]

Overall order is the sum of the exponents, or orders, of the reactants

2 + 1 = 3

The reaction is 3rd order

Determining Order withConcentration vs. Time data

The Integrated Rate Law• Zero Order: graphing time vs. concentration

produces a linear graph

• First Order: graphing time vs. (ln concentration) produces a linear graph

• Second Order: graphing time vs. 1/concentration produces a linear graph

Solving an Integrated Rate Law

Time (s) [H2O2] (mol/L)

0 1.00

120 0.91

300 0.78

600 0.59

1200 0.37

1800 0.22

2400 0.13

3000 0.082

3600 0.050

• Problem: Find the integrated rate law and the value for the rate constant, k

• A graphing calculator with linear regression analysis greatly simplifies this process!!

Let’s Graph!!

In your graphing calculator, put the data in this way:

1.) time vs. [H2O2]

2.) time vs. ln[H2O2]

3.) time vs. 1/[H2O2]

Time (s) [H2O2] (mol/L)

0 1.00

120 0.91

300 0.78

600 0.59

1200 0.37

1800 0.22

2400 0.13

3000 0.082

3600 0.050

The graph that gives you the straight line … is our winner!!

And the winner is… Time vs. ln[H2O2]

1. As a result, the reaction is 1st order

2. The (differential) rate law is:

2 2[ ]R k H O

3. The integrated rate law is:

2 2 2 2 0ln[ ] ln[ ]H O kt H O

4. But…what is the rate constant, k ?

Collision ModelKey Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why?

Other Factors Affecting Reaction Rate

• Temperature affects rate– Higher temp = more collisions– More collisions = faster rate

• However (there’s always a “however” isn’t there?)– Measured rate is much smaller than the predicted

number of collisions– Why? Not all collisions have enough “oomph” to

make a reaction go

Collision Model1.) Collisions must have sufficient energy to produce the reaction (must equal or exceed the activation energy).

2.) Colliding particles must be correctly oriented to one another in order to produce a reaction.

Activation Energy• Arrhenious proposed the concept of

an activation energy needed to react• Not all collisions occur with the

necessary energy to cause bonds to break and then reform

time

Energy

Ea

Endothermic Reactions

Exothermic Reactions

Activation Energy• Arrhenious found that the number of

collisions with the correct activation energy increase exponentially with temp

(collisions with Ea) = (total collisions)e-

Ea/RT

• However () the observed reaction rate is still smaller than the rate predicted by the above equation

• This is due to molecular orientation– The molecules must not only hit with the

right energy, but the right orientation as well

Arrhenious Equation

• Taking the orientation into account, Arrhenious proposed this equation:

k=Ae-Ea/RT

where A = frequency factor (a measure of number of collisions with

correct orientation)• Taking the natural log of the equation

gives:ln(k) = -(Ea/RT) + ln(A)

The Arrhenius Equation, Rearranged

1ln( ) ln( )aE

k AR T

Simplifies solving for Ea

-Ea / R is the slope when (1/T) is plotted against ln(k)

ln(A) is the y-intercept Linear regression analysis of a table of (1/T) vs. ln(k) can quickly yield a slope

Ea = -R(slope)

Catalysis• Catalyst: A substance that speeds up a reaction without being consumed

• Enzyme: A large molecule (usually a protein) that catalyzes biological reactions.

• Homogeneous catalyst: Present in the same phase as the reacting molecules.

• Heterogeneous catalyst: Present in a different phase than the reacting molecules.

Lowering of Activation Energy by a Catalyst

Catalysts Increase the Number of Effective Collisions

Reaction Mechanism

The reaction mechanism is the series of elementary steps by which a chemical reaction occurs.

The sum of the elementary steps must give the overall balanced equation for the reaction

The mechanism must agree with the experimentally determined rate law

Rate-Determining Step

In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction.The experimental rate law must agree with the rate-determining step

Overall Reaction: 2NO2(g) + F2(g)→ 2FNO2(g)

Experimental rate law: Rate = k[NO2][F2]

Identifying the Rate-Determining Step

For the reaction: 2H2(g) + 2NO(g) N2(g) +

2H2O(g) The experimental rate law is:

R = k[NO]2[H2] Which step in the reaction mechanism is the rate-determining (slowest) step?

Step #1 H2(g) + 2NO(g) N2O(g) + H2O(g) Step #2 N2O(g) + H2(g) N2(g) + H2O(g)

Step #1 agrees with the experimental rate law

Identifying Intermediates

For the reaction: 2H2(g) + 2NO(g) N2(g) + 2H2O(g)

Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?)

Step #1 H2(g) + 2NO(g) N2O(g) + H2O(g) Step #2 N2O(g) + H2(g) N2(g) + H2O(g)

2H2(g) + 2NO(g) N2(g) + 2H2O(g)

N2O(g) is an intermediate

The Raschig reaction produces hydrazine, N2H4, an industrially important reducing agent, from NH3 and OCl- in basic, aqueous solution. A proposed mechanism is:

Step 1: (fast) NH3(aq) + OCl-(aq) → NH2Cl(aq) + OH-

(aq)

Step 2: (slow) NH2Cl(aq) + NH3(aq) → N2H5(aq) + Cl-(aq)

Step 3: (fast) N2H5+

(aq) + OH-(aq) → N2H4(aq) + H2O(l)

a. What is the overall stoichiometric equation?b. Which step of the three is rate-determining?c. Write the rate equation for the rate-determining step.d. What reaction intermediates are involved?

An Example …

Rate Laws SummaryZero Order First Order Second Order

Rate Law Rate = k Rate = k[A] Rate = k[A]2

Integrated Rate Law

[A] = -kt + [A]0 ln[A] = -kt + ln[A]0

Plot the produces a straight line

[A] versus t ln[A] versus t

Relationship of rate constant to slope of straight line

Slope = -k Slope = -k Slope = k

Half-Life

1

[ ]versus t

A

0

1 1

[ ] [ ]kt

A A

01/ 2

[ ]

2

At

k 1/ 2

0.693t

k 1/ 2

0

1

[ ]t

k A