Ch. 12 Rates and Rate Laws

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# Ch. 12 Rates and Rate Laws - PowerPoint PPT Presentation

Ch. 12 Rates and Rate Laws. A chemical reaction depends on particles Colliding into one another. Reaction Rate. The change in concentration of a reactant or product per unit of time. Reaction Rates:. 2NO 2 (g)  2NO(g) + O 2 (g). 1. Can measure disappearance of reactants.

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Ch. 12RatesandRate Laws

A chemical reaction depends on particles

Colliding into one another.

Reaction Rate

The change in concentration of a reactant or product per unit of time

Reaction Rates:

2NO2(g)  2NO(g) + O2(g)

1. Can measure

disappearance of

reactants

2. Can measure

appearance of

products

3. Are proportional

stoichiometrically

Reaction Rates:

2NO2(g)  2NO(g) + O2(g)

4. Are equal to the

slope tangent to

that point

5. Change as the

reaction proceeds,

if the rate is

dependent upon

concentration

[NO2]

t

Rate Laws

Differential rate lawsexpress (reveal) the relationship between the concentration of reactants and the rate of the reaction.

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

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

What does a Rate Law look like?

Rate = k[A] [B]

k, is the rate law constant

Exponent is the order of each reactant

Adding exponents gives the overall rate order

Determining Order withConcentration vs. Time data

(the Integrated Rate Law)

Zero Order:

has a constant rate

First Order:

Half-life is constant, doubling conc. will double reaction rate

Second Order:

Doubling the concentration, will quadruple the rxn rate

Writing a (differential) Rate Law

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

2 NO(g) + Cl2(g)  2 NOCl(g)

Writing a Rate Law

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

R = k[NO]x[Cl2]y

In experiment 1 and 2, [Cl2] is constant while [NO] doubles.

The rate quadruples, so the reaction is second order with respect to [NO]

 R = k[NO]2[Cl2]y

Writing a Rate Law

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

R = k[NO]2[Cl2]y

In experiment 2 and 4, [NO] is constant while [Cl2] doubles.

The rate doubles, so the reaction is first order with respect to [Cl2]

 R = k[NO]2[Cl2]

Writing a Rate Law

Part 2– Determine the value for k, the rate constant, by using any set of experimental data:

R = k[NO]2[Cl2]

Writing a Rate Law

Part 3– Determine the overall order for the reaction.

R = k[NO]2[Cl2]

= 3

2

+

1

 The reaction is 3rd order

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

Classwork
• P.581 #27-28
Graphs of Orders

Value of K is given by the slope of the line.

• P.553 First order: time vsln [A]
• Linear with a negative slope
• P.558 Second order: time vs 1/[A]
• Linear with a positive slope
• P.559 Zero order: time vs [A]
• Linear with a negative slope
Solving an Integrated Rate Law

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!!

Time vs. [H2O2]

Regression results:

y = ax + b

a = -2.64 x 10-4

b = 0.841

r2 = 0.8891

r = -0.9429

Time vs. ln[H2O2]

Regression results:

y = ax + b

a = -8.35 x 10-4

b = -.005

r2 = 0.99978

r = -0.9999

Time vs. 1/[H2O2]

Regression results:

y = ax + b

a = 0.00460

b = -0.847

r2 = 0.8723

r = 0.9340

And the winner is… Time vs. ln[H2O2]

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

2. The (differential) rate law is:

3. The integrated rate law is:

4. But…what is the rate constant, k ?

Finding the Rate Constant, k

Method #1: Calculate the slope from the

Time vs. ln[H2O2] table.

Now remember:

 k = -slope

k = 8.32 x 10-4s-1

Finding the Rate Constant, k

Method #2: Obtain k from the linear regresssion analysis.

Regression results:

y = ax + b

a = -8.35 x 10-4

b = -.005

r2 = 0.99978

r = -0.9999

Now remember:

 k = -slope

k = 8.35 x 10-4s-1

Classwork
• P.582 #35, 39