Rate Laws.

Slides:



Advertisements
Similar presentations
5.2 Systems of Linear Equations in Three Variables
Advertisements

Fill in missing numbers or operations
Chapter 12 Chemical Kinetics.
KINETICS -REACTION RATES
Processing the Data: Boyle’s Law:
Half Life. The half-life of a quantity whose value decreases with time is the interval required for the quantity to decay to half of its initial value.
= (x + 6) (x + 2) 1. x2 +8x x2 +16x + 48 = (x + 12) (x + 4)
1  1 =.
Rate Laws. Rate = k [A] x [B] y [A] and [B] represent the concentrations of reactants x and y often (but not always) represent the coefficients in front.
Number Sense and Operations Exit Patterns, Functions, and Algebra Geometry and Spatial Sense.
2 pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt ShapesPatterns Counting Number.
Welcome to Who Wants to be a Millionaire
GCSE Higher Revision Starters 11 Module 3 and 5 Revision.
© University of South Carolina Board of Trustees Trial[NO] M [H 2 ] M Initial Rate, M/s x x x10 -2.
Integrated Rate Law.
Chem Ch 28/#1 Todays To Do List Chemical Kinetics: Rate Law – What is it? - Experimental methods.
Chapter Earlier, rate was shown to be a change in concentration compared to a change in time. This is also know as a. slope.
Entry Task: Dec 6th Thursday
Rate Law Learning Goals:
There are essentially two types of problems which can be analyzed by ICE tables. Type 1 A. The initial or equilibrium concentration of some substances.
Multiplication Tables Test 3 and 5 Times Tables 3 and 5 times tables There are 10 questions. Each one will stay on the screen for 15 seconds. Write down.
Enrichment - Rate Equations to Determine Reaction Order The equation for a straight line is: Compare this equation to the rearranged first order rate-law.
Percent Reaction and Q Chem 12 Page 509 &
CHEM Pharmacy Week 11: Kinetics - Half Life
The Distributive Property. The distributive property is mental math strategy that can be used when multiplying. 43 x 5 =?
Are You Smarter Than a 5 th Grader? 1,000,000 5th Grade Topic 1 5th Grade Topic 2 4th Grade Topic 3 4th Grade Topic 4 3rd Grade Topic 5 3rd Grade Topic.
Rates and Rate Laws.
16.1 Rate expression Distinguish between the terms rate constant, overall order of reaction and order of reaction with respect to a particular reactant.
Bell Schedules Club Time is available from 8:05-8:20  1 st 8:20 – 9:15  2 nd 9:20 – 10:10  3 rd 10:15 – 11:05  4 th 11:10 – 12:50 A(11:10)
Unit Rate $45 for 5 hours. Unit Rate $45 for 5 hours $45 5 H.
Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they.
Learning Goals Students will: understand the Rate Law Equation determine the Rate Law Equation given experimental data.
KINETICS.
Relating [Reactant] and Rate In this section we will assume that the [products] does not affect the rate Therefore, for the general equation aX + bY 
Unit 6: Kinetics IB Topics 6 & 16 Part 2: Reaction Order & Half Life.
Chemistry 40S Unit 3: Chemical Kinetics Lesson 4.
Chapter 13: Chemical Kinetics CHE 124: General Chemistry II Dr. Jerome Williams, Ph.D. Saint Leo University.
AP Chemistry – Chemical Kinetics Black powder burns quickly, but the reaction is much slower than nitroglycerin detonating. 4 C 3 H 5 (NO 3 ) 3 → 12 CO.
Rate Orders and Rate Laws. Reaction Rates Are measured as the change in concentration over time. ∆[reactants] Are measured as the change in concentration.
Chemical Kinetics The area of chemistry that concerns reaction rates and reaction mechanisms.
The Rate Law. Objectives: To understand what a rate law is To determine the overall reaction order from a rate law CLE
Sometimes using simple inspection of trials cannot be used to determine reaction rates Run #[A] 0 [B] 0 [C] 0 v0v M0.213 M0.398 M0.480 M/s
1 Chemical Kinetics Chapter Chemical Kinetics Kinetics is the study of how fast chemical reactions occur and how they occur. There are 4 important.
Rate Law p You have qualitatively observed factors affecting reaction rate but… There is a mathematical relationship between reaction rate and.
Dr. Harris Lecture 18 HW: Ch 17: 5, 11, 18, 23, 41, 50 Ch 17: Kinetics Pt 1.
1. Kinetics 1.1 The rate of chemical reactions 1.2 The rate expression and order of reaction 1.3 Determining the rate equation 1.4 The rate determining.
REACTION RATE LAWS MRS. NIELSEN HONORS CHEMISTRY.
Chemical Kinetics The area of chemistry that concerns reaction rates and reaction mechanisms.
Rate Laws.
Rate Laws.  Determine the rate law from experimental data.  Explain the effect of concentration on reaction rates.  Derive rate law form a reaction.
*Measuring how fast reactions occur. Lecture 2: Stoichiometry & Rate Laws.
6.5 Rate Laws & Order of Reaction Rate Law The rate (r), is proportional to the product of the initial reactant concentrations raised to some exponent.
Tutorial 10 Chemical kinetics. LEARNING OUTCOMES By the end of this session the student should be able to: 1.Define reaction rate. 2.Identify the factors.
Ch 14- Chemical Kinetics -The area of chemistry concerned with the speeds, or rates, of reactions.
Integrated Rate Law.
Rates and Rate Laws.
Rate Law.
Rate Law Main Concept: The rate law shows how the rate depends on reactant concentrations.
Chapter 17 The Reaction Process.
Rates and Rate Laws.
Rate Orders.
Integrated Rate Law.
Chemical Kinetics Method of Initial Rates
Kinetics - Reaction Rates
Chemical Kinetics Chemical Reaction Rates (13.2)
Rate Laws.
Unit 4: Solutions and Kinetics
Integrated Rate Law.
Rate Law p
Rate Laws How rate laws are determined ALWAYS experimentally
Presentation transcript:

Rate Laws

Determine the rate law and order of a chemical reaction from experimental data. Include: various reaction orders, rate versus concentration graphs. Additional KEY Terms Ratio

Rate is proportional to changes in A and B A + B C Rate is proportional to changes in A and B Rate Law helps us calculate the rate of a REACTION not a reactant.

It is an expression that shows the quantitative effect of concentration changes on reaction rate. Rate = k[A]x where: k rate constant [A] concentration of A x reaction order **The rate constant (k) and the order (x) can only be determined experimentally**

Temperature is the only factor to affect The rate constant (k) is specific for each reaction at a specific temperature. Temperature is the only factor to affect the rate constant. Rate = k[A]x

Reaction Order

Each reactant can affect the rate differently A + B Products Rate = k[A]x[B]y Each reactant can affect the rate differently The orders of a reaction (x and y) indicate how much each [reactant] affects the rate of a reaction.

Rate = k[A] 1 x First order reaction ( x = 1) The reaction rate is directly proportional to changes in reactant concentration. [A] is doubled - rate doubles 2 = 21 [A] is tripled - rate triples 3 = 31 x Rate = k[A] 1

Rate = k[A]2 Second order reaction (x = 2) The reaction rate is proportional to changes in reactant concentration squared. Rate = k[A]2 Doubling [A] - increase rate 4x 4 = 22 Tripling [A] - rate increase 9x 9 = 32

**NOT included in the rate law if determined Third order reaction (x = 3) The reaction rate is proportional to changes in reactant concentration cubed. Rate = k[A]3 Zero order reaction (x = 0) The rate does not depend on the [A]. Changing [A] does NOT change the rate. Rate = k **NOT included in the rate law if determined to be zero order**

Rate = k[A][B]2 1st order + 2nd order = 3rd order overall Overall order of reaction is the sum of the orders: x + y = overall reaction order 1st order + 2nd order = 3rd order overall

Calculating Rate Law

Several ways to determine the rate law: differential rate law - uses calculus integrated rate law – uses graphing software initial rates method – uses data tables Determining rate law: Measure the effect of changes in concentration of one reactant on rate, while keeping the other reactant constant.

Ratio Approach 8 = 2  4 2 = 1 2 

= Using ratios: A + B → products same Rate = k[A]x[B] y Trial [A] (mol/L) [B] (mol/L) Initial Rate (mol/Ls) 1 0.10 0.20 2.0 2 0.30 18.0 3 0.40 16.0 same Rate = k[A]x[B] y Rate = k[A]x[B] y Rate2 k [A]2x  [B]2y Rate1 k [A]1x  [B]1y = α 9.0 α [3]x  1 9.0 α [3]x 18.0 α 0.3 x  0.2 y 2.0 0.1  0.2 x = second order

= Using ratios: A + B → products 2 Rate = k[A]2[B] y 1 Trial [A] (mol/L) [B] (mol/L) Initial Rate (mol/Ls) 1 0.10 0.20 2.0 2 0.30 18.0 3 0.40 16.0 Rate = k[A]2[B] y 1 Rate = k[A]x[B] y Rate3 k [A]3x  [B]3y Rate1 k [A]1x  [B]1y = α 8.0 α [2]2  [2]y 8.0 α 4  [2]y 2.0 α [2]y 16.0 α 0.2 2  0.4 y 2.0 0.1  0.2 y = first order

Can’t have multiple answers…pick new trials. Using ratios: A + B → products 2 Trial [A] (mol/L) [B] (mol/L) Initial Rate (mol/Ls) 1 0.10 0.20 2.0 2 0.30 18.0 3 0.40 16.0 Rate = k[A]2[B] y You might pick bad trials Can’t have multiple answers…pick new trials. Rate2 α [A]2x  [B]2y 9.0 α [3]  [1]y 2 Rate1 [A]1x  [B]1y 9.0 α 9  [1]y 1 α [1]y 18.0 α 0.3 x  0.2 y 2.0 0.1  0.2 y = 0 or 1 or 2…..

Common Sense Approach

x y 1 1 rate = k[H2O2] [HI] rate = k[H2O2] [HI] = 2x 2 same same Rate2 k [H2O2]2x  [HI]2y Rate1 k [H2O2]1x  [HI]1y 2.0 α 1 ּ [2]y = x y 1 1 rate = k[H2O2] [HI] 2.0 α [2]y 0.0152 α 0.1 x  0.2 y 0.0076 0.1  0.1 rate = k[H2O2] [HI] x = first order

x 1 y Rate = k[A] [B] 8.0 α [2]x  [4]1 Rate α [A]x  [B]y Tl [A]i mol/L [B]i Initial Rate (mol/Ls) 1 0.0100 0.0240 1.45 x 10−4 2 0.0120 7.25 x 10−5 3 0.0200 0.0480 5.80 x 10−4 x y Rate = k[A] [B] 1 8.0 α [2]x  [4]1 Rate α [A]x  [B]y 8.0 α [2]x  4 Rate3 α [A]3x  [B]3y Rate2 [A]2x  [B]2y 2.0 α [2]x x = first order 5.8 -4 α 0.2 x  0.048 1 7.25 -5 0.1  0.012

3 A (g) + B (g) + 2 C (g) 2 D (g) + 3 E (g) a. Write the rate law for this reaction. b. Calculate the value of the rate constant (k). c. Calculate the rate for Trial #5. d. Calculate the concentration of A in Trial #6.

rate = k[A][B]2 = = 1 2 same same same same same same same same same same same same a. Write the rate law for this reaction. Rate2 k [A]2x  [B]2y  [C]2z Rate1 k [A]1x  [B]2y  [C]2z = Rate2 [A]2x Rate1 [A]1x = rate = k[A][B]2

rate = k[A][B]2 b. Calculate the value of the rate constant (k). To find the value of k, we use that data from ANY trial. Don’t include units for k.

rate = (200)(0.50 mol/L)(0.40 mol/L)2 rate = 16 mol/Ls c. Calculate the rate for Trial #5. rate = k[A][B]2 rate = (200)(0.50 mol/L)(0.40 mol/L)2 rate = 16 mol/Ls

d. Calculate the concentration of A in Trial #6. rate = k[A][B]2

Conclusion: Everything in the Rate Law must be determined experimentally: 1. Write a basic rate law with all reactants 2. Determine the order for each reactant (1, 2, 0) 3. Re-write the rate law with the determined order of reaction for each 4. Solve any problems

CAN YOU / HAVE YOU? Determine the rate law and order of a chemical reaction from experimental data. Include: various reaction orders, rate versus concentration graphs. Additional KEY Terms Ratio