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Title: Lesson 4 Determining Reaction Order and Rate Equation Learning Objectives: – Build on our understanding of reaction order to construct rate equations.

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Presentation on theme: "Title: Lesson 4 Determining Reaction Order and Rate Equation Learning Objectives: – Build on our understanding of reaction order to construct rate equations."— Presentation transcript:

1 Title: Lesson 4 Determining Reaction Order and Rate Equation Learning Objectives: – Build on our understanding of reaction order to construct rate equations. – Analyse experimental data in order to determine a rate equation.

2 Main Menu Lesson 6: Rate Equations  Objectives:  Build on our understanding of reaction order to construct rate equations.  Analyse experimental data in order to determine a rate equation.

3 Main Menu What is a rate equation?  A rate equation allows us to calculate the rate we would expect for any concentration of reactants.  For example in the reaction: A + B + C  D + E  The rate equation is: Rate = k[A] x [B] y [C] z  Where:  [A], [B] an [C] are the concentrations of each reactant  x, y, and z are the order of reaction with respect to each reactant  k is the ‘rate constant’

4 Main Menu The rate constant, k  The rate constant k is essentially a measure of how readily a reaction will take place:  Higher k  faster reaction  Lower k  slower reaction  k is dependent on temperature  As temperature increases, so does k  As temperature decreases, so does k  k is independent of concentration:  Provided the temperature is fixed, k is always the same

5 Main Menu Units of k vary depending on the overall order of reaction

6 Main Menu For example (from last lesson):  The reaction is 0 th order w.r.t reactant A, and 1 st order w.r.t reactant B  Therefore, the rate equation is:  Rate = k[B]  The value of k is given by (using values for Experiment 1): k = Rate / [B] k = 1.25x10 -2 / 1.00 = 1.25x10 -2 s -1 ExperimentInitial [A] ([A] 0 ) Initial [B] ([B] 0 ) Initial Rate (v 0 ) 11.00 M 1.25 x 10 -2 M/s 21.00 M2.00 M2.5 x 10 -2 M/s 32.00 M 2.5 x 10 -2 M/s

7 Main Menu Another example (from last lesson): ExperimentInitial [NO] / mol dm –3 Initial [H 2 ] / mol dm –3 Initial rate / mol (N 2 ) dm –3 s –1 10.100 2.53×10 –6 20.1000.2005.05×10 –6 30.2000.1001.01×10 –5 40.3000.1002.28×10 –5  The reaction is 1 st order w.r.t [H 2 ], and second order w.r.t [NO]  Therefore, the rate equation is:  Rate = k[H 2 ][NO] 2  The value of k is given by (using values for Experiment 2): k = Rate / [H 2 ][NO] 2 k = 5.05x10 -6 / (0.200x0.100 2 ) = 2.53x10 -3 mol -2 dm 6 s -1

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9 Solutions

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11 Rate Equations in Practice  In this activity you will need to determine the rate equation for the following reaction  You will use real-life data collected from the lab, which will be messy and lumpy in way that the data in exams won’t be.  Follow the instructions hereinstructions here  You will need to download the spreadsheet ‘Iodination of Propanone’ from the blog

12 Main Menu Review Rate = k[A] x [B] y [C] z  To determine k, divide the rate by the concentration of reactants.  To determine the units of k, sub the rate and concentration units into the above, and cancel units to express in their lowest terms.


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