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Chapter 13.1-13.3

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Earlier, rate was shown to be a change in concentration compared to a change in time. This is also know as a. slope

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Rate, the speed of a reaction, can be expressed as a function of concentration The term function meaning, of course, an equation… Rate = k [Reactant 1] x [Reactant 2] y [Reactant 3] z Rate Constant Concentrations of Reactants Orders Brackets = Molarity

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k is the rate constant, a function of the particular reaction depends on strength of bonds which must be broken (Activation Energy) depends on presence of catalyst depends on temperature

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[Reactants] is the concentration of one of the reactants notice [Products] do not effect the rate some reactants may not effect rate, and so they do not show up in the rate law (these are said to have an order of zero)

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the order of a reactant is the exponent x, which is called the order with respect to that Reactant the higher the order wrt R1, the greater the influence of [R1] on the rate a reactant which does not effect rate is said to be 0 th order orders do not come from the balanced equation; they must be determined experimentally by studying the reaction mechanism orders can be fractional or even negative

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example Hydrogen peroxide oxidizes iodide ion in acidic solution: H 2 O 2 (aq) + 3 I – (aq) + 2 H + (aq) I 3 – (aq) + 2 H 2 O (l) The rate law for this reaction is Rate = k [H 2 O 2 ] [I – ] What is the order of the reaction with respect to each reactant species? What is the overall order? 1 st order WRT H 2 O 2 1 st order WRT I – 1 + 1 = 2 2 nd order overall

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Notice that the rate law covers all of the factors which effect rate. concentration temperature pressure/volume of gases catalyst surface area of reactants

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So if the rate law can not be determined by the coefficients from the balanced equation, how can the orders be determined? Method of Initial Rates Or Graphical Analysis of Data

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In the Method of Initial Rates, a reaction is run several times as the concentrations of the reactants are changed and the initial rate is measured each time to solve for the orders… 1.make sure you have at least two trials for which only [x] has changed 2.comparing the initial rates of these two trials will yield the order wrt x 3.repeat for each of the other reactants

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Practice 2 NO+2 H 2 N 2 +2 H 2 O Rate = k[NO] x [H 2 ] y to solve for the orders… 1.make sure you have at least two trials for which only [NO] has changed Ex p. [NO] o [H 2 ] o Rate (mol/Lsec) 10.010 3.0 x 10 –11 20.0100.0309.0 x 10 –11 30.0200.01012.0 x 10 –11 40.0100.05015.0 x 10 –11 50.0600.010108.0 x 10 –11

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Practice 2 NO+2 H 2 N 2 +2 H 2 O Rate = k[NO] x [H 2 ] y After picking the experiments, plug the information into the reactions generic rate law as a ratio: Ex p. [NO] o [H 2 ] o Rate (mol/Lsec) 10.010 3.0 x 10 –11 20.0100.0309.0 x 10 –11 30.0200.01012.0 x 10 –11 40.0100.05015.0 x 10 –11 50.0600.010108.0 x 10 –11

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The order WRT [NO] = 2 Practice 2 NO+2 H 2 N 2 +2 H 2 O Rate = k[NO] x [H 2 ] y Ex p. [NO] o [H 2 ] o Rate (mol/Lsec) 10.010 3.0 x 10 –11 20.0100.0309.0 x 10 –11 30.0200.01012.0 x 10 –11 40.0100.05015.0 x 10 –11 50.0600.010108.0 x 10 –11

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Practice 2 NO+2 H 2 N 2 +2 H 2 O Rate = k[NO] 2 [H 2 ] y to solve for the orders… 1.make sure you have at least two trials for which only [H 2 ] has changed Ex p. [NO] o [H 2 ] o Rate (mol/Lsec) 10.010 3.0 x 10 –11 20.0100.0309.0 x 10 –11 30.0200.01012.0 x 10 –11 40.0100.05015.0 x 10 –11 50.0600.010108.0 x 10 –11

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Practice 2 NO+2 H 2 N 2 +2 H 2 O Rate = k[NO] 2 [H 2 ] y After picking the experiments, plug the information into the reactions generic rate law as a ratio: Ex p. [NO] o [H 2 ] o Rate (mol/Lsec) 10.010 3.0 x 10 –11 20.0100.0309.0 x 10 –11 30.0200.01012.0 x 10 –11 40.0100.05015.0 x 10 –11 50.0600.010108.0 x 10 –11

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The order WRT [H 2 ] = 1 Practice 2 NO+2 H 2 N 2 +2 H 2 O Rate = k[NO] 2 [H 2 ] y Ex p. [NO] o [H 2 ] o Rate (mol/Lsec) 10.010 3.0 x 10 –11 20.0100.0309.0 x 10 –11 30.0200.01012.0 x 10 –11 40.0100.05015.0 x 10 –11 50.0600.010108.0 x 10 –11

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Practice 2 NO+2 H 2 N 2 +2 H 2 O Rate = k[NO] 2 [H 2 ] 1 The rate law so far looks like… Rate = k[NO] 2 [H 2 ] 1 The Overall Order is… 2 + 1 = 3 Ex p. [NO] o [H 2 ] o Rate (mol/Lsec) 10.010 3.0 x 10 –11 20.0100.0309.0 x 10 –11 30.0200.01012.0 x 10 –11 40.0100.05015.0 x 10 –11 50.0600.010108.0 x 10 –11

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Practice 2 NO+2 H 2 N 2 +2 H 2 O Rate = k[NO] 2 [H 2 ] 1 to solve for the value of k… 1.pick any trial, plug in the initial [ ]s and the orders, and solve for k Ex p. [NO] o [H 2 ] o Rate (mol/Lsec) 10.010 3.0 x 10 –11 20.0100.0309.0 x 10 –11 30.0200.01012.0 x 10 –11 40.0100.05015.0 x 10 –11 50.0600.010108.0 x 10 –11

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What about units??? Practice 2 NO+2 H 2 N 2 +2 H 2 O Rate = k[NO] 2 [H 2 ] 1 to solve for the value of k… 1.pick any trial, plug in the initial [ ]s and the orders, and solve for k 12.0x10 –11 = k[0.020] 2 [0.010] 1 k = 3.0x10 –5 Ex p. [NO] o [H 2 ] o Rate (mol/Lsec) 10.010 3.0 x 10 –11 20.0100.0309.0 x 10 –11 30.0200.01012.0 x 10 –11 40.0100.05015.0 x 10 –11 50.0600.010108.0 x 10 –11

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Practice 2 NO+2 H 2 N 2 +2 H 2 O Rate = k[NO] 2 [H 2 ] 1 to solve for the units of k… 12.0x10 –11 = k[0.020] 2 [0.010] 1 k = 3.0x10 –5 1 / M 2 s Ex p. [NO] o [H 2 ] o Rate (mol/Lsec) 10.010 3.0 x 10 –11 20.0100.0309.0 x 10 –11 30.0200.01012.0 x 10 –11 40.0100.05015.0 x 10 –11 50.0600.010108.0 x 10 –11

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Practice 2 NO+2 H 2 N 2 +2 H 2 O Rate = k[NO] 2 [H 2 ] 1 Final Rate Law… Rate=(3.0x10 –5 1 / M 2 s )[NO] 2 [H 2 ] 1 Ex p. [NO] o [H 2 ] o Rate (mol/Lsec) 10.010 3.0 x 10 –11 20.0100.0309.0 x 10 –11 30.0200.01012.0 x 10 –11 40.0100.05015.0 x 10 –11 50.0600.010108.0 x 10 –11

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Example 2 Fe 2+ + Cl 2 2 Fe 3+ + 2 Cl – Exp[Fe 2+ ][Cl 2 ][H + ]Rate (M/sec) 10.0020 1.01.0x10 –5 20.00400.00201.02.0x10 –5 30.00200.00401.02.0x10 –5 40.0040 1.04.0x10 –5 50.0020 0.52.0x10 –5 60.0020 0.11.0x10 –4 What is the reaction order with respect to Fe 2+, Cl 2, and H + ? What are the rate law and the relative rate constant?

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Example 2 Fe 2+ + Cl 2 2 Fe 3+ + 2 Cl – Exp[Fe 2+ ][Cl 2 ][H + ]Rate (M/sec) 10.0020 1.01.0x10 –5 20.00400.00201.02.0x10 –5 30.00200.00401.02.0x10 –5 40.0040 1.04.0x10 –5 50.0020 0.52.0x10 –5 60.0020 0.11.0x10 –4 What is the reaction order with respect to Fe 2+, Cl 2, and H + ? What are the rate law and the relative rate constant? Rate=k[Fe 2+ ] x [Cl 2 ] y [H + ] z

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Example 2 Fe 2+ + Cl 2 2 Fe 3+ + 2 Cl – Exp[Fe 2+ ][Cl 2 ][H + ]Rate (M/sec) 10.0020 1.01.0x10 –5 20.00400.00201.02.0x10 –5 30.00200.00401.02.0x10 –5 40.0040 1.04.0x10 –5 50.0020 0.52.0x10 –5 60.0020 0.11.0x10 –4 What is the reaction order with respect to Fe 2+, Cl 2, and H + ? What are the rate law and the relative rate constant? Rate=k[Fe 2+ ] x [Cl 2 ] y [H + ] z How do you find the order of Fe 2+ ?

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Example 2 Fe 2+ + Cl 2 2 Fe 3+ + 2 Cl – Exp[Fe 2+ ][Cl 2 ][H + ]Rate (M/sec) 10.0020 1.01.0x10 –5 20.00400.00201.02.0x10 –5 30.00200.00401.02.0x10 –5 40.0040 1.04.0x10 –5 50.0020 0.52.0x10 –5 60.0020 0.11.0x10 –4 What is the reaction order with respect to Fe 2+, Cl 2, and H + ? What are the rate law and the relative rate constant? Rate=k[Fe 2+ ] x [Cl 2 ] y [H + ] z How do you find the order of Fe 2+ ?

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Example 2 Fe 2+ + Cl 2 2 Fe 3+ + 2 Cl – Exp[Fe 2+ ][Cl 2 ][H + ]Rate (M/sec) 10.0020 1.01.0x10 –5 20.00400.00201.02.0x10 –5 30.00200.00401.02.0x10 –5 40.0040 1.04.0x10 –5 50.0020 0.52.0x10 –5 60.0020 0.11.0x10 –4 What is the reaction order with respect to Fe 2+, Cl 2, and H + ? What are the rate law and the relative rate constant? Rate=k[Fe 2+ ] 1 [Cl 2 ] y [H + ] z

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Example 2 Fe 2+ + Cl 2 2 Fe 3+ + 2 Cl – Exp[Fe 2+ ][Cl 2 ][H + ]Rate (M/sec) 10.0020 1.01.0x10 –5 20.00400.00201.02.0x10 –5 30.00200.00401.02.0x10 –5 40.0040 1.04.0x10 –5 50.0020 0.52.0x10 –5 60.0020 0.11.0x10 –4 What is the reaction order with respect to Fe 2+, Cl 2, and H + ? What are the rate law and the relative rate constant? Rate=k[Fe 2+ ] 1 [Cl 2 ] y [H + ] z How do you find the order of Cl 2 ?

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Example 2 Fe 2+ + Cl 2 2 Fe 3+ + 2 Cl – What is the reaction order with respect to Fe 2+, Cl 2, and H + ? What are the rate law and the relative rate constant? Rate=k[Fe 2+ ] 1 [Cl 2 ] y [H + ] z How do you find the order of Cl 2 ? Exp[Fe 2+ ][Cl 2 ][H + ]Rate (M/sec) 10.0020 1.01.0x10 –5 20.00400.00201.02.0x10 –5 30.00200.00401.02.0x10 –5 40.0040 1.04.0x10 –5 50.0020 0.52.0x10 –5 60.0020 0.11.0x10 –4

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Example 2 Fe 2+ + Cl 2 2 Fe 3+ + 2 Cl – Exp[Fe 2+ ][Cl 2 ][H + ]Rate (M/sec) 10.0020 1.01.0x10 –5 20.00400.00201.02.0x10 –5 30.00200.00401.02.0x10 –5 40.0040 1.04.0x10 –5 50.0020 0.52.0x10 –5 60.0020 0.11.0x10 –4 What is the reaction order with respect to Fe 2+, Cl 2, and H + ? What are the rate law and the relative rate constant? Rate=k[Fe 2+ ] 1 [Cl 2 ] 1 [H + ] z

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Example 2 Fe 2+ + Cl 2 2 Fe 3+ + 2 Cl – Exp[Fe 2+ ][Cl 2 ][H + ]Rate (M/sec) 10.0020 1.01.0x10 –5 20.00400.00201.02.0x10 –5 30.00200.00401.02.0x10 –5 40.0040 1.04.0x10 –5 50.0020 0.52.0x10 –5 60.0020 0.11.0x10 –4 What is the reaction order with respect to Fe 2+, Cl 2, and H + ? What are the rate law and the relative rate constant? Rate=k[Fe 2+ ] 1 [Cl 2 ] 1 [H + ] z How do you find the order of H + ?

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Example 2 Fe 2+ + Cl 2 2 Fe 3+ + 2 Cl – Exp[Fe 2+ ][Cl 2 ][H + ]Rate (M/sec) 10.0020 1.01.0x10 –5 20.00400.00201.02.0x10 –5 30.00200.00401.02.0x10 –5 40.0040 1.04.0x10 –5 50.0020 0.52.0x10 –5 60.0020 0.11.0x10 –4 What is the reaction order with respect to Fe 2+, Cl 2, and H + ? What are the rate law and the relative rate constant? Rate=k[Fe 2+ ] 1 [Cl 2 ] 1 [H + ] z How do you find the order of H + ?

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Example 2 Fe 2+ + Cl 2 2 Fe 3+ + 2 Cl – Exp[Fe 2+ ][Cl 2 ][H + ]Rate (M/sec) 10.0020 1.01.0x10 –5 20.00400.00201.02.0x10 –5 30.00200.00401.02.0x10 –5 40.0040 1.04.0x10 –5 50.0020 0.52.0x10 –5 60.0020 0.11.0x10 –4 What is the reaction order with respect to Fe 2+, Cl 2, and H + ? What are the rate law and the relative rate constant? Rate=k[Fe 2+ ] 1 [Cl 2 ] 1 [H + ] –1

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Example 2 Fe 2+ + Cl 2 2 Fe 3+ + 2 Cl – Exp[Fe 2+ ][Cl 2 ][H + ]Rate (M/sec) 10.0020 1.01.0x10 –5 20.00400.00201.02.0x10 –5 30.00200.00401.02.0x10 –5 40.0040 1.04.0x10 –5 50.0020 0.52.0x10 –5 60.0020 0.11.0x10 –4 What is the reaction order with respect to Fe 2+, Cl 2, and H + ? What are the rate law and the relative rate constant? Rate=k[Fe 2+ ] 1 [Cl 2 ] 1 [H + ] –1 How do you find the value for k? 1.0x10 –5 =k[.002] 1 [.002] 1 [1] –1 2.5=k

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Example 2 Fe 2+ + Cl 2 2 Fe 3+ + 2 Cl – Exp[Fe 2+ ][Cl 2 ][H + ]Rate (M/sec) 10.0020 1.01.0x10 –5 20.00400.00201.02.0x10 –5 30.00200.00401.02.0x10 –5 40.0040 1.04.0x10 –5 50.0020 0.52.0x10 –5 60.0020 0.11.0x10 –4 What is the reaction order with respect to Fe 2+, Cl 2, and H + ? What are the rate law and the relative rate constant? Rate=k[Fe 2+ ] 1 [Cl 2 ] 1 [H + ] –1 How do you find the unit for k? M/s=k[M] 1 [M] 1 [M] –1 M 1 M –1 M –1 M 1 s –1 =k s –1 =k

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The second way, also experimental, to determine the rate law is based on the integration of the basic rate law. Let us begin by introducing the integrated rate laws… Let us assume that… the Reactant of interest is A the reaction which contains A has been run, allowing the concentration of A to be measured as time changes

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Lets begin with the simplest… Zeroth-Order Rate Law: Rate = k [A] 0 Rate = k The instantaneous rate law: By combining these two rates and rearranging them… This results in the integrated rate law: [A] t – [A] o = – kt

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Lets rearrange this integrated rate law into something more useful… such as the form… y = m x + b [A] t = – kt + [A] o [A] t [A] o Slope = –k

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First-Order Rate Law: Rate = k [A] 1 The instantaneous rate law: By rearranging these… This results in the integrated rate law: ln [A] t – ln [A] o = – kt

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Lets rearrange this integrated rate law into something more useful… such as the form… y = m x + b ln [A] t = – kt + ln [A] o ln[A] t ln[A] o Slope = –k

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Second-Order Rate Law: Rate = k [A] 2 The instantaneous rate law: By rearranging these… This results in the integrated rate law:

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Lets rearrange this integrated rate law into something more useful… such as the form… y = m x + b t Slope = +k

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Chapter 14: Rates of Reaction Chemistry 1062: Principles of Chemistry II Andy Aspaas, Instructor.

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