Presentation is loading. Please wait.

Presentation is loading. Please wait.

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 

Similar presentations


Presentation on theme: "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 "— Presentation transcript:

1 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  products The rate depends on the [reactants] Rate  [X] m [Y] n This relationship can be expressed in a general equation called the Rate Law Equation

2 The Rate Law Equation Rate = k[X] m [Y] n Where, – k is the rate constant Depends on the temperature Is constant under constant conditions – m and n are the rate law exponents (usually 1 or 2) Do not change with temperature Must be found by experiment

3 Reaction Order Determined by the value of the rate law exponents If the exponent is 1 for a reactant, that reactant is 1 st order If the exponent is 2 for a reactant, that reactant is 2 nd order The overall reaction order, is the sum of the reaction orders (m + n) In this case the overall reaction order is 3 rd order (1 + 2)

4 The Rate Constant The magnitude of k indicates the speed of a rxn. The smaller k is, the slower the rxn. k is NOT affected by the [reactants] k  as temperature  (k is unique for each temperature)

5 Defining First Order Reactions Has an overall order of 1 Eg. For the reaction: 2N 2 O 5(g)  4NO 2(g) + O 2(g) Experiment has shown that this is a first-order reaction Therefore, Rate = k[N 2 O 5 ] 1 The overall reaction order = 1, so this is a first order reaction

6 Defining First Order Reactions Reactions with more than one reactant can also be first order, if the concentration of one reactant does not affect the rate ie. The overall reaction order is (0 + 1) With first order reactions, – As [reactant] doubles  rate doubles

7 Defining Second Order Reactions Overall reaction order = 2 Can have one reactant that is second order Can have 2 reactants that are each 1 st order Can have one reactant that is second order and one that does not affect the rate. In second order reactions, rate increases by the square of the concentration or the reactant

8 Using a Rate Equation to Predict Rate Example: Given the following reaction, 2 C 4 H 6  C 8 H 12 If the reaction is second order with respect to C 4 H 6. If the initial rate of reaction was 0.25 mol C 4 H 6 /(L·s), what would be the initial rate of reaction if the initial [C 4 H 6 ] was doubled?

9 Solution The rate equation is: r = k[C 4 H 6 ] 2 If the initial [] is doubled (multiplied by 2), then the rate will be multiplied by 2 2 = 4 Therefore, the new rate of reaction is: 4 x 0.25 mol/(L  s) = 1.0 mol/(L  s)

10 Determining Exponents for the Rate Law Equation Must be done experimentally Initial Rates Method – Reaction is performed several times – Each time with different starting concentrations for the reactants – All other factors are kept constant – The initial rate of reaction is recorded and compared

11 Ex. Initial Rates Method ExperimentInitial [N 2 O 5 ] (mol/L) Initial Rate (mol/(L  s)) x x x Value for m can be found by inspection Looking at the exp. data above As the initial [N 2 O 5 ] doubles the rate doubles As the initial [N 2 O 5 ] triples the rate triples Therefore the reaction is first order

12 Ex. Initial Rates Method Value for m can be found by comparing the rate law equation for each trial Rate 1 = k(0.010 mol/L) m = 4.8 x mol/(L  s) Rate 2 = k(0.020 mol/L) m = 9.6 x mol/(L  s) Compare the 2 rates using a ratio K is a constant so we can cross it out k(0.010 mol/L) m = 4.8 x mol/(L  s) k(0.020 mol/L) m = 9.6 x mol/(L  s) (0.5) m = 0.5 (m = 1 to make this true)

13 Determining the Rate Constant If you know the order Use Rate = k[reactant] m Isolate for k to determine the rate constant Use the rate of reaction and [reactant] from one of the trials See Sample Problem on Pg. 375 Complete Practice Problems Pg. 377 #2, 3, 4, 6

14 The Half Life of a Rxn Half life = the time needed for the [reactant] or mass of reactant to decrease by ½ Half life for a first-order reaction is constant Since the breakdown of radioactive isotopes and many chemical rxns are 1 st order we can use the following rxn. The half life for a first-order reaction can be calculated using: t ½ = / k Where, k is the rate constant (s -1 ) Sample Problems: Pg. 380 Practice Problems: Pg. 381 # 7a, 8


Download ppt "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 "

Similar presentations


Ads by Google