Presentation on theme: "16.1 Rate expression 16.1.1 Distinguish between the terms rate constant, overall order of reaction and order of reaction with respect to a particular reactant."— Presentation transcript:
1 16.1 Rate expressionDistinguish between the terms rate constant, overall order of reaction and order of reaction with respect to a particular reactant.Deduce the rate expression for a reaction from experimental data.Solve problems involving rate expressionSketch, identify and analyze graphical representation for zero-, first- and second-order reactions.
2 Expression that shows how rate depends on concentration. A + B => C Rate Law ExpressionExpression that shows how rate depends on concentration.A + B => Crate = k [A]m [B]nm and n are called reaction orders. Their sum is called the overall reaction order.k is the rate constant. It is specific to a reaction at a certain temperature.
3 Rate = k [A]m [B]nThe exponents in a rate law must be determined by experiment. They are not derived from the stoichiometric coefficients in a chemical equationThe values of exponents establish the order of a reaction for each species and overall order for the reactionThe proportionality constant, k, is the rate constant and its value depends on the reaction, the temperature, and the presence or absence of a catalyst.
4 Exercise 1: 2NO + 2H2 N2 + 2H2O Experiment Number Conc. of NO (M) Conc. of H2Rate of N2 forming (Ms-1)10.2100.1220.033920.2440.067830.4200.1356
5 1. Write the rate law expression; 2. To find the order of [NO], keep the other one constant.
6 Study the data Rate = k [NO]2[H2] Exp #1 and #2, [NO] is unchanged, [H2] is doubled and this causes the rate to double (0.0678/ =2), the H2’s rate order is 1.Exp #1 and #3, [H2] is unchanged, [NO] is doubled and this causes the rate to quadruple (0.1356/ =4), NO’s rate order is 2.Rate = k [NO]2[H2]The overall rate order for this reaction is 3 (1+2)
7 Suppose…If when we ran exp #1 and #3 , the rate didn’t change, what rate law would we expect?Since it didn’t change when we doubled [NO], the rate order is 0, meaning the rate doesn’t depend on the concentration of NO at all, so rate= k[H2] and the overall reaction order is 1.
8 Sample Exercise 2:Use the kinetics data to write the rate law for the reaction. What overall reaction order is this? NO + O2 2NO2Exp #[NO][O2]Rate forming NO2 (M/s)10.0150.04820.0300.19230.09640.384
9 exp #1 and exp #2, [O2] remained constant, where [NO] is doubled exp #1 and exp #2, [O2] remained constant, where [NO] is doubled. The rate is quadrupled. The rate order for [NO] is 2.Exp #1 and exp #3, [NO] remained constant, where [O2] is doubled. The rate is doubled. The rate order for [O2] is 1.Overall reaction order (1+2) = 3
10 Rate data for the reaction: Sample Exercise 3Rate data for the reaction:CH3Br + OH- CH3OH + Br-Exp #CH3BrOH-Rate of forming CH3OH10.2000.01520.4000.03030.060Use the data to find the experimental rate law.
11 Distinctions Between Rate And The Rate Constant k The rate of a reaction is the change in concentration with time, whereas the rate constant is the proportionality constant relating reaction rate to the concentrations of reactants.The rate constant remains constant throughout a reaction, regardless of the initial concentrations of the reactants.The rate and the rate constant have the same numerical values and units only in zero-order reactions.
12 Method Of Initial Rates SummaryThe effects of doubling one initial concentration:For zero-order reactions, no effect on rate.For first-order reactions, the rate doubles.For second-order reactions, the rate quadruples.For third-order reactions, the rate increases eightfold.The value of k for the reaction can be calculated.
18 First orderConc. shows an exponential decrease, that is the time for the conc. to fall from its initial value to half its initial value, is equal to the time required for it to fall from half to one quarter to one eighth, etc…This is known as half-life, t ½Look at next slide for graphical representation
21 Radioactive decay First order exponential decay Half life is important and can be found from a graph or the equationt ½ = ln2kExample: if the rate constant of a first order reaction is s-1, then the half life will bet ½ = ln = 139 s0.005