Presentation on theme: "Chapter 13 Chemical Kinetics CHEMISTRY. Kinetics is the study of how fast chemical reactions occur. There are 4 important factors which affect rates of."— Presentation transcript:
Chapter 13 Chemical Kinetics CHEMISTRY
Kinetics is the study of how fast chemical reactions occur. There are 4 important factors which affect rates of reactions: –reactant concentration, –temperature, –action of catalysts, and –surface area. Goal: to understand chemical reactions at the molecular level. Factors that Affect Reaction Rates
Change of Rate with Time Consider: C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) Reaction Rates
Change of Rate with Time C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) –We can calculate the average rate in terms of the disappearance of C 4 H 9 Cl. –The units for average rate are mol/L·s or M/s. –The average rate decreases with time. –We plot [C 4 H 9 Cl] versus time. –The rate at any instant in time (instantaneous rate) is the slope of the tangent to the curve. –Instantaneous rate is different from average rate. –We usually call the instantaneous rate the rate. Reaction Rates
For the reaction A B there are two ways of measuring rate: –the speed at which the products appear (i.e. change in moles of B per unit time), or –the speed at which the reactants disappear (i.e. the change in moles of A per unit time). Reaction Rates
Reaction Rate and Stoichiometry In general for: aA + bB cC + dD Reaction Rates
Rate is (-) if reagent is consumed. Rate is (+) if compound is produced. Rate will ultimately be (+) because change in concentration will be negative. Two (-)’s become (+).
Take Note: Rate must ALWAYS be a positive value!
Rate Comparison 2NO 2 2NO + O 2 *If main interest is on the consumption of starting reagent, then: Rate = - [NO 2 ] = t t t Since 2 NO 2 molecules are consumed for every O 2
Take Note! Since the direction of equilibrium changes as more product is produced, rates have to be determined as soon as the experiment has begun.
2N 2 O 5 4NO 2 + O 2 [N 2 O 5 ] (mol/L)Time (sec)
Sample Problem A. How is the rate at which ozone (O 3 ) disappears related to the rate at which O 2 appears in the reaction: 2 O 3 (g) 3 O 2 (g)? B. If the rate at which O 2 appears, [O 2 ]/ t, is 6.0 x M/s at a particular instant, at what rate is O 3 disappearing at this same time, - [O 3 ]/ t?
Answers A. “Related to” means compare, so write the rate expression comparing the compounds. B. 4.0 x M/s
Sample Problem The decomposition of N 2 O 5 proceeds according to the following equation: 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) If the rate of the decomposition of N 2 O 5 at a particular instant in a reaction vessel is 4.2 x M/s, what is the rate of appearance of: (a) NO 2, (b) O 2 ?
Differential Rate Law - is a rate law that expresses how rate is dependent on concentration Example: Rate = k[A] n
Differential First Order Rate Law First Order Reaction – Rate dependent on concentration – If concentration of starting reagent was doubled, rate of production of compounds would also double
Using Initial Rates to Determines Rate Laws A reaction is zero order in a reactant if the change in concentration of that reactant produces no effect. A reaction is first order if doubling the concentration causes the rate to double. A reacting is nth order if doubling the concentration causes an 2 n increase in rate. Note that the rate constant does not depend on concentration. Concentration and Rate
Differential Rate Law For single reactants: A C Rate = k[A] n For 2 or more reactants: A + B C Rate = k[A] n [B] m Rate = k[A] n [B] m [C] p
Exponents in the Rate Law For a general reaction with rate law we say the reaction is mth order in reactant 1 and nth order in reactant 2. The overall order of reaction is m + n + …. A reaction can be zeroth order if m, n, … are zero. Note the values of the exponents (orders) have to be determined experimentally. They are not simply related to stoichiometry. Concentration and Rate
In general rates increase as concentrations increase. NH 4 + (aq) + NO 2 - (aq) N 2 (g) + 2H 2 O(l) Concentration and Rate
Rate law: The constant k is the rate constant. Concentration and Rate
Problem NH 4 + +NO 2 - N 2 +2H 2 O Give the general rate law equation for rxn. Derive rate order. Derive general rate order. Solve for the rate constant k.
To Determine the Orders of the Reaction (n, m, p, etc….) 1. Write Rate law equation. 2. Get ratio of 2 rate laws from successive experiments. Ratio = rate Expt.2 = k 2 [NH 4 + ] n [NO 2 - ] m rate Expt.1 k 1 [NH 4 + ] n [NO 2 - ] m 3. Derive reaction order. 4. Derive overall reaction order.
Experimental Data Expt.[NH 4 ] initial [NO 2 - ] initial Initial Rate M M1.35 x M0.010 M2.70 x M0.010 M5.40 x 10 -7
Experiment Number [A] (mol·L -1 )[B] (mol·L -1 )Initial Rate (mol·L -1 ·s -1 ) x x x A + B C (a) Determine the differential rate law (b) Calculate the rate constant (c) Calculate the rate when [A]=0.050 mol·L -1 and [B]=0.100 mol·L -1
Use the data in table 12.5 to determine 1) The orders for all three reactants 2) The overall reaction order 3) The value of the rate constant
Experiment Number [NO] (mol·L -1 )[H 2 ] (mol·L -1 )Initial Rate (mol·L -1 ·s -1 ) x x x (a) Determine the differential rate law (b) Calculate the rate constant (c) Calculate the rate when [NO]=0.050 mol·L -1 and [H 2 ]=0.150 mol·L -1 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O(g)
Consider the general reaction aA + bB cC and the following average rate data over some time period Δt: Determine a set of possible coefficients to balance this general reaction. Sample Problem:.
Problem Reaction: A + B C obeys the rate law: Rate = k[A] 2 [B]. A. If [A] is doubled (keeping B constant), how will rate change? B. Will rate constant k change? Explain. C. What are the reaction orders for A & B? D. What are the units of the rate constant?
You now know that…. The rate expression correlates consumption of reactant to production of product. For a reaction: 3A 2B - 1 [A] = 1 [B] 3 t2 t The differential rate law allows you to correlate rate with concentration based on the format: Rate = k [A] n
You also know that… 1. Rate of consumption of reactant decreases over time because the concentration of reactant decreases. Lower concentration equates to lower rate. 2. If a graph of concentration vs. time were constructed, the graph is not a straight line
Experiment 23 You varied concentration of KIO 3 and held the concentration of NaHSO 3 constant. From Expt. A E, you increased the amount of KIO 3. Observed result: The time it took for the reaction to occur DECREASED.
Conclusion Higher concentration of KIO 3 lead to faster rate of reaction.
How can we make the line straight? What is the relationship between concentration and time?
By graphing concentration vs. 1/time?
The Integrated Rate Law makes this possible!
Integrated Rate Law Expresses the dependence of concentration on time
First Order Reactions Goal: convert rate law into a convenient equation to give concentrations as a function of time. For a first order reaction, the rate doubles as the concentration of a reactant doubles. The Change of Concentration with Time
Integrated Rate Laws Zero Order: [A] t = -kt + [A] o First Order:ln[A] t = -kt + ln[A] o Second Order: 1 = kt + 1 [A] t [A] o where [A] o is the initial concentration and [A] t is the final concentration.
Integrated First-Order Rate Law ln[A] t = -kt + ln[A] 0 Eqn. shows [concn] as a function of time Gives straight-line plot since equation is of the form y = mx + b
Zero Order Reactions A plot of [A] t versus t is a straight line with slope -k and intercept [A] 0. The Change of Concentration with Time
First Order Reactions A plot of ln[A] t versus t is a straight line with slope -k and intercept ln[A] 0. The Change of Concentration with Time
Second Order Reactions A plot of 1/[A] t versus t is a straight line with slope k and intercept 1/[A] 0. The Change of Concentration with Time