Chapter : CHEMICAL KINETICS Study of the rate of a chemical reaction. Reaction Rate (fast or slow?) Igniting gasoline? Making of oil? Iron rusting? We know about speed (miles/hr). Speed Rate = changes that occur within an interval of time. Instead of distance, amount of substance is measured. Ex. 5 mole Fe/year What controls the rate of a chemical reaction? Properties of the atoms, ions, and moles.
Reaction Rates The rate of a reaction is a quantity that expresses how the concentration ([ ]) of a reactant or product changes ( ) with time. For the general reaction, aa + bb cc + dd a moles of A decompose b moles of B decompose c moles of C form d moles of D form Rate of the reaction is: Rate = c C t d D t a A t b B t * The minus takes into account the fact that the [ ] is decreasing as time passes.
For the reaction, H (g) + ICl(g) I (g) + HCl(aq) Rate = c C t d D t a A t b B t Below is data for the reaction of.0m ICl and.0m H at 30 C. How is the rate of a chemical reaction measured? Follow the progress of the reaction. For example, the reaction: Mg(s) + HCl(aq) H (g) + MgCl (aq) 3
A graph of amount of H (g) vs time can be made. The amount of H (g) formed is measured as time passes. If we again consider the reaction: H (g) + ICl(g) I (g) + HCl(aq) 4
As the reaction proceeds the amount of: [H ] decreases [I ] increases [ICl] decreases [H ] increases The rate of decomposition (or formation) of a substance can be determine at any given period of time during the reaction. The slope of a tangent drawn to the curve of [ ] vs Time is equal to the instantaneous rate of that substance. Slope of tangent = instantaneous rate 5
Reaction Rate and Concentration Simply stated: Chemical reactions (generally) occur when reactant particles collide. The higher [ ] of reactant the greater the number of collisions occur per unit time. The greater number of collision = the rate of the reactions increases. The opposite is also true. When the number of reactant molecules decreases, the number of collisions decrease = the reaction rate decreases. 6
Rate Expression and the Rate Constant For the reaction: N O 5 (g) NO (g) + ½ O (g) The dependence of rate on the [N O 5 ] is plotted below. It simply shows that as the [N O 5 ] increases the rate increases. A STRAIGHT LINE! The equation of this line is simply: Rate = k[n O 5 ] (k = rate constant) 7
This equation is called the Rate Law (or the Rate Expression) * Important: The rate is dependent only on the nature of the reactants involved in the chemical reaction. * Order of Reaction: Single Reactant Rate Laws have been determined for many reactions. For the reaction A Products The general rate expression is: rate = k[a] m (m = order of the reaction) m = 0 m = m = Zero Order First Order Second Order, etc 8
The order of the reaction must be determined EXPERIMENTLY. HOW??? Answer: Measure the initial rate as a function of [ ] of reactant. For condition #: rate = k[a] m For condition #: rate = k[a] m Divide rate by rate (k cancels!) So, Rate Rate Rate Rate m m m m 9
Consider the following data: [A] (M) 0.0 0.0 0.30 0.40 Rate (M/s) 0.085 0.34 0.76.4 ) Determine the order of the reaction. ) Write the Rate Law. Answer: Using Rate Rate.4.76.40.30 m m.8 = [.3] m m = Second Order Rate = k[a] What is the value of k if the rate is has the units of M/s & [A] has the units of M??? 0
Note: Once the order of the reaction is known, the rate constant, k, can be determined. Order of Reaction: More than one Reactant Many reactions involve more than one reactant. For the reaction involving species A and B, A + B Products The general rate rate is Rate = k[a] m x [B] n The overall order = m + n
The order can be determined by holding the [ ] of one reactant constant while varying the [ ] of the other reactant. (The process is similar to that of a single reactant.) Consider the following example: 3 A + B C + D Rate data: Exp. Exp. Exp. 3 [A].00 x 0 -.00 x 0 -.00 x 0 - [B].00 x 0-3.00 x 0 -.00 x 0 - Rate 6.00 x 0-3.44 x 0 -.0 x 0 - The general rate expression is: Rate = k[a] m x [B] n
Part : In Expt. and 3, [A] is constant and since k cancels,.0 6.00 Rate Rate x x 3 Rate Rate 0 0 3 3 becomes 3 m [ B] [ B].00 3.00 [ B] [ B] x x n 3 0 0 n n.00 =.00 n n =, The reaction is st order in [B]. Part : The same process is then done for the other reactant. 3
However, in this situation, no two values of [B] are constant. Simply plug in the order for [B] into the formula below for rates and. (Again k cancels out.) Rate Rate m [ B] [ B] n.0 x 0.00 x 0 m 3.00 x 0 6.00 x 0 3.00 x 0.00 x 0 4.0 =.0 m [3.0] 8.00 =.0 m m = 3, the reaction is 3rd order in [A] The overall order is 4. The overall rate expression is Rate = k[a] 3 [B] 4
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[REACTANT] AND TIME Knowing the relationship rate to rate constant, k, and [reactant] is very useful. However, the relationship between concentration and time is often more useful. Integrated Rate Equations provide us this relationship. Using calculus, this relationship can be determined. ) First Order Reactions Consider the first order reaction, A B Rate t Rate = k[a] d[ A] dt 6
SO, d[ A] dt Rearrangement results in: d[ A] k[ A] kdt d[ A] kdt [] [] t d[ A] k o 0 t dt -(ln[a] t ln[a] o )= k(t - 0) or ln[a] t ln[a] o = -kt 7
where ln t o kt [A] o = original [ ] of the reactant [A] t = [ ] of the reactant at time t. k = the first order constant. If this equation is rearranged in the following way: ln[a] t ln[a] o = -kt to ln[a] t = -kt + ln[a] o It is in the form of the equation of a line: y = m x + b Where, y = ln[a] t m = -k x = t b = ln[a] 0 8
For the first order decomposition of [N O 5 ] a plot of ln[a] t vs. time produces a straight line. The equation in the form of ln t o kt can be rearranged to solve for time, t. This can of course be done with the other form of the equation: ln[a] t ln[a] o = -kt The half-life, t /, of a reactant can be also determined. 9
At t /, ½ of the [ ] of a reactant has been converted in product, therefore, [A] t = ½ [A] o. As a result, t n k (Independent of [ ] For st Order Rxn only!) 0
Second Order Reactions A Products Rate = k[a] If [A] t vs. time is plotted, a straight line results with the form t o kt The half-life of a second order reaction can be determined. For [A] t with [A] = ½ [A] o [A] o [A] o [A] o = = - = [A] o [A] o [A] o [A] o [A] o = kt + kt + kt + kt = kt
t / = Zero Order Reactions k o A Products Rate = k[a] 0 (Independent of [ ] - * RARE *) If [A] t vs. time is plotted, a straight line results with the form y = b + m x [A] = [A] o + (-k)t The half-life of a zero order reaction can be determined. For [A] t = [A] o kt, with [A] t = ½[A] o So, ½ [A] o = [A] o kt kt = [A] o - / [A] o
t k o 3
If asked to determine the order of a reaction bases on data, one should plot: [A] vs. time ln[a] vs. time Zero Order First Order [ ] A vs. time Second Order The graph which produces a straight line indicates the order of the reaction. 4
Collision Theory Helps to understand the rate of chemical reactions. Collision Theory states that:. Atoms, ions, & molecules can form a chemical bond when they collide.. However, the particles must have enough KE. 5
Question: What is it that determines if a chemical reaction will or will not occur? Answer: The amount of KE of the atoms, ions, or molecules. Question: At what point is it determined if a chemical reaction will occur? Answer: The transition state. The particles in the transition state must have enough KE. This energy is call Activation Energy. Activation Energy (E a ) = minimum energy colliding particles must have in order to react with each other. 6
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The Arrhenius Equation In 889, the Swedish chemist, Svante Arrhenius showed that based on Kinetic Theory, the fraction of molecules having an energy = or > E a is f e E a RT e = base of natural logarithms E a = activation energy R = gas constant = 8.34 J/K mole T = Temperature (Kelvin) If f is directly proportional to the rate constant, k (*assumption*) then, k Af Ae E a RT A = Constant 8
Reaction Rate and Temperature If one wants to cook food faster, the temperature of the food is increased. The effect of Temp. & Reaction Rate can be explained with Kinetic Theory of Gases. If Temperature, the fraction of molecules having high KE. Result: # molecules with enough E a needed to react 9
After some mathematical yoga, the Arrhenius Equation is lnk ln A E a RT If a plot of ln k vs. /T is constructed, a linear graph results. Following the form y = mx + b y = n k b = n A m = -E a /R Similar to the Clausius Clapeyron equation, rate constants at two different temperatures can be evaluated. 30
For condition #: lnk ln A For condition #: lnk ln A E a RT E a RT Subtracting # from # gives: ln k a ln k ln A ln RT E A Ea RT Resulting in the Arrhenius Equation: ln k ln k or E R a T T ln k k E a R T T 3
Reaction Mechanisms A reaction mechanism is simply a description of the sequence of steps that a reaction goes through at the molecular level. When CO and NO react at 600 C, the reactions occurs in -step: CO(g) + NO (g) CO (g) + NO(g) With the Rate = k[co] [NO ] At low temperatures, the mechanism changes into a -step process: NO (g) + NO (g) CO(g) + NO 3 (g) CO(g) + NO (g) NO 3 (g) + NO(g) CO (g) + NO (g) CO (g) + NO(g) With the Rate = k[no ] 3
IMPORTANT: The reaction rate, the rate law, and the reaction order all depend upon the mechanism of the chemical process. Elementary Steps The individual steps in a reaction mechanism are called elementary steps. They can be: ) Unimolecular A B + C Rate = k[a] ) Bimolecular A + A B + C Rate = k [A] [A] or Rate = k [A] 3) Termolecular (very rare) A + B + C D + E Rate = k [A] [B] [C] 33
Slow Steps For a multi-step reaction mechanism, one step is usually slower than the others. The slow step determines the rate of the reaction. Therefore, the slow step of a mechanism is called the rate determining step. The slow step determines the overall rate of the reaction. For example, Step : A B fast Step : B C slow Step 3: C D fast A D The rate for converting A to D is approximately the rate of the slow step, B C. 34
Determining Rate Law from Reaction Mechanisms Use a -step process, ) Find the slowest step in the mechanism. ) Determine the rate law. NO (g) + NO (g) CO(g) + NO 3 (g) CO(g) + NO (g) NO 3 (g) + NO(g) slow CO (g) + NO (g) fast CO (g) + NO(g) Applying the process, the rate is: Rate = k[no ] 35
Eliminating Intermediates Sometimes, the rate expression obtained by the process just described involves reactive intermediates. Reactive Intermediates A species produced in one step of the mechanism but consumed in a later step. Since reactive intermediates are short lived, they often cannot be measured. They are eliminated from the rate expression. Therefore, the final rate law must include only those species that appear in the balanced chemical equation for the overall reaction. 36
Consider the following example: k NO + Br k - NOBr (fast, equilibrium) NOBr + NO k NOBr (slow) NO + Br NOBr Overall Important facts The nd rxn is the slow and rate determining step. Rate = k [NOBr ] [NO] The st rxn is at equilibrium (i.e., rate of forward = rate of reverse). k [NO] [Br ] = k - [NOBr ] The [NOBr ] cannot be in the rate law - [NOBr ] reactive intermediate. From above, [NOBr ] = k k [NO] [Br ] Substitute [NOBr ] in the rate expression. 37
The rate expression becomes: Rate = k k k [NO] [Br ] [NO] Rate = k k k [NO] [Br ] Experimentally, a rate constant is determined to be "k", however, the following is really true: k k k k So, the rate law is Rate = k[no] [Br ] which matches experimental results. 38
Factors Affecting Reaction Rates ) Surface Area & Size Greater surface available more area for collisions. Chemical reaction to occurs faster. Chances for collisions between particles increases. ) Concentration Higher concentration, more reactants present. More collisions. 3) Temperature Higher temperature, higher KE, more collisions. 39
4. Catalyst A substance that increases the rate of the reaction without being consumed in the chemical reaction. Example: No reaction occurs at room temperature. H (g) + O (g) H O(l) Reaction quickly occurs at room temperature with platinum metal. H (g) + O (g) Pt H O(l) Pt = Platinum metal Water forms quickly. Catalyst 40
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Example #: In biology ENZYMES (biological catalyst) The enzyme that breaks down protein in food, occurs at your body s temperature (37 C) within a day. Without the enzyme, it would take ~ 50 years. Example #: Chemical Inhibitors Substance that interferes with the catalysts. AIDS drugs (AZT) 4