Chemical Kinetics (PDF)

Topics under Physical Chemistry for this semester  Chemical Kinetics  Simple rate equations, orders of reactions and rate constants  Effect of temperature on rate constants & activation energy  Introduction to catalysis  Chemical equilibrium  Forward and reverse rea...

Topics under Physical Chemistry for this semester  Chemical Kinetics  Simple rate equations, orders of reactions and rate constants  Effect of temperature on rate constants & activation energy  Introduction to catalysis  Chemical equilibrium  Forward and reverse reactions  Expressions for equilibrium constants  Le Chatelier’s Principle  pH and pOH  Acids and bases  Buffers  Electrochemistry  Redox processes Dr. Nelson  Galvanic cells  Electrolysis Chemical Kinetics Lecture 1 Mark A.W. Lawrence  Introduction  When a reaction occurs a chemist will have 3 fundamental questions:  What happened? (reactants and products sometimes a mechanism)  To what extent did it happen? (Thermodynamics)  How fast did it happen? (Kinetics)  All 3 processes are related though we may study them “independently”.  Why study chemical kinetics?  Environmental (atmospheric, commercial effluent etc) and economic importance (catalyst ($$$$$$), structural integrity, manufacturing processes etc)  Chemical Kinetics  The rates of chemical reactions are tremendously different.  Adding HCl to NaOH: As you mix them the reaction is finished  Some organic synthesis: Can take hours, even with refluxing   How do we describe the rate of a reaction?  We can specify how fast the concentration of a reactant or product changes per unit time. concentration  i.e. rate  time  A+BC  One can monitor the decay of A or B or the formation of C.  If we should plot a concentration-time profile it would have a general trend shown below: A + B  C ∆[𝑐𝑜𝑛𝑐] 𝑐2 − 𝑐1 𝑀 C ∆𝑡 = 𝑡2 − 𝑡1 𝑠 𝑎𝑣𝑒 ∆[𝐴] 0.39 − 0.61 𝑀 = A or ∆𝑡 𝑎𝑣𝑒 12.5 − 5.0 𝑠 = −𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 B ∆[𝐶] 0.58 − 0.35 𝑀 = ∆𝑡 𝑎𝑣𝑒 12.5 − 5.0 𝑠 = +𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟  There is average rate and instantaneous rate (slope of tangent to that point).  Notice how the [A] or [B] decreases with time thus there is a “negative” change in the concentration of the reactants and d[A]/dt = -ve value.  Likewise the [C] (the product of the reaction) increases with time and as such d[C]/dt = +ve value.  If we are monitoring the same reaction, the rate should be a measurable value and be the same irrespective of whether we monitor a product or a reactant. d [ A] d [ B] d [C ] rate     dt dt dt  Chemical Kinetics  Notice the form of the expression for the rate and how it affects the units for the rate of a reaction: conc mol dm3   mol dm3 s 1 time s  How does the stoichiometry influences the expression for the rate of the reaction?  E.g. A + 2B  3C + ½ D d [ A] 1 d [ B] 1 d [C ] d [ D] rate     2 dt 2 dt 3 dt dt  i.e. the rate is divided by the stoichiometric coefficient (normalized with respect to each species.  Next we look at the rate constant for a reaction:  Chemical Kinetics—rate constants  Generally the rate of a chemical reaction depends on the concentration of at least one of the reactants (rate a [reactants]x)  If we have: a A + b B  products  The rate law is usually written in the form: rate a [ A]m [ B]n d[ A] d[ B] d[ product]     rate  k[ A]m [ B]n dt dt dt  Or simply: rate = k[A]m[B]n  Where k is the rate constant.  NOTE: The only way to write the rate law (or equation) is to measure the rate! The orders m & n are in NO WAY related to the stoichiometric coefficients (a and b) of the equation! Chemical Kinetics—order of reactions  The indices (or orders) m + n determine the overall order of the reaction.  m &/or n can be +ve, -ve, 0, fractions or integers (whole #).  So if in the rate expression: d [ A]   k [ A]m [ B]n dt  If m = 1 and n = 0 and: d [ A] d [ A]   k[ A] or just simply  1  k[ A] dt dt  Thus the order for the reaction is just 1—a 1st order reaction.  Now let us look at the units for this first order reaction. Chemical Kinetics—order of reactions  If we say for a 1st order reaction: rate = k[A]  mol dm-3 s-1 = k × mol dm-3 rate concentration  Thus: mol dm3 s 1 1 k  s mol dm3  Likewise if m & n = 0 (a zero order reaction) then: rate = k  Thus k => mol dm-3 s-1  Now using this simple procedure, what would be the units for a second order reaction?  E.g. rate = k [A][B] or rate = k [A]2.  k = dm3 mol-1 s-1  Important Points  DO NOT CONFUSE rate and rate constants!  Rate can be written in terms of reactants or products and the stoichiometric coefficient affects the differential expression for the rate.  The rate constant, however, is not affected by stoichiometric coefficients or concentrations.  DO NOT CONFUSE the differential expressions of rate and the rate law/equation/expression!  The rate can be written as a differential with respect to reactants or products.  The rate equation is only written with respect to the reactants!  The units of the rate constants are determined by the overall order of the rate expression.  The unit for rate is always mol dm-3 s-1 due to its differential nature.  Important Points  Rate (except for a zero order reaction) is dependent on the concentration of the reactants.  Rate constants are INDEPENDENT of the concentration of the reactants.  Rate constants are only affected by temperature. Graphical Treatment of Kinetic Data Zero Order reaction: d [ A]  rate   k dt  If we rearrange the above expression to yield a linear expression of the form y = mx + a: -d[A] = kdt or d[A] = -kdt  [A]t = [A]0 – kt (when we solve the diff. eqn)  Thus a plot of [A] vs time should yield a straight line with slope = -k. d[A] dt  If we plot d[A]/dt (rate) vs [A] we should get zero slope since the rate expression states that rate is independent of [A] (i.e. zero order). [A] First-order reactions First Order: d [ A]    k[ A] dt  Again we can treat the data as we did with the 0 order:  The expression as is, is in a linear form and as such a plot of d[A]/dt agaisnt [A] should yield a straight line with slope = -k (difficult plot).  Likewise if we multiply the expression by –dt: d[A] = -k[A]dt:  A plot of a plot of [A] vs t should result in an exponential decrease in A with time (figure a).  If we plot ln[A] vs t we get a straight line (figure b) First-order reactions  See derivation in class for the integrated form of d[A] = -k[A]dt.  The integrated form of a 1st order reaction is:  ln[A]t = ln[A]0 – kt  y =a - mx  Where [A]t = concentration at some time t and [A]0 = initial concentration.  Thus a plot of ln[A]t vs t should give a straight line with slope = -k.  Half-life for 1st order reactions  The half life is defined as: The time taken for the reactants to reach half of the initial concentration. (A general definition)  For a 1st order reaction the ½ life is independent of the initial concentration.  The ½ life for a 1 order st reaction is given by:  ln 2 0.693 t1/ 2   k k  Where k is the rate constant  Likewise the concentration after a given number of ½ lives can be found by the x expression: 1 [ A]t  [ A]0   2  Where x = number of ½ lives. Second-order reactions  Starting with our generalized rate law:  rate = k [A]m[B]n  If m = 2 and n = 0 (or if [A] = [B] and m = n = 1)  rate = k[A]2 (i.e. a second order rate expression) d [ A]   k[ A]2 dt 1 1  And the solution to this expression is:   kt [ A]t [ A]0  (Note some text will have the solution as 1/[A]t = 1/[A]0 +2kt)  Nevertheless:  A plot of 1/[A]t vs t is linear for these second-order reactions and the slope give the rate constant. 1  The half-life t1 / 2  k[ A]0  SUMMARY Reaction Rate law Integrated Rate Law Plot Half life order Zero r=k [A] = [A] - kt [A] vs t t = ½[A] k t 0 ½ 0 First r = k [A] ln[A] = ln[A] - kt ln[A] vs t t = ln 2/k t 0 ½ Second r = k [A]2 1/[A] = 1/[A] + kt 1/[A] vs t t = 1/(k[A] ) t 0 ½ 0 Chemical Kinetics Lecture 2 Mark A.W. Lawrence  Calculating the order of reactions  Initial rate method: (A + B  C)  Measure the slope at start of the reaction— which gives the initial rate.  We vary the concentration of one species while keeping the others fixed and see how the rate in affected by each [reactant].  Now recall that for the reaction:  aA + bB  cC  Rate = k[A]m[B]n where m & n are the reaction orders.  If we wish to determine m and n from a data set where [A] and the [B] were systematically varied and the initial rate determined, what is the approach?  Calculating the order of reactions  Let us answer this question with a simple example at fixed temp: [A] / mol dm-3 [B] / mol dm-3 Initial rate / mol dm-3 s-1 0.001 0.005 2.0 x 10-4 (r1) 0.001 0.010 2.0 x 10-4 (r2) 0.002 0.010 4.0 x 10-4 (r3)  Since rate = k[A]m[B]n  r1 = k[A1]m[B1]n and r2 = k[A2]m[B2]n etc. r1 k[ A1 ]m [ B1 ]n  And  …….etc. r2 k[ A2 ]m [ B2 ]n  Recall the rate constant is independent of the conc. of the reactants.  Thus they cancel in the ratio since the experiments are done at the same temperature  m n m Thus: r1  [ A1 ] [ B1 ] n r1  [ A1 ]   [ B1 ]   or      r2 [ A2 ]m [ B2 ]n r2  [ A2 ]   [ B2 ]   At this point it should be clear that if we are interested in n for e.g., then the coefficient to m must be 1 in order to simplify the calculations.  Under what conditions will the coefficient of [A1]m/[A2]m = 1?  When [A1] = [A2] [A] / mol [B] / mol Initial  Now lets plug in some numbers: dm-3 dm-3 rate / mol  r1 2.0 104 (0.001)m (0.005)n dm-3 s-1  4  r2 2.0 10 (0.001)m (0.010)n 0.001 0.005 2.0 x 10-4 2.0  104 (0.005)n (r1)   2.0  104 (0.010)n 0.001 0.010 2.0 x 10-4  Thus: 1 = (½)n (r2) 0.002 0.010 4.0 x 10-4  Taking the log of the expression: (r3)  log 1 = n log (½) Thus n = 0  Likewise if we wish to find m: 4 r2 2.0 10 m  0.001     r3 4.0 104  0.002   Thus: log (½) = m log(½)  And m = 1  And the rate equation is given by:  rate = k[A]1[B]0 i.e. rate = k[A]  We can also find the rate constant: k = rate/[A]  i.e it is a 1st order reaction  k = (2.0 x 10-4 / 0.001) = 0.2 s-1  We can also determine the ½ life for the (1st order) reaction: ln 2 0.693 t1/ 2   k 0.2 s 1  t1/2 = 3.5 s  Worked Example  The initial rate data at 25 oC are listed in the table below for the reaction: NH4+ (aq) + NO2- (aq)  N2 (g) + 2 H2O (l) Initial rate of consumption of NH4+ / Experiment # Initial [NH4+] / M Initial [NO2-] / M M s-1 1 0.24 0.10 7.2 × 10-6 2 0.12 0.10 3.6 × 10-6 3 0.12 0.15 1.22 × 10-5  What is the rate law? [6 Marks]  What is the value of the rate constant? [2 Marks] Solution Initial rate of consumption Experiment # Initial [NH4+] /M Initial [NO2-] /M of NH4+ / M s-1 1 0.24 0.10 7.2 × 10-6 2 0.12 0.10 3.6 × 10-6 3 0.12 0.15 1.22 × 10-5  Write a general rate law: rate = k[NH4+]x[NO2-]y  First, we have to find the order wrt both reactants, NH4+ and NO2- i.e. x and y  The order wrt NH4+ (x) can be found by comparing the rates wrt reactions 1 and 2 (since [NH4+] changes and [NO2-] is constant in exp 1 & 2). 𝑟 3.6×10−6 0.12 𝑥 0.10 𝑦  2 = = 𝑟1 7.2×10−6 0.24 0.10  0.5 = 0.5x; therefore x = 1  Initial rate of consumption Experiment # Initial [NH4+] / M Initial [NO2-] / M of NH4+ / M s-1 1 0.24 0.10 7.2 × 10-6 2 0.12 0.10 3.6 × 10-6 3 0.12 0.15 1.22 × 10-5  The order wrt NO2- can be found by comparing the rates wrt reactions 2 and 3. 𝑦 𝑟3 1.22×10−5 0.12 𝑥 0.15  = = 𝑟2 3.6×10−6 0.12 0.10  3.39 = 1.5y;  y log 1.5 = log 3.39  therefore, y = (log3.39/log1.5) = 3  Thus x = 1 and y = 3  Since rate = k[NH4+]x[NO2-]y  Rate law: rate = k[NH4+][NO2-]3  (b) Rate constant:  Since rate = k[NH4+][NO2-]3  k = rate /(NH4+)(NO2-)3 = 7.2 × 10-6 Ms-1/(0.24M) (0.10M)3  k = 0.030M-3 s-1  Reaction Mechanisms  Thus far the discussion on chemical kinetics has been centred around reaction rates and rate constant.  Equally important to chemical kinetics is the reaction mechanism.  The reaction mechanism is defined as the sequence of molecular events or reaction steps that defines the pathway from reactants to products.  For e.g. the reaction A  E  Each step is called an elementary step. k1 A B  Each step has its own rate and k2 B C Reaction rate constant. k3 C D Mechanism  The slowest step is called the D k4 E rate determining step (r.d.s).  E.g. If k2