CHAPTER 5 : CALCULUS

Transcription

1 Dr Roger Ni (Queen Mary, University of Lonon) - 5. CHAPTER 5 : CALCULUS Differentiation Introuction to Differentiation Calculus is a branch of mathematics which concerns itself with change. Irrespective of whether it is the change of concentration in a chemical reaction or the change of slope of a graph, calculus provies the necessary techniques. Consier a simple eample, the graph of y. Fig. y y Tangent at - 3 At any given point the slope of the curve will be given by the slope of the tangent at that point (the tangent is the straight line which just coincies with the curve at the point of concern as illustrate for ). From an inspection of the figure above, it is clear that the value of the slope for this function is always changing with the value of. The slope can also be measure by assuming the curve to be mae up of very large number of very short straight lines. Any one such straight line can be imagine as the hypotenuse of a right angle triangle as shown in Fig. below, the other two sies being small increments along the an y aes, an y respectively. Fig. ( +,y + y) y y (,y) y

2 Dr Roger Ni (Queen Mary, University of Lonon) - 5. The slope of the hypotenuse (the otte line), given by of the tangent at. y, is then approimately equal to the slope As y then, for this particular eample, it must be the case that : y + y ( + ) y + y + + ( ) Subtracting y ( ) from both sies gives: y + ( ) It is now necessary to imagine that the triangle in Fig. becomes smaller an smaller so that y, an the length of the hypotenuse become inefinitely small. This process will lea, in the limit, to the slope of the hypotenuse in Fig. being eactly equal to the slope of the tangent at point. Uner these circumstances, when is very small then ( ) is etremely small an can be neglecte. When this limit is approache it is usual to write y an as an. So y can be written as an the slope of the tangent at is : We can generalise this approach to consier a curve representing the function y n. Then: Noting y + ( + ) n ( + ) + n n ( + ) + y n + n (from the binomial epansion - see 6.) The higher terms involving (), () 3 etc. (represente by... ) may be ignore because is so very small. Thus: ( y + ) n n n n + n. + n + n n n n n n or to put it another way : ( n ) n n i.e. the erivative of the function n is n n

3 Dr Roger Ni (Queen Mary, University of Lonon) This then is a general equation for ifferentiating raise to any power. Note that if a preceing constant k, ha been inclue in the above calculations it woul still be present in the answer, i.e. If y k n then ( k n ) kn n Special cases :. if n, as in y k, then k - the slope of a straight line is a constant.. if n, as in y k, then - the function correspons to a horizontal straight line with zero slope. Stanar Derivatives You shoul aim to remember the following stanar erivatives - in the following table, the symbol k is use to represent a constant. Function Derivative y k ( constant) y k f() ( k f ( )) k ( f ( )) y f() + g() ( f ( ) + g( )) ( f ( )) + ( g( )) y n n n ( ) n y e y ln y sin y cos ( e ) e (ln ) (sin ) cos (cos ) sin

4 Dr Roger Ni (Queen Mary, University of Lonon) Derivatives of More Comple Functions Α situation often arises when rather more complicate functions have to be ifferentiate. There are various proceures for ealing with these situations, which reuce the problem to the point where we can use the stanar erivatives given on the preceing page. Differentiation by Substitution (Chain Rule) This metho is use for functions such as, for eample, ep (k) or sin θ, which is the usual way of writing (sin θ). Thus: u i) y ep (k) ; let u k then y ep (u) an k y ep (u) ep(u) u We now make use of the Chain Rule relationship : u u ep( u) k k ep(k) ii) y sin θ ; let u sin θ then y u u an cosθ θ y u u u Again using the Chain Rule : u θ u θ u cosθ θ sinθ cosθ θ

5 Dr Roger Ni (Queen Mary, University of Lonon) Differentiation of Proucts (Prouct Rule) Another common situation is to fin that it is necessary to ifferentiate the prouct of two simpler functions. Let the simpler functions be enote by u an v, then y u.v. Consier now small increments to each function, so that (y + ) (u + u) (v + v) y + u.v + v.u + u.v + u.v Subtraction of y ( u.v) from each sie, an neglecting the prouct, u.v, of two very small terms u an v yiels v.u + u.v If the small change in each of the functions arises from a small change,, in the variable then u v v. + u. i.e. ( uv) u v v. + u. ( Prouct Rule ) Differentiation of Quotients (Quotient Rule) Suppose u y i.e. y u v v Using the prouct rule : u v ( ) + v u Using the metho of substitution : ( v ) ( v v v ). v v. Combining the two previous results gives : u v v + v u u v v v + v u u v v u ( Quotient Rule ) v

6 Dr Roger Ni (Queen Mary, University of Lonon) Secon Derivatives y The secon erivative, enote by y, is obtaine by ifferentiating the first erivative, i.e. Eample If y then an y () Partial Derivatives Partial erivatives are obtaine by ifferentiating with respect to one variable whilst treating all other inepenent variables as constants, For eample, suppose z y + y then z z y z z y y. y + y ( y is treate as a constant). + y + y ( is treate as a constant) Note the use of a curly for the partial erivative. Partial erivatives occur frequently in thermonamics e.g. the heat capacity at constant pressure, C P, is the (partial) erivative of enthalpy with respect to temperature at constant pressure, i.e. H CP T P

7 Dr Roger Ni (Queen Mary, University of Lonon) Chemical Applications of Differentiation : Kinetics The ieas of calculus which have been introuce in the preceing sections fin one of their most important applications in the stu of the rates at which chemical reactions procee. A brief introuction to this topic, that of chemical reaction kinetics, follows. Representation of Reaction Rates Suppose we have a reaction between two species, A an B, of the type : A + B Proucts The rate of a reaction is a measure of the rate of change of the concentration (or partial pressure in the gas phase) of one of the reactants or proucts. e.g. c A Rate or t Rate [ Proucts] t If the reaction procees in the inicate irection then the concentration of A, enote by c A or [A], will ecrease over time, whereas the concentration of proucts will increase over time c - consequently A [ Proucts] will be negative whereas will be a positive quantity. t t In general, the rate may epen on the concentration of any or all of the various substances present. A rate epression relates the rate to the concentrations of the reactants (an in some cases also to the proucts). Often a simple algebraic representation is possible, for eample c t A k c n A c m B where n an m are integers. The reaction is then sai to be of orer n with respect to substance A an of orer m with respect to substance B. k is the rate constant (or rate coefficient), so calle because it is inepenent of concentration. However, k oes epen on temperature, i.e. its value varies when the temperature is change. Note that the kinetic orers ( n an m ) are not, in general, relate to the stoichiometric coefficients of reagents in the chemical equation (which were assume to be : in the above eample).

8 Dr Roger Ni (Queen Mary, University of Lonon) Maima an Minima The function 3, y. y plotte below has a maimum at, y 4 / 3 an a minimum at 3 y / / > / < / > / 3 4 At both the maimum an the minimum the slope is zero an. Thus the positions of any maima an minima ( turning points ) can be foun by ifferentiating a function an equating the result to zero. Eample 3 y At a turning point ( )(3 ) i.e. turning points occur at an 3 One can istinguish between a maimum an a minimum by noting the tren of in the region of the turning point.. Consier, the maimum (at ) in the above eample. when <, the slope,, is positive when > (but less than 3 ), the slope,, is negative. Thus in the region of the maimum is ecreasing as increases.

9 Dr Roger Ni (Queen Mary, University of Lonon) Consequently the erivative of the slope,, with respect to is negative, y i.e. the secon erivative,, must be negative. Check: y (3 4 + ) 4 + y At, 4 + ( ). Consier, the minimum (at 3 ) in the above eample. when < 3 (but greater than ), the slope,, is negative when > 3, the slope,, is positive. Thus in the region of the maimum is increasing as increases. Consequently the secon y erivative,, must be positive. Check: y (3 4 + ) 4 + y At 3, 4 + ( 3) + y If both an are zero at a given point, that point is sai to be a horizontal point of infleion. An eample of such a point is the point, y 3 on the curve 3 y , shown below. y 8 6 / > y/ > 4 / / > y/ < y/

10 Dr Roger Ni (Queen Mary, University of Lonon) - 5. Summary : y Minimum +ve Maimum ve Pt. of Infleion These mathematical techniques are wiely use in chemistry since one often wishes to preict the conitions uner which some quantity has a maimum or minimum value, e.g., the attainment of thermonamic equilibrium may correspon, uner ifferent conitions, to the maimisation of the entropy or to the minimisation of the Gibbs free energy of the system.

11 Dr Roger Ni (Queen Mary, University of Lonon) - 5. Integration Integration can be consiere as both. the inverse (reverse) operation to ifferentiation, an. a metho for etermining the area uner a plotte function. Integration as the Inverse of Differentiation If ( f ( )) f ( ) then f f + C ( ) ( ) where C is the so-calle constant of integration. Differentiation e.g.. + C Integration The constant of integration arises because any function of the form ( + constant ) will yiel upon ifferentiation so there is ambiguity when it comes to consiering the reverse operation. This type of integral where there is an ill-efine constant of integration is known as an inefinite integral. From our stu of ifferentiation we know: ( n ) n n n n n + C an n+ n + n n n+ + C n + The relationship in the bo is the stanar form for the integration of n : it works for all values of n ecept, a special case to be consiere later. We can use this same approach to erive a set of stanar integrals as liste on the net page.

12 Dr Roger Ni (Queen Mary, University of Lonon) - 5. Stanar Inefinite Integrals You shoul aim to remember the following stanar integrals - in the following table, the symbol k is use to represent a constant. Function y k f() Integral k f ( ). k f ( ). y f() + g() ( ( ) g( )). f ( ). + f + g( ). y n n+ n + C n + (provie n ) y e e e + C y + C. ln y sin y cos sin. cos + C cos. sin + C

13 Dr Roger Ni (Queen Mary, University of Lonon) Definite Integrals an Integration as a Metho of Determining Areas The following integral is an eample of a efinite integral. f ( ). ( f ( ) + C) ( f ( ) + C) f ( ) f ( ) The integration is now carrie out between two limits: is the lower limit, is the upper limit. This type of integral calculation is commonly abbreviate in the following fashion: [ f ( ) ] f ( ) f ( ) f ( ). Graphically the integral is equal to the area between the curve (forme by plotting f() against ) an the -ais, between the limits an. Eample What is the area uner the curve y + between an? Area, A ( + ). 3 [ ] 3 + y y + ( 8 / 3 + ) ( + ) 4 / (to 4 s.f.) A More Comple Integrals As with ifferentiation, there are stanar approaches which may be use to hanle more comple functions.. Integration by Substitution This metho may be use when the function to be integrate consists of a prouct of two parts, where one part is the ifferential of the other. Eample sin cos. Note that cos is the erivative of sin, so this prouct meets the requirement for this approach.

14 Dr Roger Ni (Queen Mary, University of Lonon) Let u sin u cos u cos. Hence sin cos. u. u u + C sin + C. Integration by the Metho of Partial Fractions This metho may be use when the function consists of a fraction, an the enominator of this fraction is a polynomial which may be factorize. Eample 8 I 3 4 The fraction requiring integration may be split into a series of partial fractions (see.8): 3 4 ( 4) ( + )( ) ( + )( ) Consequently, I ln + ln( + ) + ln( ) + C ln + ln( + ) + ln( ) + C ( + )( ) ln + C ( 4) ln + C

15 Dr Roger Ni (Queen Mary, University of Lonon) Chemical Applications of Integration. Reaction Kinetics - Integrate Rate Epressions We can now erive an apply the concentration-time relationships for reactions ehibiting various kinetic orers. (a) Zero-orer Chemical reaction : A Proucts Rate epression : Integration : [A] k[a] t [A] k. t k. [A] k. t [A] kt + C If [A] [A] o when t, then [A] o + C C [A] o Hence, [A] kt + [A] o [A] o kt Thus the concentration of the reactant ecreases linearly with time in a zero-orer reaction. To test for zero-orer kinetics, plot [A] against time, t ; if the graph is linear over a time corresponing to several half-lives then the reaction is zero-orer an the rate constant, k, can be evaluate from the slope of the graph (which is equal to k ). The units of k will be those of the slope which are (concentration / time), e.g. mol m 3 s. [A] [A] o ZERO ORDER [A] k t + [A] o slope k Time, t

16 Dr Roger Ni (Queen Mary, University of Lonon) (b) First orer Chemical reaction : Rate epression : Integration : A Proucts [A] k[a] k[a] t [A] k. t [A]. [A] k. t [A] ln [A] kt + C If [A] [A] o when t, then Hence, n [A] o + C C n [A] o ln [A] kt + n [A] o ln [A] n [A] o kt [A] ln kt or [A] [A] o e kt [A] In a first-orer reaction, the natural logarithm of the concentration of the reactant ecreases linearly with time, an the concentration itself falls eponentially with time. [A] [A] o FIRST ORDER [A] [A] o e k t Time, t To test for first-orer kinetics, plot ln [A] against t ; if the graph is linear over a time corresponing to several half-lives then the reaction is first-orer an the rate constant, k, can be evaluate from the slope of the graph (which is equal to k ). The units of k will be those of the slope which are (time).

17 Dr Roger Ni (Queen Mary, University of Lonon) ln [A] ln [A] o FIRST ORDER ln [A] k t + ln [A] o slope k Time, t. Beer-Lambert Absorption Law This law lies at the heart of quantitative spectroscopy since it escribes how, in an overall way, raiation is absorbe by matter. It is assume that when raiation, of initial intensity I passes through a thin slice of matter of thickness, then the small amount absorbe, I, is a certain fraction of the orginal intensity, provie is sufficiently small. Epresse mathematically, this means that: I k.i. where k is a constant epening upon the material an the wavelength. I is negative because I is ecreasing as increases. The equation can be arrange to give: an then integrate to give: I I I I k k k In this case the integration can be carrie out between efinite limits. These limits are the initial values of I an, I o, an respectively, an the final values of I an, i.e. I an, where I is the intensity of light emerging from a sample of macroscopic thickness. As note previously, these limits are written at the bottom an top of the integral sign an the subsequent working is set out as shown below. I I I I k o [ ln I ] k[ ] I I o ln I ln I k( ) o I I o ln k I or ln k o I

18 Dr Roger Ni (Queen Mary, University of Lonon) In solution k is relate to two properties of the absorbing species: ε the molar absorption coefficient (etinction coefficient) an c the concentration. As much spectroscopy is carrie out in solution the Beer-Lambert law is therefore often epresse in the form: I ln ε. c. I o Note on units - the right han sie is a logarithm an has no units. By convention, c is usually given in units of mol m 3 ( mol L ) an is the length of the sample cell in cm. The units of ε are therefore typically L mol cm. 3. Raioactive Decay It is foun that the rate of raioactive ecay is proportional to the amount of raioactive material present thus Rate of ecay t λ where is the number of raioactive atoms present at any time t an λ is the ecay constant. Note - t is negative since nuclei are ecaying an their number ecreasing. Rearrangement yiels : λ t λ t, an integration gives If o atoms were present at the beginning (i.e. at t ) an after time t - then, with limits, the integration yiels: [ ln ] λ[ t] t o ln ln o λ ( t ) ln λt or o ln λ t or N N o e λt o A special case is when ½ o, i.e. when the number of atoms has been reuce to half its initial value. The time taken for this to happen is known as the half-life ( τ ½ ). In this case ln o o λτ λτ ln. 693 Thus the ecay constant can be foun if the half-life is known.