CHEM 240 Bioanalytical Chemistry

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Lecture 22. Chemical kinetics: Temperature effects

Wednesday 17 March 2010

The temperature dependence of reaction rates. The Arrhenius equation. Activation energy and its relation to reaction rate. Finding activation energy from the values of the rate constant k at two different temperatures. The transition state and reaction coordinate diagrams. Kinetics wrap-up: Calculation of the Arrhenius factor, A.

22. Summary

Lecture 22 Summary

We now finish our exploration of chemical kinetics with a quantitative treatment of the temperature dependence of reaction rates, followed by consideration of catalysis and enzyme kinetics. The study of biological catalysis and enzymes is something near and dear to many biochemists. We'll have a look at how kinetic principles we've learned so far apply to catalysts, including enzymes. We'll find we can construct a simple mechanism for an enzyme-catalyzed reaction.

Activation energy and the temperature dependence of reaction rates

All of our investigation of rate laws and rate constants so far have related to conditions of constant temperature. Yet we know that temperature is an important factor in determining reaction rates. In general, as temperature increases, so does reaction rate, which is in accord with our experience. For example, we know that a mixture of fuel and air does not react until an initiating spark is provided. The concept of activation energy explains qualitatively this kind of observation, and the quantitative nature of the temperature dependence of reaction rates shows that the value of the rate constant increases with increasing temperature. The form of the rate law remains unchanged. Based on experiments performed near the end of the 19th century, the Swedish chemist Svante Arrhenius developed an equation that expressers the key relationship between temperature and rate constant that we now know as the Arrhenius equation.

The first equation (eq. 1, the Arrhenius equation) shows the exponential dependence of rate constant k on activation energy Ea and temperature, T. The negative sign in the exponent means that as Ea increases, the value of k greatly decreases, and that as T increases, k greatly increases. Equation 2 is obtained by taking the logarithm of both sides of eq. 1. It shows that a plot of ln k versus 1/T should yield a straight line with slope –Ea/R (see graph below). Equations 3 are just a slight variation of eq. 2. Equation 4 shows that if we measure k at two different temperatures T1 and T2, we can then estimate the value of Ea.

 At left is a graphical representation of the logarithmic form of the Arrhenius equation. Experimental data for chemical reactions generally show this behavior, at least over a certain range of temperature.

Activation energy and the temperature dependence of reaction rates

Example: Using rate constants at two different temperature to determine activation energy.

For the reaction

2 HI(g)  =  H2(g)  +  I2(g)

the rate constant is measured at two different temperatures:

@ T = 500 K, k = 9.51 × 10−9 L mol−1 s−1  and  @ T = 600 K, k = 1.10 × 10−5 L mol−1 s−1

Using this information, calculate (a) the activation energy, Ea and (b) the value for the pre-exponential Arrhenius factor, A.

Solution: For part (a), it is convenient to use equation (4). The equation can be solved algebraically for Ea:

Ea  =  − R ln(k2/k1) / (T2−1T1−1)

We can plug in numbers from above, as well as the value R = 8.31447 × 10−3 kJ mol−1 K −1. The result is

Ea  =  1.76 × 102 kJ mol−1

(b) We can solve equation (1) for A and plug in numbers, using one of the pairs of k and T:

A = k / exp{−Ea / RT }
= ( 9.51 × 10−9 L mol−1 s−1) / exp{ −1.76 × 102 kJ mol−1/ (8.31447 × 10−3 kJ mol−1 K −1)(500 K)}
= 2.31 × 1010 L mol−1 s−1

Catalysis and enzymes

The following are the key points concerning catalysts

¶ Catalysts speed up reaction rates and are not reactants or products of the net reaction

Catalysts can be participants in a reaction mechanism, combine with reactants to form intermediates, but free catalyst is regenerated, which can then undergo another round of catalysis.

¶ Usually the catalyst provides an alternate mechanism for the reaction, one with a lower activation energy ( kcat > kuncat)

This relates primarily - again - to the participation of the catalyst. A catalyst can speed a bimolecular reaction by bringing the reacting molecules together - providing adjacent, properly-oriented binding sites for the reactants, for example - but it may also provide a completely different mechanistic pathway by reacting to form intermediates that are not accessible in an uncatalyzed mechanism. The lowered activation energy achieved by a mechanism including intermediates reached via lower-energy transition states shows up as a much higher rate constant for the catalyzed reaction: kcat > kuncat.

¶ The position of equilibrium is not affected, but the approach to equilibrium is more rapid

This is a thermodynamic principle. Catalysts do not affect the thermodynamics of a reaction, which determine the equilibrium constant ("Big K") for the reaction. Catalysts affect the reaction kinetics by increasing the rates (little k's) for both the forward and reverse reactions.

Enzymes are biological catalysts and are most commonly protein molecules adapted to speed up a specific reaction

Enzymes are amazing from the standpoint of how greatly they accelerate reaction rates in biological systems, their exquisite specificity (no unintended side products, as in synthetic organic chemistry!), and their ability to do so under mild physiological conditions.

 A reaction coordinate diagram for a catalyzed reaction. An explanation for the ability of a catalyst to speed up a reaction is that it can lower the activation energy of the reaction. Let us consider the diagram at left to represent an elementary reaction that can take place with or without catalysis. The red curve shows the energy profile for the uncatalyzed reaction. The activation energy for uncatalyzed conversion to products is much greater than that for the catalyzed reaction (indigo curve). This means that the rate constant for the catalyzed reaction, kcat, will be much greater than kuncat, the rate constant for the uncatalyzed reaction. For both the uncatalyzed and the catalyzed reaction, the potential energy change, ΔErxn, is the same. This means that while a catalyst does not alter the conditions under which the reaction is at equilibrium, it greatly speeds up the approach to equilibrium. The reaction coordinate diagram shows that the energy of activation for the reverse reaction is lowered by the catalyst as well.

 A simple mechanism for an enzyme-catalyzed reaction. Consider a reaction, S → P. The reactant S is referred to in enzymology as the substrate. P, of course, stands for product. Catalysts, and enzymes in particular, are notable for their ability to greatly speed up a reaction despite being present in substoichiometric amounts. This means, for example, that one enzyme molecule may catalyze the conversion of thousands of substrate molecules to product per second. However, in order for an enzyme to catalyze the reaction, the enzyme E must form a complex with the substrate. The enzyme-substrate complex, ES, is thus an intermediate in the mechanism. The enzyme binds substrate in a way that creates a low-activation energy path for its conversion to product. (1)  E  +  S  =  ES (2)  ES  →  E  +  P Does such a mechanism explain experimental observations of enzyme-catalyzed reactions? The figure at left shows the idealized results of measurements of a typical enzyme-catalyzed reaction. Initial rates are measured at different starting substrate concentrations, and when these are graphed, it is found that the measured values are well-fit by a hyperbolic curve given by the equation shown. Two parameters describe this equation. One parameter is Vmax, which represents a limiting maximum velocity, the rate approached as the substrate concentration increases. The other is usually called KM, which corresponds to the substrate concentration at which V = Vmax /2. The experimental data and the rate vs. substrate equation indicate a zero-order reaction when substrate concentration becomes much greater than KM, while when substrate concentration is low ([S] << KM), the rate is that of a first-order reaction, with V0 = (Vmax / KM ) [S]. We can show that the mechanism above leads to the given initial velocity versus substrate concentration equation. One relatively simple way to do this is to make the assumption that step (1), formation of the ES complex, is significantly faster than step (2), production of products from ES, i.e. the mechanism can be represented as (1)  E  +  S  =  ES        (fast) (2)  ES  →  E  +  P      (slow) This is not true for all enzymes, but it is true enough for some, and those cases where this assumption does not hold can be dealt with using an alternate assumption that leads ultimately to the same result. Now if step (2) is the much slower step, it is the rate-limiting step, and its rate determines the overall reaction rate. Writing the rate law for this elementary step, we have V0  =  k2 [ES] where k2 is the first-order rate constant for step (2) in the forward direction. Since ES is an intermediate and the rate law for a reaction is expressed in terms of a rate constant and reactant concentrations, we must find an expression for [ES] in terms of [S] (strictly speaking, we mean [S]0, the initial substrate concentration). In this case, we will also make use of the total enzyme concentration, [E]T, since this is a quantity that we can control in the experimental setting. We can see that at high substrate concentrations, essentially all enzyme at any instant has substrate bound, and therefore [ES] = [E]T, and the enzyme is being driven as fast as it can possibly go, given its limited amount. Under such conditions, we say that the enzyme is saturated with substrate, and we have Vmax  =  k2 [E]T The parameter Vmax is thus set by the catalytic rate constant and total enzyme concentration. To get [ES] at less than saturating substrate concentrations, we use both the relation [ES] = [E]T – [E], where [E] is the concentration of free enzyme, and the relation we get from assuming that step (1) is at equilibrium: k1/ k–1  =  [ES] / [E][S] where k1 and k–1 are the forward and reverse rate constants, respectively, for step (1). Solving the latter for [E] and substituting into the above expression for [ES], we obtain [ES]  =  [E]T – [E]  =  [E]T – (k–1/ k1)( [ES] / [S] ) Solving this for [ES], we have [ES]  =  [E]T – [E]  =  [E]T /{(k–1/k1)(1 / [S]) + 1} =  [E]T [S] /{(k–1/k1) + [S]} If we now define KM = k–1/k1 (i.e. KM is the equilibrium constant for the reverse of step 1, ES →  E + S) and combine the expression for [ES] we just obtained with the rate equation V0  =  k2 [ES], we have V0  =  k2 [ES]  =  k2[E]T [S] /{KM + [S]}  =  Vmax [S] /{KM + [S]} where we have substituted Vmax  =  k2 [E]T to obtain the last equality, which gives us the equation we were after. This shows that the simple mechanism we proposed above as a model for at least some enzyme-catalyzed reactions does indeed give us an initial rate vs. substrate concentration curve consistent with what is typically observed in enzyme kinetics experiments.

Learning objectives

Page update in progress, Jan 2010 (01-01-10)

References

1. Chemistry: Science of Change (4th edition) Oxtoby, Freeman, Block

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 [ E-mail: cronk@gonzaga.edu ]