Geology 560, Prof. Thomas Johnson Unit IV, Part 2: Activation energies and the responses of reaction rates to temperature and compositional changes.

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1 Geology 560, Prof. Thomas Johnson Unit IV, Part 2: Activation energies and the responses of reaction rates to temperature and compositional changes. eading: White, Chapter 5; Walther, Chapter 13 Excellent additional eference: Kinetics Theory in the Earth Sciences, Antonio C. Lasaga, Princeton Univ. Press, So far in our study of kinetics, we have looked at collision theory and at how concentrations and reaction rates are related. Now we will delve more into the rate constants. Some reactions have very large rate constants because they happen readily, i.e., the activated complexes are easily formed and/or do not require very high energies to form. Other reactions are very rare and thus slow) because they require formation of high-energy, unlikely collisions or activated complexes. Here, we examine those issues. 1) Activation energy and reaction rate a. Diagram of an activation energy barrier: The reaction coordinate represents some variable that changes as the reaction happens. For the NaCl example given in the last set of notes, it might be the distance between the Cl - ion being removed and the next ion in the crystal. As this distance increases, the energy increases at first. The maximum energy point would be the most awkward point where the Cl - is squeezing past a water molecule and about to change places with it. If the energy is not great enough, the top of the hill is not attained and the reactants go back to the way they started If the activation barrier is surpassed, the reaction proceeds and as it does so, the energy decreases as the system achieves a new, more stable configuration. For a demonstration of this, try: b. eaction rates depend on, qualitatively you MUST know these general ideas well): a. Activation energy: Larger activation energy ==> slower rate b. Temperature: Higher T ==> greater rate because greater temperature increases the likelihood that the needed activation energy will be attained c. Other details of the reaction. The activated complex can be very simple or it could be a very odd, unlikely collision/combination in some unlikely geometry. This leads to slower rates c. Mathematically, the rate constant, k, for the forward reaction is given by: a. k = Ae This the Arrhenius Equation 1. A is called the frequency factor ; you can think of this as expressing the number of fruitful collisions/combinations per second for a given concentration

2 ! E A 2. e is a Boltzmann factor. Boltzmann factors appear in many places in physical chemistry. ecall that vibrational energy is quantized; just like electron energies, molecular bonds have certain allowed energy levels. An individual molecule cannot have any arbitrary amount of energy. It can be in the ground state, or it can be in one of the quantized higher energy levels. In essence, the Boltzmann factor tells us the fraction of molecules that will exceed the activation energy. a. BEWAE there are two s in rate equations: i. When you see this is the gas constant x temperature ii. We also use for ate: 1. is the rate for the forward reaction 2. - is the rate for the reverse reaction 3. is the net rate of reaction 3. This plot shows you the value of the Boltzmann factor for E A = 10 kcal/mol: 4.5E-12 4E E-12 Boltzmann factor 3E E-12 2E E-12 1E-12 5E T deg-c) This shows that the fraction of bonds exceeding the activation energy is very small but increases greatly with increasing T, as expected. 4. For an activation energy of 10 kcal/mol, the rate constant near room T doubles for every 10 degrees increase in T, roughly. d. Determining E A from experimental data Arrhenius plots): a. Starting with k = Ae b. lnk = ln A 1 T, take ln of both sides: c. Plot lnk) vs. 1/T an Arrhenius plot) 1. Slope is -E A / here is the gas constant) 2. Intercept is A d. Note this assumes A is not T dependent. It actually is somewhat T dependent, but not as strongly as the exponential term. e. T dependence of rates. Note that the Arrhenius plot gives us a way to extrapolate data we have for a certain T range to determine the rate of a reaction at some other T. a. Very helpful because experiments are often impossible at low T might take years), but we may be able to measure rates at greater T and extrapolate down to lower T 2) Multiple reaction mechanisms are possible for many of the reactions we care about. Each mechanism will have a unique E A value. If you are lucky, one mechanism is MUCH faster than the other so you an ignore the others. 3) Catalysis. E A can be lowered if a catalyst modifies the reaction mechanism and makes the reaction happen more readily. a. Example: eduction of CrVI) to CrIII) by certain organic reductants happens very slowly in a simple solution. But when gibbsite is added, the reactants adsorb onto the solid and the reaction rates increases greatly. The surface lowers the activation energy by weakening bonds of the reactants or holding them together on the surface.

3 b. Solid surfaces, enzymes, and dissolved ions can act as catalysts 4) Net rate of reaction can be related to ΔG. First, we look at this qualitatively: a. - - b. We know that the net rate of reaction approaches zero as the reaction approaches equilibrium- that s what equilibrium means. Thus, as ΔG approaches zero, we know the net rate must also approach zero, so = - c. AND if ΔG is negative, we know net rate will be positive net forward reaction, > - ) d. AND if ΔG is positive, net rate will be negative met backward reaction; < - ). 5) elationship between enthalpy of reaction and the activation energy of the reverse reaction, E A-, to that of the forward reaction, E A. E A ΔH E A- If ΔH is negative, the reverse reaction s activation energy is greater. In the diagram of an activation energy barrier above, you can see that the reverse reaction has a larger energy gain needed to get it to the top of the activation energy barrier. Thus, the reverse rate tends to be slower which tends to make the net rate positive. Note that we have ignored entropy here; this is not quite the complete story. White has an almost complete development of the theory, but we will skip that to get to the most important result: 6) Quantitative relationship between and ΔG. a. Starting with k = Ae b. k k! = A e e!e!e! = A e! for the forward and reverse reactions, we can write E A! c. but from 5 above we know: E A E A- = ΔH 0 d. but also, A = e S 0 a. This is not obvious and comes from some detailed derivations that I m skipping; refer to White, Walther or Lasaga for complete derivations. But this does make some sense, in that A is a frequency factor, with units of time -1 and the frequency with which the activated complex is formed depends strongly on how ordered it is. An activated complex which is complicated and needs to be put together in an exacting configuration is a highly ordered, low entropy complex and also happens less frequently. e. Next we need to take into account the concentrations of the reactants and products: a. = k [reac tants] assume eality for now) k! [ products] b. note that the second term on right side is close to Q and recall that!g =!G 0 lnq f. When we put all this together, we come up with a very simple and reasonable result:

4 ! G g. =! e h. Note that the activities = concentrations here, as this theory assumes ideality) are taken into account here, so the ΔG here is NOT ΔG 0, the standard state ΔG for unit activities, it is the ΔG for the reaction in the actual solution with whatever concentrations are actually present, and at any T and P. i. A little algebra gives us, the net reaction rate: G 1! e j. This is a very important relationship. It says that the net reaction rate is related to ΔG of the reaction by an exponential relationship. a. Far from equilibrium, the net reaction rate is simply the forward rate. This makes sense, as the backward rate is tiny in comparison. b. But as a reaction approaches equilibrium, the net rate slows as the reaction products build up and back reaction increases. The following diagram is for 298 K: net/ Delta G kcal/mol) k. Note that the ΔG used in this equation is the actual free energy change of the reaction, with all the activities of the reactants and products incorporated into it 7) How can I calculate ΔG? ecall from several weeks ago: ΔG = ΔG 0 lnq obtain ΔG 0 from thermo tables; calculate Q from product and reactant activities) a. Note that, for a negative ΔG, as a reaction proceeds, the reactants concentrations decrease, the products concentrations increase, Q increases, and ΔG moves toward zero. 8) Another useful form of the net rate equation: a. Start with the net rate equation from before: 1! e!g b. But we can replace the e with something different if we do the following c. the ecall that ΔG = ΔG 0 lnq d. Also recall ΔG 0 = -lnk e. So ΔG = -lnk lnq= -lnq/k) f. earrange to get e g. 1! Q K!G = Q K = K! Q K h. In plain English this says, The net reaction rate is equal to the forward rate multiplied by the deviation of Q from K relative to K). 9) Approximation for a reaction near equilibrium When ΔG < 0.4, roughly) We can invoke the approximation: G

5 1- e x x works for small x only) to get!g 10) ates of multi-step reactions. All this theory applies, strictly speaking, to elementary reactions only. However, it works reasonably well for many overall reactions, provided me make one change 1! e n G where n is any real number. You can think of n as a fudge factor that corrects for the fact that the ΔG for the overall reaction could be quite different from that of the rate-limiting step but the two differ by a factor of n regardless of concentrations of reactants and products). 11) Kinetics of enzymes: Michaelis-Menten Kinetics. For the case where we have a two step reaction: E is Enzyme, S is substrate, P is product) and the following two assumptions are valid: a. The total concentration of enzyme is constant b. The concentration of the ES complex changes slowly compared to S and P c. When you work out the kinetics for this case, you find the rate is given by: = v max S K M [ ][ E ] [ S] d. This equation has the expected properties: a. ate reaches a constant value, independent of [S], when [S] is high [ ] 1. = v max E 2. Makes sense- enzyme is saturated b. ate is proportional to both [S] and [E] at low [S]- looks like a simple kinetic expression, first order with respect to [S] and [E]