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### Lecture_26

Course: MBB 323, Fall 2009
School: Sveriges...
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323 Lecture MBB 26 Kinetics (Chapter 7) 1 Reaction order We will work in this section with concentrations. We will label the first derivative of a particular concentration as follows: cB(t) : concentration of molecule B at time t. vB(t) = dcB/dt : first derivative of concentration of B Will approach the math from the perspective of understanding the reaction order of a particular reaction. Plot your data...

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323 Lecture MBB 26 Kinetics (Chapter 7) 1 Reaction order We will work in this section with concentrations. We will label the first derivative of a particular concentration as follows: cB(t) : concentration of molecule B at time t. vB(t) = dcB/dt : first derivative of concentration of B Will approach the math from the perspective of understanding the reaction order of a particular reaction. Plot your data and determine the reaction order. Perform further experiments... 2 Reaction order Zero order (p321-322): A -> B vB(t) = -vA(t) = k Integration gives, cB(t) = cB(0) + kt (k, is the slope) cA(t) = cA(0) - kt First order (p322-329) C -> D Important: radioactive decay and many other processes. vD(t) = -vC(t) = kcC Integration (see overhead) gives: cD(t) = cC(0)(1-e-kt) (this boundary condition assumes zero D present at time zero, which does not need to be true) -kt 3 Reaction order Second order (p330-338), A + A -> P (Class I, 330-331) vP(t) = kc2A(t) A + B -> P (Class II, 331-332) vP(t) = kcA(t)cB(t), the solution of this diff. eqn. gives: 1/(cA(0)-cB(0))ln((cB(0)cA(t))/(cA(0)cB(t))) = kt Spse the concentration of B is very large compared to A. -ln (cA(t)/cA(0))/cB(0) =kt ln (cA(t)/cA(0)) = -kcB(0)t cA(t) = cA(0)e-kcB(0)t (this is a first order solution!) 4 Parallel reactions Suppose a substrate reacts by two different processes. A -> B, and A -> C, each with distinct rates. This is a question on your assignment. Read p343-344 carefully. Can be confusing, but for first order processes, the rate of decay depends on the total decay rate by all process. This is obvious if you think about the rate that A is converted into products. Not so obvious if you think about the accumulation of say only B. This accumulation also occurs at the rate of decay times a constant factor (p344). 5 Transition State Theory Any reaction can in principle be broken down into a number of intermediates (0 or more) separated by transition states (1 or more). An intermediate can in principle be purified, but could be very unstable (short lived). A transition state lasts for only several bond vibrations. How quickly does an atom vibrate in a Carbon molecule? monoxide has a fundamental vibrational frequency of about: 6.6 x 1013 Hz (see p467) Time scales are therefore ~10-14 s Light would only travel about 3 m in this time! Changing atom masses changes vibrational frequencies. 6 Transition State Theory Reactions take place along a reaction surface defined by reaction coordinates and the potential energy at each point. An actual reaction can be thought of as a trajectory over this space (starting from the initial state and arriving at the final state). These trajectories must be treated quantum mechanically QM and not classically. This is because QM, allows tunneling, or processes that are classically forbidden to occur. See overhead The most interesting paths however are the ones that are the minimum energy paths (think of a valley through a mountain, cars do not drive over the mountain, but rather travel through the valley). This minimum energy path defines a reaction coordinate (a 1-D trajectory that is most likely for the reaction. 7 1 TS, 0 intermediates: G!Transition State (virtual, use double dagger, here !) MN! G GInitial State (real) M+N GFinal State (real) P+Q Reaction coordinate 8 Transition State Theory (p357-360) From these free energies we can define for a reaction: M + N <-> MN! -> P + Q K! k! G = GFin...

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