num_kin_eq1 - 5.68J/10.652J Feb 13, 2003 Numerical Solution...

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5.68J/10.652J Feb 13, 2003 Numerical Solution to Kinetic Equations Lecture 1 The Problem We want to solve l H 1 dn i = ± ν r l ( T n ) Transport V of out i of Eq.(1) , ' ' V dt i Where n i is the number of moles of species ‘i’, ν is the matrix of stoichiometric coefficients, and r l = Rateof Step ' l ' (units: moles/volume-second) For a typical bimolecular elementary step reaction A+B = C+D: T A B T C r l = k forward ] ][ )[ ( k ][ )[ ( D ] reverse For a typical gas-phase dissociation reaction A= C + D: , C r l = k ( P T ] )[ k ( P T ][ )[ D ] forward , A reverse For a typical gas-phase isomerization reaction A = C: , A , C r l = k forward ( P T ] )[ k ( P T ] )[ reverse The rate constants also usually depend (weakly) on the chemical environment (e.g. dielectric constant), so they strictly are not really constant but (very weak) functions of all the n’s. Note that the forward and reverse rates are normally related by the equilibrium constant, related to the free energy of reaction (thermochemistry). It is important to appreciate that except in the simplest cases, there is no analytical solution to this system of differential equations, so we MUST use the computer to solve them numerically. Approximations like quasi-steady-state extend the range of systems where analytical (approximate) solutions are possible, but you don’t have to get very complicated before even the steady-state-approximation solutions are intractable.
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± ² ³ ´ µ i i More About the Fundamental Equation Chemistry textbooks usually give instead of Eq. (1) the equation: i d l H , ] [ = ± ν r l ( T n ) Eq.(2) dt i But Eq.(2) is true only if dV/dt = 0 (“isochoric”) and there is no transport (i.e. perfectly homogeneous, no convective flow). These textbook approximations are good for dilute reactions in solution or for constant volume apparatus (e.g. bomb calorimeter, shock tube). In many important cases the system is isobaric, i.e. dP/dt = 0, so if the temperature increases or there is a change in the number of moles of gas V will change. It is convenient to express the equation in terms of intensive quantities rather than the extensive variables n and V. The intensive variables that work best are mass fractions y ι : W n i = ] [ y = i W i i ρ M total dy W i 1 dn i V dt Transport V of out i of ' ' = dt You can plug in the same expression as above for the term in parentheses. Note that mole fractions x i are not such a good choice for kinetics as mass fractions, since the number of moles can change in chemical reactions, but the mass does not. However one writes the chemical reaction equations, one must be explicit about the
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num_kin_eq1 - 5.68J/10.652J Feb 13, 2003 Numerical Solution...

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