University:
University of Pennsylvania
Course:
CHEM 2220 | Physical Chemistry II
Academic year:

2020
Views:

150

Pages:

3
Author:

rehinhea0c

Arrhenius Equation The rate constant “k” for a reaction is related to the absolute temperature T by the equation: k = Ae-B/T A and B are constants and this equation can be also as k= Ae-E/RT R is gas constant and is equal to 8.3145JK-1mol-1 and E is the activation energy for the reaction. This equation was arrived at by Van’f Hoff and it is often referred to as the Arrhenius equation. Van’t Hoff argued as follows: The temperature dependence of a standard equilibrium constant kco is given by dLnkco/dT = ∆Uo/RT2 ………………………….1 where, ∆Uo is the standard internal energy change in the reaction. For the reaction such as k1 A + B ↔ Y + Z, k-1 the equilibrium constant kc is equal to the ratio of the rate constant k1 and k-1, kc = k1/k-1. ………………………………………….2 Substituting for kc in equation 1 gives, dlnk1/dT – dlnk-1/dT = ∆U/RT2 …………………………3 Equation 3 can be split into two equations dlnk1/dT= constant + E1/RT2………………………………4 and dlnk-1/dT = constant + E-1/RT2 …………………………….5 In this case, E1-E-1 is equal to ∆Uo .Experimentally, it is found that the constants in equations 4 and 5 can be set to zero and on integrating each of these equations give k1= A1e-E1/RT…………………………………………..6 k-1= A-1e-E-1/RT …………………………………………7 The quantities A1 and A-1 are known as prexponential factors of the reactions and E1 and E-1 are known as activation energies. Arrhenius pointed out that the fraction of molecules colliding with sufficient energy to react is small, and according to Boltzmann principle, “the fraction of collisions in which the energy is in excess of a particular E is e-E/RT” .This fraction is larger the higher the temperature T and the lower the energy E.The rate constant kά e-E/RT . To test Arrhenius equation, take natural logs on both sides lnk = lnA – E/RT…………………………………………8 The data below is for the experiment done in 1889 by Hecht and Conrad on ethoxide and methyl iodide 1 A graph of rate against absolute temperature is exponential as shown below. Temp. o ( C) Temp.T (K) 0 6 12 18 24 30 273.15 279.15 285.15 291.15 297.15 303.15 Rate k 168 354 735 1463 3010 6250 To test Arrhenius equation, take logs on both sides of the equation k1= A1e-E1/RT gives the equation below lnk = lnA – E/RT If the law applies, a plot of lnk against 1/T will be a straight line. The slope will be -E/R. 2 Temp.T (K) 273.15 279.15 285.15 291.15 297.15 303.15 Rate 168 354 735 1463 3010 6250 1/T 0.00366 0.00358 0.00351 0.00343 0.00337 0.0033 ln( Rate) k 5.124 5.869 6.6 7.288 8.01 8.74 (Source: http:webserver.lemoyne.edu/faculty/giunta) The slope (which has the units of K) will be –E/R. Alternatively, taking common logarithms, Log10k= log10A-E/2.303RT From the equation, a plot of log10k against 1/T will give a straight line. The slope of this plot of log10k against 1/T will be –E/2.303R.Proving Arrhenius equation. The Arrhenius equation is applicable to various reaction systems. Once activation energy for a reaction is known; it is easy to predict the value for the rate constant k at a different temperatureT1 from its value k at another temperature T. Equation 8 becomes:ln( k1/k)=Ea/R*(1/T-1/T1) for a reaction taking place at T and T1, temperatures. 3

Arrhenius Equation

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