We could find out more information about ths reaction by studying the temperatur

Question

We could find out more information about ths reaction by studying the temperature dependence of the rate constant, k. These same experiments could be carried out at a lower temperature (ice bath) and a second rate constant, k2 could be determined for this temperature. Using the Arrhenius equation, these data could then be used to determine the activation energy, Ea, and the pre-exponential factor, A, for the reaction. The general form of the Arrhenius equation is shown below: For measurements at two different temperatures, this reduces to: ln(k2/k1) = (Ea/R)(1/T1 - 1/T2) A plot of ln k v.s. (1/T) for rate constants (k) measured at different temperatures (T) will have a slope equal to Ea/R. Substituting a value of k and the Ea into the first equation would allow you to solve for ln A. At room temperature (298 K) the rate constant for this reaction is 2.5 M-1s-1. When the experiment was repeated at 0.0oC, the rate constant was determined to be 1.0 M-1s-1. What is the activation energy, Ea for this reaction, in kJ/mol?

Explanation / Answer

ln(2.5) = Ea/R*(3.07*10^-4) So Ea = 24.8 KJ

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