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Chemistry, 23.01.2020 19:31 mayhy100

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?

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