11: Determination of an Activation Energy

INTRODUCTION

As noted in your previous experiment, two of the factors that affect the rate of a chemical reaction are temperature and a catalyst. In this experiment, we will investigate in more detail just how those factors affect the rate of a chemical reaction.

From considerations of kinetic theory, we know that only a small proportion of the many molecular collisions result in a reaction. In most collisions, the molecules simply rebound like billiard balls, without change. Only the more energetic molecules collide with sufficient energy for their electron orbitals to interpenetrate, thus forming an activated complex. These then react to form more stable products. The proportion of such energetic molecules depends upon the absolute temperature and is in accord with the Maxwell-Boltzmann distribution law, which defines the increasing proportion of such more energetic molecules at higher temperatures. The minimum net energy that the colliding molecules must possess to react is called the activation energy, E_a. The situation is somewhat analogous to an attempt to roll a ball up an incline, over a hump, and down to a lower, more stable level. The extra energy necessary to get over the hump is the activation energy. The activation energy for a chemical reaction can be found experimentally by measuring the reaction rate at different temperatures.

A catalyst is a substance that permits the formation of the activated complex more easily, with lower activation energy. At a given temperature, a larger proportion of the molecules will possess enough energy to react, and thus the catalyst increases the rate of the reaction.

How can we calculate the activation energy? The relationship between the reaction rate constant, k, the absolute temperature, T, and the activation energy, E_a, is given by the equation

\[ k = A \exp\left(-\frac{E_\text{a}}{RT}\right) \label{EARR} \]

Where

\( R \) is the gas constant (8.314 J/mol·K),

\( A \) is a constant for the reaction.

The factor \( \exp\left(-\frac{E_\text{a}}{RT}\right) \) comes from the Boltzmann principle, which states that, for the equilibrium at any absolute temperature T, the ratio of the number of molecules with large energy to the number with small energy is a function of the factor \( \exp\left(-\frac{\Delta E}{RT}\right) \), where \(\Delta E\) is the difference in energies. We may rewrite equation \(\ref{EARR}\) in the logarithmic form

\[ \ln k = -\frac{E_\text{a}}{R}\left(\frac{1}{T}\right) + \ln A \label{LINARR} \]

Since the rate constant, k, is inversely proportional to the time, t, for a given amount of reaction, we may rewrite equation \(\ref{LINARR}\) as

\[ \ln \frac{c}{t} = -\frac{E_\text{a}}{R}\left(\frac{1}{T}\right) + \ln A \]

where c is some constant. Manipulating this equation gives

\[ \ln c - \ln t = -\frac{E_\text{a}}{R}\left(\frac{1}{T}\right) + \ln A \]

or

\[ \ln t = \frac{E_\text{a}}{R}\left(\frac{1}{T}\right) + constant \]

Thus, if we graph the natural log of time (\(\ln t\)) against the reciprocal of the absolute temperature (\(1/T\)), the slope, m, of the straight line thus obtained, multiplied by R will give us the activation energy E_a.

\[ m = \frac{E_\text{a}}{R} \ \ \ \text{or} \ \ \ E_\text{a} = mR \]

In this experiment, you will determine the activation energy, both with and without a catalyst, for the oxidation of iodide (\(\ce{I^-}\)) to iodine (\(\ce{I2}\)) by peroxydisulfate (\(\ce{S2O^{2-}_8}\)), also known as persulfate. In the process, the peroxydisulfate is reduced to sulfate (\(\ce{SO^{2-}_4}\)).

\[ \ce{I^-}(aq) + \ce{S2O^{2-}_8}(aq) \rightarrow \ce{I2}(aq) + 2\ce{SO^{2-}_4}(aq) \label{I-toI2} \]

This reaction is rather slow, and you can easily measure the time required for a given amount of reactants by having a large excess of iodide and peroxydisulfate and a small amount of thiosulfate (\(\ce{S2O^{2-}_3}\)), with starch as an indicator. Thiosulfate reacts rapidly to reduce the iodine back to iodide. In the process, the thiosulfate is oxidized to tetrathionate (\(\ce{S4O^{2-}_6}\)).

\[ \ce{I2}(aq) + 2\ce{S2O^{2-}_3}(aq) \rightarrow 2\ce{I^-}(aq) + \ce{S4O^{2-}_6}(aq) \label{I2toI-} \]

Reaction \(\ref{I2toI-}\) ensures that no iodine remains in the solution until the thiosulfate is exhausted. At this point, the iodine formed from reaction \(\ref{I-toI2}\) suddenly turns the starch blue.