The Effect of Temperature

Submitted by jwmoore on Thu, 02/17/2011 - 17:09

The fact that a small increase in temperatureA physical property that indicates whether one object can transfer thermal energy to another object. can double the rate of a reaction is primarily dependent on an observation we made in the section on molecular speed distribution. A seemingly minor temperature rise causes a major increase in the number of molecules whose speeds (and hence molecular energies) are far above average. This happens because the distribution of molecular speeds (figure 2 in the section on molecular speed distribution) is lopsided and tails off very slowly at the high-speed end. Since only those molecules whose energyA system's capacity to do work. exceeds the activation energyThe energy barrier over which a reaction must progress in order for reactants to form products; the minimum energy that reactants must have if they are to be converted to products. can react, a significant increase in the fraction of molecules having high energy causes a significant increase in rate.

The fraction of molecules which have enough kinetic energy to react depends on the activation energy E, the temperature T, and the gas constantA proportionality constant between the product of the pressure and volume of a gas and the product of the chemical amount (moles) and temperature of the gas. R in the following way:


Fraction of molecules having

enough energy to react = eE/RT      (1a)

                                               =10E/2.303RT      (1b)


Although a derivation of this result is beyond the scope of our current discussion, we can look at this equation to see that it agrees with common sense. The larger the E, the more negative the exponent, and the smaller the fraction of molecules which can react. Thus if two similar reactions are studied under the same conditions of temperature and concentrationA measure of the ratio of the quantity of a substance to the quantity of solvent, solution, or ore. Also, the process of making something more concentrated., the one with the larger activation energy is expected to be slower. As a corollary to this, if we can find some way to reduce the activation energy of a particular reaction, then we can get that reaction to go faster. Equation (1) also shows why changing the temperature has such a large effect on reaction rates. Increasing T decreases the magnitude negative exponent in the equation. This increases the fraction of molecules which can react, as shown by the following example:



EXAMPLE The reaction


CO + NO2 → CO2 + NO


has an activation energy of 116 kJ mol. Calculate the fraction of molecules whose kinetic energies are large enough that they can collide with sufficient energy to react (a) at 298 K, and(b) at 600 K.


SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. In both cases we use Eq. (1).


a) At 298 K, the power of 10 will be

\frac{-E^{\ddagger }}{\text{2}\text{.303}RT}=\frac{-\text{116 kJ mol}^{-\text{1}}}{\text{2}\text{.303 }\times \text{ 8}\text{.314 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\text{  }\times \text{ 298 K}}
=\frac{-\text{116  }\!\!\times\!\!\text{  10}^{\text{3}}\text{ J}}{\text{2}\text{.303 }\times \text{ 8}\text{.314 J }\times \text{ 298}}=-\text{20}\text{.3}


Then the fraction of molecules is

Fraction = 10–20.3 = 5.0 × 10–21

b) At 600 k,

Fraction = 10E‡/2.303RT = 10–10.1 = 8.0 × 10–11


Note: In an alternative way, the baseIn Arrhenius theory, a substance that increases the concentration of hydroxide ions in an aqueous solution. In Bronsted-Lowry theory, a hydrogen-ion (proton) acceptor. In Lewis theory, a species that donates a pair of electrons to form a covalent bond. e instead of 10 may be used. Using part b as an example we have

Fraction = 10E/2.303RT = eE/RT = e–23.25 = 8.0 × 10–11


In this case approximately doubling the temperature causes the fraction of molecular collisions which can result in reaction to increase by a factor of


\frac{\text{8}\text{.0 }\times \text{ 10}^{-\text{11}}}{\text{5}\text{.0 }\times \text{ 10}^{-\text{21}}}=\text{1}\text{.6 }\times \text{ 10}^{\text{10}}


that is, by about 16 billion. This means that the reaction is expected to be about 16 billion times faster at the higher temperature. This corresponds to slightly more than doubling the rate for each of the 30 intervals of 10 K between 298 and 600 K.

Equation (1) is also useful if we want to measure the activation energy of a reaction experimentally because it shows how the rate constantIn a differential rate equation, the proportionality constant that relates the rate with the concentrations of reactants and other species that affect the rate. The rate constant is the rate of reaction when all concentrations are 1 M. should vary with temperature. We have already seen that the rate of a unimolecular processAn elementary step in a reaction mechanism in which a single molecule reacts to form products; formally no species collide in such a step, but the energy needed to overcome an activation barrier is provided to the single molecule through prior collisions. should be proportional to the concentration of reactantA substance consumed by a chemical reaction.. If only a fraction of those reactant molecules have enough energy to overcome the activation barrier, then the rate should also be proportional to that fraction. Thus we can write


Rate = k(cA) = k′10E/2.303RT(cA)


From this equation it is clear that


k = k′10E/2.303RT


If we divide both sides by the units per second so that we can take logs, we obtain


\text{log }\frac{k}{\text{1 s}^{-\text{1}}} = \text{log }\frac{k'}{\text{1 s}^{-\text{1}}}E/2.303RT


or          \text{log }\frac{k}{\text{1 s}^{-\text{1}}} = \left( \frac{-E^{\ddagger }}{\text{2}\text{.303}R} \right)\text{ }\left( \frac{\text{1}}{T} \right) + \text{log }\frac{k'}{\text{1 s}^{-\text{1}}}

                      y         =             m        ×      x      +     b


This is the standard form for the equation of a straight line whose slope m is –E/2.303RT and whose intercept b is log (k′/s–1). Thus if we plot log (k′/s–1) versus 1/T and a straight line is obtained, we can determine E.

A similar situation applies to the second-order rate equationAn equation that describes the rate of a reaction as a function of the rate constant and the concentrations of reactants (and any other substances that affect the rate, such as products or catalysts); also called rate law. associated with a bimolecular processIn a reaction mechanism, an elementary step in which two atoms, molecules, or ions must collide in order for a reaction to occur., where


Rate = k(cA)(cB) = k′10E/2.303RT(cA)(cB)


In this case the rate constant has units cubic decimeter per mole per second, and so we must divide by these before taking logs. The equation that is obtained is


\text{log }\frac{k}{\text{dm}^{\text{3}}\text{ mol}^{-\text{1}}\text{ s}^{-\text{1}}}=\left( \frac{-E^{\ddagger }}{\text{2}\text{.303}R} \right)\text{ }\left( \frac{\text{1}}{T} \right)\text{ + log }\frac{k'}{\text{dm}^{\text{3}}\text{ mol}^{-\text{1}}\text{ s}^{-\text{1}}}


Again E can be obtained from the slope, provided a straight-line plot is obtained. An example of such a plot for the reaction


2HI → H2 + I2


is shown in Fig. 1. From its slope we can determine that E = 186 kJ mol–1. This means of determining activation energy was first suggested in 1889 by Svante Arrhenius and is therefore called an Arrhenius plot.


Figure 1 Arrhenius plot for the reaction 2HI → H2 + I2. From the slope of the graph an activation energy of 186 kJ mol–1 may be calculated.


It should be pointed out that the equations just derived for Arrhenius plots apply strictly to unimolecular and bimolecular elementary processes. If a reaction mechanismThe set of elementary steps by which a reaction is thought to occur; each individual step corresponds to either a bimolecular collision or unimolecular reaction of a single molecule. involves several steps, there is no guarantee that the same elementary process will be rate limiting at several widely different temperatures. The observed rate constant may also be a productA substance produced by a chemical reaction. of several constants, as in the H2 + I2 reaction (Equation 5 in the section on Reaction Mechanisms). If either a different rate-limiting stepThe step in a reaction mechanism that by its relatively slow rate limits the overall rate of a reaction; also called rate-determining step. or the product of constants has a different temperature dependence, then an Arrhenius plot may not be linear. This also makes it dangerous to use such a plot to predict reaction rates at temperatures quite different from the conditions under which experimental measurements were made.


The slope and y-intercept of an Arrhenius plot can be used to determine the values for the Arrhenius EquationAn equation that expresses the logarithmic relationship between the rate constant of a reaction and the reciprocal of the absolute temperature; ln k = exp(Ea/RT)..

k = Ae-Ea/RT

where A is called the frequencyThe rate at which a periodic event occurs; specifically, the rate at which the waves of electromagnetic radiation pass a point. factor, Ea is the activation energy, R is the gas law constant and T is the temperature in KelvinA unit of temperature equal to 1/273.16 of the thermodynamic temperature of the triple point of water; the kelvin is the same size as the degree Celsius. The thermodynamic temperature scale (Kelvin scale) has absolute zero as its zero point.. The frequency factor value depends on how often molecules collide and how orientation of the molecules is related to the reaction. A benefit of the Arrhenius equation is that it gives a direct method for computing the dependence of the rate constant, k, on temperature.