4.6: Collision Theory

Activation Energy and the Arrhenius Equation

The minimum energy necessary to form a product during a collision between reactants is called the activation energy ([latex]E_a[/latex]). The kinetic energy of reactant molecules plays an important role in a reaction because the energy necessary to form a product is provided by a collision of a reactant molecule with another reactant molecule. (In single-reactant reactions, activation energy may be provided by a collision of the reactant molecule with the wall of the reaction vessel or with molecules of an inert contaminant.) If the activation energy is much larger than the average kinetic energy of the molecules, the reaction will occur slowly: Only a few fast-moving molecules will have enough energy to react. If the activation energy is much smaller than the average kinetic energy of the molecules, the fraction of molecules possessing the necessary kinetic energy will be large; most collisions between molecules will result in reaction, and the reaction will occur rapidly.

Figure 4.6.3 shows the energy relationships for the general reaction of a molecule of [latex]A[/latex] with a molecule of [latex]B[/latex] to form molecules of [latex]C[/latex] and [latex]D[/latex]:

[latex]A+B⟶C+D \nonumber[/latex]

The figure shows that the energy of the transition state is higher than that of the reactants [latex]A[/latex] and [latex]B[/latex] by an amount equal to [latex]E_a[/latex], the activation energy. Thus, the sum of the kinetic energies of [latex]A[/latex] and [latex]B[/latex] must be equal to or greater than E_a to reach the transition state. After the transition state has been reached, and as [latex]C[/latex] and [latex]D[/latex] begin to form, the system loses energy until its total energy is lower than that of the initial mixture. This lost energy is transferred to other molecules, giving them enough energy to reach the transition state. The forward reaction (that between molecules [latex]A[/latex] and [latex]B[/latex]) therefore tends to take place readily once the reaction has started. In Figure 4.6.3, [latex]ΔH[/latex] represents the difference in enthalpy between the reactants ([latex]A[/latex] and [latex]B[/latex]) and the products ([latex]C[/latex] and [latex]D[/latex]). The sum of [latex]E_a[/latex] and [latex]ΔH[/latex] represents the activation energy for the reverse reaction:

[latex]C+D⟶A+B \nonumber[/latex]

We can use the Arrhenius equation to relate the activation energy and the rate constant, k, of a given reaction:

[latex]k=Ae^{−E_a/RT} \label{Arrhenius}[/latex]

In this equation,

• [latex]R[/latex] is the ideal gas constant, which has a value 8.314 J/mol/K,
• [latex]T[/latex] is temperature on the Kelvin scale,
• [latex]E_a[/latex] is the activation energy in joules per mole,
• [latex]e[/latex] is the constant 2.7183, and
• [latex]A[/latex] is a constant called the frequency factor, which is related to the frequency of collisions and the orientation of the reacting molecules.

Both postulates of the collision theory of reaction rates are accommodated in the Arrhenius equation. The frequency factor A is related to the rate at which collisions having the correct orientation occur. The exponential term, [latex]e^{−E_a/RT}[/latex], is related to the fraction of collisions providing adequate energy to overcome the activation barrier of the reaction.

At one extreme, the system does not contain enough energy for collisions to overcome the activation barrier. In such cases, no reaction occurs. At the other extreme, the system has so much energy that every collision with the correct orientation can overcome the activation barrier, causing the reaction to proceed. In such cases, the reaction is nearly instantaneous.

The Arrhenius equation (Equation \ref{Arrhenius}) describes quantitatively much of what we have already discussed about reaction rates. For two reactions at the same temperature, the reaction with the higher activation energy has the lower rate constant and the slower rate. The larger value of [latex]E_a[/latex] results in a smaller value for [latex]e^{−E_a/RT}[/latex], reflecting the smaller fraction of molecules with enough energy to react. Alternatively, the reaction with the smaller [latex]E_a[/latex] has a larger fraction of molecules with enough energy to react. This will be reflected as a larger value of [latex]e^{−E_a/RT}[/latex], a larger rate constant, and a faster rate for the reaction.

An increase in temperature has the same effect as a decrease in activation energy. A larger fraction of molecules has the necessary energy to react (Figure 4.6.4), as indicated by an increase in the value of [latex]e^{−E_a/RT}[/latex]. The rate constant is also directly proportional to the frequency factor, [latex]A[/latex]. Hence a change in conditions or reactants that increases the number of collisions with a favorable orientation for reaction results in an increase in [latex]A[/latex] and, consequently, an increase in [latex]k[/latex].

A convenient approach to determining [latex]E_a[/latex] for a reaction involves the measurement of [latex]k[/latex] at different temperatures and using of an alternate version of the Arrhenius equation that takes the form of linear equation:

[latex]\begin{align*}
\ln k&=\left(\dfrac{−E_a}{R}\right)\left(\dfrac{1}{T}\right)+\ln A\\
y&=mx+b
\end{align*} \nonumber[/latex]

Thus, a plot of [latex]\ln k[/latex] versus [latex]\dfrac{1}{T}[/latex] gives a straight line with the slope [latex]\dfrac{-E_\ce{a}}{R}[/latex], from which E_a may be determined. The intercept gives the value of [latex]\ln A[/latex]. This is sometimes call an Arrhenius Plot.

Example 4.6.1

Determination of E_a The variation of the rate constant with temperature for the decomposition of HI(g) to H₂(g) and I₂(g) is given here. What is the activation energy for the reaction?

[latex]\ce{2HI}(g)⟶\ce{H2}(g)+\ce{I2}(g) \nonumber[/latex]

variation of the rate constant with temperature for the decomposition of HI(g) to H2(g) and I2(g)

T (K)	k (L/mol/s)
555	3.52 × 10−7
575	1.22 × 10−6
645	8.59 × 10−5
700	1.16 × 10−3
781	3.95 × 10−2

Solution

Values of [latex]\dfrac{1}{T}[/latex] and ln k are:

Solutions to Example 12.5.1

[latex]\mathrm{\dfrac{1}{T}\:(K^{−1})}[/latex]	ln k
1.80 × 10−3	−14.860
1.74 × 10−3	−13.617
1.55 × 10−3	−9.362
1.43 × 10−3	−6.759
1.28 × 10−3	−3.231

Figure 4.6.5 is a graph of ln k versus [latex]\dfrac{1}{T}[/latex]. To determine the slope of the line, we need two values of ln k, which are determined from the line at two values of [latex]\dfrac{1}{T}[/latex] (one near each end of the line is preferable). For example, the value of ln k determined from the line when [latex]\dfrac{1}{T}=1.25×10^{−3}[/latex] is −2.593; the value when [latex]\dfrac{1}{T}=1.78×10^{−3}[/latex] is −14.447.

The slope of this line is given by the following expression:

[latex]\begin{align*}
\ce{Slope}&=\dfrac{Δ(\ln k)}{Δ\left(\dfrac{1}{T}\right)}\\
&=\mathrm{\dfrac{(−14.447)−(−2.593)}{(1.78×10^{−3}\:K^{−1})−(1.25×10^{−3}\:K^{−1})}}\\
&=\mathrm{\dfrac{−11.854}{0.53×10^{−3}\:K^{−1}}=2.2×10^4\:K}\\
&=−\dfrac{E_\ce{a}}{R}
\end{align*} \nonumber[/latex]

Thus:

[latex]\begin{align*} E_\ce{a} &=\mathrm{−slope×\mathit R=−(−2.2×10^4\:K×8.314\: J\: mol^{−1}\:K^{−1})} \[latex]4pt] &=\mathrm{1.8×10^5\:J\: mol^{−1}} \end{align*} \nonumber[/latex]

In many situations, it is possible to obtain a reasonable estimate of the activation energy without going through the entire process of constructing the Arrhenius plot. The Arrhenius equation:

[latex]\ln k=\left(\dfrac{−E_\ce{a}}{R}\right)\left(\dfrac{1}{T}\right)+\ln A \nonumber[/latex]

can be rearranged as shown to give:

[latex]\dfrac{Δ(\ln k)}{Δ\left(\dfrac{1}{T}\right)}=−\dfrac{E_\ce{a}}{R} \nonumber[/latex]

or

[latex]\ln\dfrac{k_1}{k_2}=\dfrac{E_\ce{a}}{R}\left(\dfrac{1}{T_2}−\dfrac{1}{T_1}\right) \nonumber[/latex]

This equation can be rearranged to give a one-step calculation to obtain an estimate for the activation energy:

[latex]E_\ce{a}=−R\left( \dfrac{\ln k_2−\ln k_1}{\left(\dfrac{1}{T_2}\right)−\left(\dfrac{1}{T_1}\right)}\right ) \nonumber[/latex]

Using the experimental data presented here, we can simply select two data entries. For this example, we select the first entry and the last entry:

First and Last Entry

T (K)	k (L/mol/s)	[latex]\dfrac{1}{T}\:(K^{-1})[/latex]	ln k
555	3.52 × 10−7	1.80 × 10−3	−14.860
781	3.95 × 10−2	1.28 × 10−3	−3.231

After calculating [latex]\dfrac{1}{T}[/latex] and ln k, we can substitute into the equation:

[latex]E_\ce{a}=\mathrm{−8.314\:J\:mol^{−1}\:K^{−1}\left(\dfrac{−3.231−(−14.860)}{1.28×10^{−3}\:K^{−1}−1.80×10^{−3}\:K^{−1}}\right)} \nonumber[/latex]

and the result is E_a = 185,900 J/mol.

This method is very effective, especially when a limited number of temperature-dependent rate constants are available for the reaction of interest.

Summary

Chemical reactions require collisions between reactant species. These reactant collisions must be of proper orientation and sufficient energy in order to result in product formation. Collision theory provides a simple but effective explanation for the effect of many experimental parameters on reaction rates. The Arrhenius equation describes the relation between a reaction’s rate constant and its activation energy, temperature, and dependence on collision orientation.