\( aA + bB \rightarrow cC + dD \)

\[-r_A = kC_A^{\alpha}C_B^{\beta}\]

\(\alpha \text{ order in A}\)

\(\beta \text{ order in B}\)

\(\text{Overall Reaction Order} = \alpha + \beta\)

The following table contains rate laws that were found experimentally.

Note: The rate constant, \(k\), and activation energies for a number of the reactions in these examples are given in the Data Base on the CD-ROM and Summary Notes.

k is the specific reaction rate (constant) and is given by the Arrhenius Equation:

\(T \to \infty\) | \(k \to A\) | \(T \to 0\) | \(k \to 0\) | \(A \approx 10^{13}\)

\(\mathbf{k = A e^{-\frac{E}{RT}}}\)

Where:

E = activation energy (cal/mol)
R = gas constant (cal/mol*K)
T = temperature (K)
A = frequency factor (units of A, and k, depend on overall reaction order)

\(\ln k = \ln A - \frac{E}{R} \left(\frac{1}{T}\right)\)

The larger the activation energy, the more temperature sensitive k and thus the reaction rate.

\(k(T_2) = k(T_1) \exp\left[\frac{E}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right]\)

Why is there an Activation Energy?

1. the molecules need energy to distort or stretch their bonds in order to break them and to thus form new bonds

2. as the reacting molecules come close together they must overcome both steric and electron repulsion forces in order to react

In our development of collision theory we assumed all molecules had the same average energy. However, all the molecules don’t have the same energy, rather there is distribution of energies where some molecules have more energy than others. The distraction function f(E,T) describes this distribution of the energies of the molecules. The distribution function is read in conjunction with dE

One such distribution of energies is in the following figure.

By increasing the temperature we increase the kinetic energy of the reactant molecules which can in turn be transferred to internal energy to increase the stretching and bending of the bonds causing them to be in an activated state, vulnerable to bond breaking and reaction.

We see that as the temperature is increased we have greater number of molecules have energies E A or greater and hence the reaction rate will be greater.

You can tell the overall reaction order by the units of k

The activation energy is a "measure" of the energy a that the reacting molecules must have in order for the reaction to occur.

The reaction of AB + C to form A + BC is shown above along the reaction coordinate. One way to think of the reaction coordinate is the linear distance between the AB molecule for a fixed linear distance between the AC molecule. At the start of the reaction the AB distance is small and the BC distance is large. As the reaction proceeds, the A–C distance remains fixed but B moves away for A and closer to C and the energy of system increases. At the top of the barrier molecule B is equal distance from A and C. But as it crosses the barrier it moved close to C to form the BC molecule and the A molecule alone.

A reaction follows an elementary rate law if and only if the (iff) stoichiometric coefficients are the same as the individual reaction order of each species. For the reaction in the previous example \(\left( A + B \rightarrow C + D \right)\), the rate law would be: \(-r_A = kC_A C_B\)

Example: If the reaction is \(2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2\), then elementary the rate would be \(-r_{\text{NO}} = k_{\text{NO}} (C_{\text{NO}})^2 C_{\text{O}_2}\)

Example: If the rate law for the non-elementary reaction \(\left( A + B \rightarrow C + D \right)\) is found to be \(-r_A = kC_A^2C_B\)

Then the reaction is said to be 2nd order in A, 1st order in B, and 3rd order overall. 

The net rate of formation of any species is equal to its rate of formation in the forward reaction plus its rate of formation in the reverse reaction:

\(\text{rate}_{\text{net}} = \text{rate}_{\text{forward}} + \text{rate}_{\text{reverse}}\)

At equilibrium, \(\text{rate}_{\text{net}} \stackrel{\text{eq}}{\approx} 0\) and the rate law must reduce to an equation that is thermodynamically consistent with the equilibrium constant for the reaction.

Example: Consider the exothermic, heterogeneous reaction \(A + B \rightarrow C\)

At low temperature, the rate law for the disappearance of A is

\(-r'_A = \frac{k_A P_A P_B}{1 + K_A P_A}\) (Recall \(P_A = C_A R T\))

At high temperature, the exothermic reaction is significantly reversible: \(\text{A} + \text{B} \rightleftharpoons \text{C}\)

What is the corresponding rate law? Let's see.

If the rate of formation of A for the forward reaction (\(\text{A} + \text{B} \longrightarrow \text{C}\)) is

\[r'_{A\text{for}} = \frac{-k_A P_A P_B}{1 + K_A P_A}\]

then we need to assume a form of the rate law for the reverse reaction that satisfies the equilibrium condition.

If we assume the rate law for the reverse reaction (\(\text{C} \rightarrow \text{A} + \text{B}\)) is:

\[r'_{{A,\text{rev}}} = \frac{k_{-A} P_C}{1 + K_A P_A}\]

then: \[r_A = r_{A,\text{net}} = r_{A,\text{for}} + r_{A,\text{rev}}\]

and: \[-r'_A = \frac{k_A}{1 + K_A P_A} \left(P_A P_B - \frac{P_C}{K_P}\right)\]

Does this rate law satisfy our requirement that, at equilibrium, it must reduce to an equation that is thermodynamically consistent with K_P? Let's see.

From Appendix C we know that for a reaction at equilibrium:

\[K_P = \frac{P_{Ce}}{P_{Ae} P_{Be}}\]

At equilibrium, \(r_{\text{net}} \cong 0\), so: \[-r_A \cong 0 = \frac{k_A}{1 + K_A P_{Ae}} \left( P_{Ae} P_{Be} - \frac{P_{Ce}}{K_P} \right)\]

Solving for K_P gives:

\[K_P = \frac{P_{Ce}}{P_{Ae} P_{Be}}\]

^* All chapter references are for the 1st Edition of the text Essentials of Chemical Reaction Engineering .

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