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Activation Energy

>> Parts of this equation/concept include:

A -	Energy Profiles
B -	Arrhenius Equation, Equation 14.21, and Equation 14.22

Energy profiles describe energy changes over the course of a reaction. Energy (as H or G) is graphed on the y-axis; reaction progress (reactants to products) is graphed on the x-axis.

Reaction profiles are shown in several figures in the text (Figure 14.17, for example). Key features of reaction profiles are the hills. There is one hill for each step in the mechanism. The leftmost point represents the potential energy of the reactants. The top of the hill is the transition state. (If there is more than one hill, there is more than one transition state.) The energy difference between the reactants and the top of the hill is the activation energy (E_a).

The rightmost point is the potential energy of the products. The energy difference between reactants and products is the enthalpy (H) or the free energy (G) of the reaction. If the energy of the reactants is higher than the energy of the products, the reaction is exothermic (H) or exergonic (G). If the energy of the reactants is lower than the energy of the products, the reaction is endothermic (H) or endergonic (G).

If there is more than one hill, the slow step is the one with the highest hill. Any valley between the hills represents the energy of the intermediate. The transition state is the energy of the activated complex.

Raising the temperature increases the rate of reaction because the added kinetic energy allows a larger fraction of reactants to "go over" the hill. Adding a catalyst actually lowers the activation energy (size of the hill).

>> Example 1

Label the following energy profile with the H, E_a, and transition state.

Solution:

The H is the energy difference between products and reactants. E_a is the energy difference between reactants and transition state. The transition state is the highest energy between reactants and products.

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The Arrhenius equation relates temperature, rate constant, and activation energy. Solving these equations requires measuring the rate constant at more than one temperature. Activation energy can be determined from temperature (T) and rate constant (k) by graphing ln k versus 1/T. The slope is equal to –E_a/R, where R is the gas constant. If the gas constant value of 8.314 J/mol•K is used, the units of activation energy will be in joules.

In this chapter both time (t) and temperature (T) are relevant. To keep the two variables from being mixed up, a lowercase t is used to represent time and an uppercase T for temperature. This is standard practice in chemistry (i.e., that uppercase and lowercase symbols have different meanings). If you carefully use the same conventions, you are less likely to make errors or be misunderstood.

>> Example 2

What is the activation energy for a reaction with the following temperature and rate constant data?

Temperature (K)	Rate constant (1/s)
200	0.236
300	0.301
400	0.340
500	0.366
600	0.385
700	0.399
800	0.409

Solution:

The linear relationship between rate constant and temperature is Equation 14.22,

ln k

=

–Ea R

1 T

+

ln A

Therefore a line would be a graph of ln(rate constant) on the y-axis and 1/temperature on the x-axis. The data for the graph is

Temperature (K)	Rate constant (1/s)	1/T	ln k
200	0.236	0.00500	–1.445
300	0.301	0.00333	–1.200
400	0.340	0.00250	–1.078
500	0.366	0.00200	–1.004
600	0.385	0.00167	–0.955
700	0.399	0.00143	–0.920
800	0.409	0.00125	–0.894

The graph is shown in Figure 14.6. The equation of the line is

ln k

=

–147

1 T

–

0.71

The activation energy is determined from the slope of this line, –147. The relationship between slope and activation energy is

slope

=

–Ea R

When R is the gas constant in units of J/mol•K, the activation energy is in units of joules. So

–147

=

–Ea 8.314

1222 J = E_a

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