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The collision model of reaction rate assumes that the rate constant

k = p*Z*e-Ea/RT
where Z is the collision rate, p is the steric factor, Ea is the activation energy, R is the gas constant and T is the temperature. If we consider this equation in terms of changing temperature, the steric factor clearly doesn't depend on temperature. Z turns out to be only weakly dependant on temperature: changing T from 500 to 600 K changes Z by less than 10%.

It is therefore a reasonable approximation to assume that the p*Z part of the above equation is a constant, and we can write

k = A*e-Ea/RT
or
ln(k) = ln(A) - Ea/RT
where A is a constant. This is known as the Arrhenius equation

Comparing the latter form of the Arrhenius equation to the equation for a straight line, y = mx + b, it is obvious that if we plot ln(k) vs. 1/T, we will get a plot where the slope is -Ea/R and the intercept in ln(A).

We can also come up with a form of the Arrhenius equation that relates the rate constant at two different temperatures in a way very similar to the Clausius-Clapeyron equation. Simply set the equation up at two different temperatures and subtract the second from the first

ln(k2) = ln(A) -Ea/RT2 -
ln(k1) = ln(A) -Ea/RT1 =
ln(k2) -ln(k1) = -Ea/R(1/T2 - 1/T1)

Example: For a given reaction, the rate constant goes up by a factor of 1.65 when you increase the temperature from 200C to 40oC. What is the activation energy for this reaction?

Solution: Use the form of the Arrhenius equation that relates the rate constant at two different temperatures.

ln(k2/k1)) = -Ea/R(1/T2 - 1/T1)
The ration of k2/k1 is 1.65, and we must remember to convert the temperatures into kelvin
ln(1.65) = -Ea/8.31*10-3 kJ/molK*(1/313 K - 1/293)
0.501 = -Ea/8.31*10-3 kJ/molK*(-2.18*10-4)
Ea = 19.1 kJ/mol


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