Notes on Kinetics
Temperature Effects

How Are Reaction Rates Affected By Temperature?

The rates of most reactions increase with a rise in temperature.

example

• the use of a pressure cooker to cook foods faster
• the use of a refrigerator to store food and slow its spoilage

The general rule is than an increase in temperature by 10 oC doubles the reaction rate.

The Kinetic Theory can be used to explain the effect of temperature on reaction rates.

Raising the temperature increases the fraction of molecules having very high kinetic energies. These are the ones most likely to react when they collide. The higher the temperature, the larger the fraction of molecules that can provide the activation energy needed for reaction.
The rate constant, k, becomes larger as the temperature increases.

What is the relation between k and the temperature, T?

f = e-Ea/RT, where f is the fraction of molecules having an energy equal to or greater than Ea.
• e is the base of natural logs
• R is the gas constant
• T is the temperature in K

If we assume that the rate constant, k, is directly proportional to "f", then,
k = cf = ce-E/RT, where c is a proportionality constant.

Now, if we do a bit on math on this equation we get:

ln k = ln c - Ea/RT

log10 k = log10 c - Ea/2.30 RT

Next, substitute R = 8.31 J/mol.K and let log10 c = A

log10 k = A - Ea / (2.30)(8.31)T

The equation in this form is called the Arrhenius Equation.

If you look at it, you will notice that it is in the form y = a + bx, with y being log10 k and x = 1/T.

A plot of log10 k vs 1/T should be a straight line. Ea for a reaction can be obtained from the slope of this line.

slope = -Ea / (2.30)(8.31)

The value of k at a particular temperature can be calculated if it is known for some other temperature

example:

The Ea for a reaction is 9.32 x 104 J. At 27 oC, k = 1.25 x 10-2 L/mol.s. Calculate the value of k at 127 oC.

Solution: Use this form of the Arrhenius Equation  log10 (k2 / k1) = (Ea / (2.30)(8.31))(T2 - T1) / T2T1 This form is sometimes called the "Two-Point" Equation

Solve for the value of log10 (k2 / k1:

log10 (k2 / k1) = (9.32 x 104 / (2.30)(8.31))(400 - 300)/(400)(300) = 4.06
log10 k2 - log10 1.25 x 10-2 = 4.06
log10 k2 - (-1.90) = 4.06
log10 k2 = 2.16
The antilog of 2.16 = 1.44 x 102 L /mol.s = k2

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 Send questions, comments or suggestions to Gwen Sibert, at the Roanoke Valley Governor's School gsibert@rvgs.k12.va.us Back to Notes Menu