# Maths in action - hard question, please help

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• Mar 15th 2010, 07:06 PM
AJShaw
Maths in action - hard question, please help
I have no idea. Can you show me the answer step by step, please.

The Arrhenius equation is given by

k = A exp (-E/RT),

where k is the rate constant, A is called the pre-exponential factor, E is called the activation energy, R is the ideal gas constant and T is the temperature , measured in Kelvins.

By taking logs and simplifying show that this equation can be written in the form
1n k = m/T + n,

identifying the value of the parameters m and n.
• Mar 15th 2010, 07:17 PM
tonio
Quote:

Originally Posted by AJShaw
I have no idea. Can you show me the answer step by step, please.

The Arrhenius equation is given by

k = A exp (-E/RT),

where k is the rate constant, A is called the pre-exponential factor, E is called the activation energy, R is the ideal gas constant and T is the temperature , measured in Kelvins.

By taking logs and simplifying show that this equation can be written in the form
1n k = m/T + n,

identifying the value of the parameters m and n.

$\ln(Ae^{-E\slash RT}) = \ln A - \frac{E}{RT}$

Tonio
• Mar 15th 2010, 07:19 PM
AJShaw
Could you elaborate more, please. I really don't get it.
• Mar 15th 2010, 07:20 PM
sa-ri-ga-ma
k = A exp (-E/RT),
ln(k) = ln(A) + (-E/R)*1/T
Now can you find the values of m and n?