logarithm of concentration

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March 13th 2008, 03:32 PM
DavePercy

logarithm of concentration

When dealing with reaction rates for first order reactions, my textbook has this formula:

$\ln [A_f] = -kt + \ln [A_0]$

where $[A_f],[A_0]$ are the final and initial concentrations or pressures of a reactant, and k is a constant with units of 1/sec. This is derived from

$\frac{[A_f]}{[A_0]} = e^{-kt}$

Is there any meaning in asking what the logarithm of a unit of concentration or pressure is? In this case it looks like the inputs to ln are actually dimensionless since the units cancel in the second equation above. In general though, what does a logarithm or exponential function output when given a unit, like $\ln(20 mi/hr)$ , $e^{12 Joules}$ , or $\ln(5 ArbitraryUnits)$ ? Do we have to leave it as something like a real number plus ln(mi) - ln(hr) ?
March 13th 2008, 04:37 PM
o_O

Yes, generally when dealing with logarithms, they deal with unitless operations (exception is when looking at pH where $pH = -log[H^{+}]$ . The connotations in doing these operations are implied).

Looking at first-order reactions, you know that the equilibrium constant k has a unit of $s^{-1}$ . So, rearranging as you did:

$\ln\left(\frac{[A]_{t}}{[A]_{0}}\right) = -kt$

Inside the operation of the logarithm, the units cancel out and on the right hand side, the units cancel out as well (given that time is in seconds).
March 13th 2008, 05:15 PM
DavePercy

Is there any situation where real-world units are passed to logarithms or exponents?