Logarithms and the rate equation

[Iota]

I've worked out a rate equation:

 

rate=k'[A]^a

 

the nice and easy bit.

 

Now I have to take the logs of both sides of the equation. The thing I'm confused about is that I have no values, it just wants me to show the equation with the logs of both sides taken, so would that literally be shown just as:

 

ln rate= ln k'[A]^a ?

I'm not sure how to do that or if the answer would be as basic as writing log next to both sides?

[John Cuthber]

What do you know about ln(a^b)?

[Iota]

Is it where: a to the power of x = b?

 

Otherwise I'm afraid I do not.

[John Cuthber]

http://en.wikipedia.org/wiki/List_of_logarithmic_identities#Using_simpler_operations

[Iota]

Thanks, I need to improve my maths skills, a point in the right direction is always good.

Right, would it simply be this:

log rate= log k' + log [A]^a

I'm unsure what to do with [A]^a, if anything at all.

Edited February 16, 2013 by Iota

[John Cuthber]

It's slightly confusing that they use a twice

If the expression was

....+ ln ([a]^b)

what would you do to it?

[Iota]

I was thinking it would possibly be this:

 

log rate= log k' + b log([A])

[John Cuthber]

You could look at the experimental data and plot the log of the rate versus the log of the concentration.

Looks like a straight line to me, and physical chemists like straight lines on their graphs.

From the slope of the line you can calculate a.

[Iota]

I'll go do that, thanks for your help.

[mississippichem]

and physical chemists like straight lines on their graphs.

From the slope of the line you can calculate a.

Nothing more true has ever been posted here.

[John Cuthber]

Indeed, and they are quite fond of taking logs of things.

I suspect that the chocolate log was invented by a slightly muddled physical chemistry professor.