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Answer 1 · 2017-03-25T18:10:32

You can do it like this:

I will confine the answer to a 1st order reaction.

Suppose we have:

#sf(Ararr"products")#

For a 1st order reaction we have:

#sf("Rate"=k[A]^1)#

Where k is the rate constant.

This can be expressed in terms of the rate of disappearance of A :

#sf(-(d[A])/(dt)=k.dt)#

Rearranging and applying integration between 0 and t gives:

#sf(int_([A]_0)^([A]_t)(d[A])/([A])=-kint_0^tdt)#

This gives:

#sf(ln[A]_(t)-ln[A]_0=-kt)#

#:.##sf(ln[A]_t=ln[A]_0-kt)#

You can see that this is a straight line graph of the form #sf(y=mx+c)#.

This means if we plot #sf(ln[A]_t)# against t the gradient will be equal to -k and the intercept will be #sf(ln[A]_0)#.

Here is an example for the breakdown of ozone by light:

i.ytimg.com

See if you can measure the gradient yourself to give you -k.

If you read off the intercept i.e the point where the graph cuts the y axis, this will give you the value of #sf(ln[A]_0)#.

You can get #sf([A]_0)# from that by selecting the #sf(e^x)# function on your calculator.