Question

Question #e6605

Answer 1

You can do it like this:

Explanation:

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:

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.