Question

Question #68ca2

Answer 1

Here's what's going on here.

Explanation:

That equation represents a variation of the integrated rate law for a first-order reaction.

For a generic first-order reaction

`color(blue)("A " -> " products")`

you know that the rate of the reaction depends linearly on the concentration of the reactant

`"rate" prop ["A"]`

Now, you can express the rate of the reaction in terms of the rate of change in the concentration of the reactant with time. If we take `A_0` to be the concentration of `"A"` at the beginning of the reaction, we can say that after a time `t`, the concentration of `"A"` will be

`["A"]_t = A_0 - x" "`, where

`x` - how much of the reactant is consumed in a period of time `t`

This means that you can write

`(dx)/dt prop A_0 - x`

The rate of formation of the product, `(dx)/(dt)`, is proportional to how much reactant remains unconsumed.

You can write the rate law for this reaction, which establishes a relationship between the concentration of the reactant and the rate of the reaction, by introducing a rate constant `k`

`(dx)/dt = k * (A_0 - x)`

To find a relationship between the concentration of the reactant and the time of the reaction, you need to integrate this equation. Rearrange to get

`(dx)/(A_0 - x) = k * dt`

`int 1/(A_0 - x) dx = k * int(dt)`

This will be equal to

`-ln(A_0 -x ) = k * t + c" " " "color(red)("(*)")`

Here `c` is an integration constant.

To get rid of `c`, use the fact that at `t=0`, i.e. before the reaction takes place, you have `["A"] = A_0` and `t=0`.

This means that you have

`-ln(A_0 - 0) = k * 0 + c implies c = - ln(A_0)`

Plug this into equation `color(red)("(*)")` to get

`-ln(A_0 - x) = k * t -ln(A_0)`

Rearrange and use the property of logs to find

`k * t = ln(A_0) - ln(A_0 - x)`

`k * t = ln(A_0/(A_0 - x))`

Now, use the conversion factor between the natural log, `"ln"`, and the common log, `"log"`

`ln(x) = 2.303 * log(x)`

to write

`color(blue)(k = 2.303/t * log(A_0/(A_0 - x)))`

The idea now is that you can express concentration as volume.

For a reaction that features two gases, you can say that the volume at infinity, `V_(oo)`, which is simply the volume of the gaseous product collected at a time after the reaction is complete, will depend on the initial amount of reactant present.

`V_(oo) prop A_0`

Likewise, the volume at time `t`, `V_(t)`, will depend on how much product was produced as a given time time `t`

`V_(t) prop x`

And so you can thus say that

`color(green)(k = 2.303/t * log(V_(oo)/(V_(oo) - V_t)))`