# Question #68ca2

##### 1 Answer

#### Answer:

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 **beginning of the reaction**, we can say that after a time

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

**consumed** in a period of time

This means that you can write

#(dx)/dt prop A_0 - x#

The rate of formation of the product,

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

#(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

To get rid of **before** the reaction takes place, you have

This means that you have

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

Plug this into equation

#-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(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*, **after the reaction is complete**, will depend on the initial amount of reactant present.

#V_(oo) prop A_0#

Likewise, the *volume at time* **produced** as a given time time

#V_(t) prop x#

And so you can thus say that

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