Question #68ca2

1 Answer
Feb 7, 2016

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 #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)))#