How to derive the following formula? [Pressure in relation to kinetics]

The equation given to you looks like this

#color(blue)(|bar(ul(color(white)(a/a)k = 2.303/tlog[ (P_0(n-1))/(nP_0-P_t)]color(white)(a/a)|)))#

To make things more general, I'll use concentration instead of pressure. Keep in mind that when volume and temperature are kept constant, concentration is proportional to pressure.

So, you're dealing with a first-order reaction

#"A"_ ((g)) -> color(purple)(n)"B"_ ((g))#

You know that the rate of the reaction is *proportional to the concentration of the reactant, #["A"]#.

You can express the rate of the reaction in terms of the change in the concentration of the reactants with time. If you take #["A"_0]# to be the initial concentration of #"A"#, you can say that you have

#"rate" prop ["A"_0] - x#

Here #x# represents the amount of #"A"# consumed in a period of time #t#

You can thus use a rate constant #k# to write

#(dx)/(dt) = k * (["A"_0] - x)#

Rearrange to get

#dx/(["A"_0] - x) = k * dt#

You need to find the relationship that exists between the concentration of #"A"# and the time of the reaction, so integrate both sides to get

#dx/(["A"_0] - x) = k * dt " "| int#

#int dx/(["A"_0] - x) = k int dt#

This will get you

#-ln(["A"]_0 - x) = k * t + c" " " "color(orange)(("*"))#

Here #c# is simply an integration constant. You can determine the value of #c# by using the fact that at #t=0# you have #x=0#

#-ln(["A"_0]) = k * 0 + c#

#c = - ln(["A"_0])#

Plug this back into equation #color(orange)(("*"))# to get

#-ln(["A"_0] - x) = k * t - ln(["A"_0])#

Rearrange to get

#ln((["A"_0])/(["A"_0] - x)) = k * t#

Use the fact that

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

and isolate #k# on one side of the equation to get

#color(blue)(|bar(ul(color(white)(a/a)k = 2.303/t log( (["A"_0])/(["A"_0] - x))color(white)(a/a)|)))#

Now for the interesting part.

You must think pressure and moles here. Pressure and concentration are directly proportional when temperature and volume are kept constant because pressure and number of moles are proportional under those conditions.

#P/n = "constant" -># when temperature and volume are kept constant

If you use pressure instead of concentration, you will have

#k = 2.303/t log( P_0/(P_0 - x))" " " "color(orange)(("* *"))#

Here's how I think the equation given to you came about. If you take

#P_0# - the initial pressure in the reaction vessel, i.e. the pressure before the reaction begins

#P_t# - the pressure inside the reaction vessel at a given time #t#

#P_f# - the pressure inside the reaction vessel after the reaction is complete

You can say that

#x prop P_t - P_0 -># the change in pressure at a given time #t# is proportional to the difference between the initial pressure and the pressure at #t#

This is equivalent to saying that the change in the number of moles of #"A"# is proportional to the difference between the total number of moles present in the reaction vessel at time #t# and the initial number of moles of #"A"#.

#P_0 prop P_f - P_0 -># the initial pressure is proportional to the difference between the pressure when the reaction is complete and the initial pressure

This is equivalent to saying that the number of moles of #"A"# that remain in the vessel after the reaction is complete will be proportional to the initial number of moles of #"A"#.

This means that

#(P_0 - x) prop (P_f - P_0) - (P_t - P_0)#

#(P_0 - x) prop P_f - color(red)(cancel(color(black)(P_0))) - P_t + color(red)(cancel(color(black)(P_0)))#

#(P_0 - x) prop P_f - P_t#

Plug this into equation #color(orange)(("* *"))# to get

#k = 2.303/t log( (P_f - P_0)/(P_f - P_t))#

But since

#P_f = nP_0#

you will end up with

#k = 2.303/t log( (nP_0 - P_0)/(nP_0 - P_t))#

which is of course equal to

#color(blue)(|bar(ul(color(white)(a/a)k = 2.303/tlog[ (P_0(n-1))/(nP_0-P_t)]color(white)(a/a)|)))#

#color(white)()#
SIDE NOTE

This equation would also make sense if pressure and temperature were kept constant. In that case, the concentration would be proportional to volume.

For the starting equation

#color(blue)(|bar(ul(color(white)(a/a)k = 2.303/t log( (["A"_0])/(["A"_0] - x))color(white)(a/a)|)))#

you'd now have

#["A"_ 0] prop V_ (oo) - V_0#

#x prop V_t - V_0#

This would get you

#("A"_ 0 - x) prop V_ (oo) - V_t#

and so the equation would look like this

#k = 2.303/t log( (V_(oo) - V_0)/(V_(oo) - V_t))#

This is the equation given to you in your Chemical Kinetics course

http://www.etoosindia.com/sites/default/files/studymaterials/IIT-JEE-Main-Advanced-Physical-Chemistry-12th-Chemical-Kinetics.pdf