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

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Let this be a #1^"st"# order reaction:

#"A"(g) -> n"B"(g)#

#k = 2.303/t log\frac(P_0(n-1))(nP_0 - P_t)#

where the final total pressure as a function of time is #P_f = nP_0# .

The formula is not applicable when #n = 1# . #n# can be fractional.

Let this be a

#"A"(g) -> n"B"(g)#

where the final total pressure as a function of time is

The formula is not applicable when

##### 1 Answer

#### Answer:

Here's my best shot at deriving this equation.

#### Explanation:

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,

You can express the rate of the reaction in terms of the **change** in the concentration of the reactants with time. If you take

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

Here **consumed** in a period of time

You can thus use a **rate constant**

#(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 **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

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

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

Plug this back into equation

#-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

#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" -># whentemperatureandvolumeare 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 pressurebefore the reaction begins

#P_t# - the pressure inside the reaction vesselat a given time#t#

#P_f# - the pressure inside the reaction vesselafterthe reaction is complete

You can say that

#x prop P_t - P_0 -># thechange in pressureat a given time#t# isproportionalto 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 **total number of moles** present in the reaction vessel at time

#P_0 prop P_f - P_0 -># the initial pressure isproportionalto 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 **remain** in the vessel after the reaction is complete will be proportional to the initial number of moles of

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

#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)|)))#

**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