# An 80.0 sample of #N_2O_5# is allowed to decompose at 45°C, how long does it take for the quantity of #N_2O_5# to be reduced to 2.5 g?

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First-order reaction,

#N_2O_5\toN_2O_4+1/2O_2#

**EDIT:** #k=6.2xx10^-4" sec"^-1#

Second part: **How many liters of **#O_2# , measured at 745 mmHg and 45°C, are produced up to this point?

*Hint: convert the mmHg to atm.*

For the first part, I got #\approx5590# but my classmate kept getting #5.59xx10^-5# for some reason. After fiddling with the calculator, I would sometimes get that, but I mostly got #5590# minutes.

First-order reaction,

**EDIT:**

Second part:

How many liters of#O_2# , measured at 745 mmHg and 45°C, are produced up to this point?

Hint: convert the mmHg to atm.

For the first part, I got

##### 2 Answers

It took

The first order **rate law** is:

#r(t) = k["N"_2"O"_5] = -(Delta["N"_2"O"_5])/(Deltat)#

If we take the change in concentration and examine a small enough interval in time,

#-(Delta["N"_2"O"_5])/(Deltat) -> -(d["N"_2"O"_5])/(dt)#

#=> k["N"_2"O"_5] = -(d["N"_2"O"_5])/(dt)# and we can then treat this with a bit of Calculus.

You can skip to the result if you don't think you need to know this. **Separation of variables** gives:

#-kdt = 1/(["N"_2"O"_5])d["N"_2"O"_5]#

Integration from time zero to time

#-kint_(0)^(t)dt = int_(["N"_2"O"_5]_0)^(["N"_2"O"_5])1/(["N"_2"O"_5])d["N"_2"O"_5]#

#-kt = ln["N"_2"O"_5] - ln["N"_2"O"_5]_0#

As a result, we obtain the **first-order integrated rate law**:

#color(green)(ln["N"_2"O"_5] = -kt + ln["N"_2"O"_5]_0)#

*You can proceed to read starting here.*

For a half-life,

#ln(1/2["N"_2"O"_5]_0) - ln["N"_2"O"_5]_0 = -kt_(1//2)#

#ln(1/2) = -kt_(1//2) = -ln2#

Therefore, the **first-order half-life** is given by:

#t_(1//2) = (ln2)/k#

Since you want the mass to drop from

- you have multiplied the mass by
#2.5//80.0 = 1//32# . - that means that
#5# half-lives have passed, since#1//32 = (1//2)^(5)# .

First we find the half-life to be:

#t_(1//2) = (ln2)/(6.2 xx 10^(-4) "s"^(-1)) = "1118 s"#

As a result, it will take this long to drop by a factor of

#color(blue)(t) = 5t_(1//2) = 5("1118 s") = color(blue)("5590 s")#

And

**PART TWO**

Since we want five half-lives passed, it means this much reacted:

#"80.0 g" - "2.5 g" = "77.5 g N"_2"O"_5#

So all we do is do a unit conversion from reactant to product, then use the ideal gas law. The reaction was:

#"N"_2"O"_5(g) -> "N"_2"O"_4(g) + 1/2"O"_2(g)#

And the unit conversion:

#77.5 cancel("g N"_2"O"_5) xx cancel("1 mol N"_2"O"_5)/(108.009 cancel("g N"_2"O"_5)) xx ("0.5 mols O"_2)/cancel("1 mol N"_2"O"_5)#

#= "0.3588 mols O"_2(g)#

From the ideal gas law

#PV = nRT# ,

we can then convert this to liters at

#745 cancel"torr" xx ("1 atm")/(760 cancel"torr") = "0.980 atm"#

#45 + 273.15 = "318.15 K"#

Therefore:

#color(blue)(V_(O_2)) = (nRT)/P#

#= ("0.3588 mols" cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot "318.15 K")/("0.980 atm")#

#=# #color(blue)("9.56 L")#