Question

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?

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.

Answer 1

It took `"5590 s"`.

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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 `t` on the left means the initial concentration `["N"_2"O"_5]_0` is allowed to proceed over time until `["N"_2"O"_5]` is reached.

`-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, `["N"_2"O"_5] = 1/2["N"_2"O"_5]_0`, so:

`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 `"80.0 g"` to `"2.5 g"`,

• 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 `2^5`:

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

Answer 2

And `"9.56 L"` of `"O"_2` gets produced after `5` half-lives of `"N"_2"O"_5`.

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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 torr"` and `45^@ "C"`.

`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")`