An 80.0 sample of $N_{2} O_{5}$ $N_2O_5$ is allowed to decompose at 45°C, how long does it take for the quantity of $N_{2} O_{5}$ $N_2O_5$ to be reduced to 2.5 g?

Question

An 80.0 sample of $N_{2} O_{5}$ $N_2O_5$ is allowed to decompose at 45°C, how long does it take for the quantity of $N_{2} O_{5}$ $N_2O_5$ to be reduced to 2.5 g?

First-order reaction,
$N_{2} O_{5} \to N_{2} O_{4} + \frac{1}{2} O_{2}$ $N_2O_5\toN_2O_4+1/2O_2$
EDIT: $k = 6.2 \times 10^{- 4} {sec}^{- 1}$ $k=6.2xx10^-4" sec"^-1$

Second part: How many liters of $O_{2}$ $O_2$ , 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 $\approx 5590$ $\approx5590$ but my classmate kept getting $5.59 \times 10^{- 5}$ $5.59xx10^-5$ for some reason. After fiddling with the calculator, I would sometimes get that, but I mostly got $5590$ $5590$ minutes.

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Answer 1 · 2018-07-18T14:34:12

It took $5590 s$ $"5590 s"$ .

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 · 2018-07-18T14:41:31

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

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:

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

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

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

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An 80.0 sample of $N_{2} O_{5}$ $N_2O_5$ is allowed to decompose at 45°C, how long does it take for the quantity of $N_{2} O_{5}$ $N_2O_5$ to be reduced to 2.5 g?

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An 80.0 sample of $N_{2} O_{5}$ $N_2O_5$ is allowed to decompose at 45°C, how long does it take for the quantity of $N_{2} O_{5}$ $N_2O_5$ to be reduced to 2.5 g?