# 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, ${N}_{2} {O}_{5} \setminus \to {N}_{2} {O}_{4} + \frac{1}{2} {O}_{2}$ EDIT: $k = 6.2 \times {10}^{-} 4 {\text{ sec}}^{-} 1$ 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 $\setminus \approx 5590$ but my classmate kept getting $5.59 \times {10}^{-} 5$ for some reason. After fiddling with the calculator, I would sometimes get that, but I mostly got $5590$ minutes.

Jul 18, 2018

It took $\text{5590 s}$.

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,

$- \frac{\Delta \left[{\text{N"_2"O"_5])/(Deltat) -> -(d["N"_2"O}}_{5}\right]}{\mathrm{dt}}$

=> 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 ${\left[{\text{N"_2"O}}_{5}\right]}_{0}$ is allowed to proceed over time until $\left[{\text{N"_2"O}}_{5}\right]$ 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]

$- k t = \ln {\left[{\text{N"_2"O"_5] - ln["N"_2"O}}_{5}\right]}_{0}$

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

$\textcolor{g r e e n}{\ln {\left[{\text{N"_2"O"_5] = -kt + ln["N"_2"O}}_{5}\right]}_{0}}$

You can proceed to read starting here.

For a half-life, ${\left[{\text{N"_2"O"_5] = 1/2["N"_2"O}}_{5}\right]}_{0}$, so:

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

$\ln \left(\frac{1}{2}\right) = - k {t}_{1 / 2} = - \ln 2$

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

${t}_{1 / 2} = \frac{\ln 2}{k}$

Since you want the mass to drop from $\text{80.0 g}$ to $\text{2.5 g}$,

• you have multiplied the mass by $2.5 / 80.0 = 1 / 32$.
• that means that $5$ half-lives have passed, since $1 / 32 = {\left(1 / 2\right)}^{5}$.

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 ${2}^{5}$:

$\textcolor{b l u e}{t} = 5 {t}_{1 / 2} = 5 \left(\text{1118 s") = color(blue)("5590 s}\right)$

Jul 18, 2018

And $\text{9.56 L}$ of ${\text{O}}_{2}$ gets produced after $5$ half-lives of ${\text{N"_2"O}}_{5}$.

PART TWO

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

${\text{80.0 g" - "2.5 g" = "77.5 g N"_2"O}}_{5}$

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

${\text{N"_2"O"_5(g) -> "N"_2"O"_4(g) + 1/2"O}}_{2} \left(g\right)$

And the unit conversion:

$77.5 \cancel{{\text{g N"_2"O"_5) xx cancel("1 mol N"_2"O"_5)/(108.009 cancel("g N"_2"O"_5)) xx ("0.5 mols O"_2)/cancel("1 mol N"_2"O}}_{5}}$

$= {\text{0.3588 mols O}}_{2} \left(g\right)$

From the ideal gas law

$P V = n R T$,

we can then convert this to liters at $\text{745 torr}$ and ${45}^{\circ} \text{C}$.

$745 \cancel{\text{torr" xx ("1 atm")/(760 cancel"torr") = "0.980 atm}}$

$45 + 273.15 = \text{318.15 K}$

Therefore:

$\textcolor{b l u e}{{V}_{{O}_{2}}} = \frac{n R T}{P}$

$= \left(\text{0.3588 mols" cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot "318.15 K")/("0.980 atm}\right)$

$=$ $\textcolor{b l u e}{\text{9.56 L}}$