# What is the initial partial pressure, in mmHg, of N2O5(g)?

## At 65∘C, the half-life for the first-order decomposition of N2O5 is 2.38 min. ${N}_{2} {O}_{5} \left(g\right)$→$2 N {O}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right)$ 1.50 g of N2O5 is introduced into an evacuated 14 −L flask at 65∘C.

Feb 24, 2017

$\text{21 mmHg}$

#### Explanation:

I suspect that there is more to this question than what you've added here since all you have to do to find out the initial partial pressure of dinitrogen pentoxide is use the ideal gas law equation.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Convert the mass of dinitrogen pentoxide to moles by using the compound's molar mass

1.50 color(red)(cancel(color(black)("g"))) * ("1 mole N"_2"O"_5)/(108.01color(red)(cancel(color(black)("g")))) = "0.01389 moles N"_2"O"_5

Rearrange the ideal gas law equation to solve for $P$

$P V = n R T \implies P = \frac{n R T}{V}$

make sure to convert the temperature to Kelvin and plug in your values to find

$P = \left(0.01389 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 65) color(red)(cancel(color(black)("K"))))/(14color(red)(cancel(color(black)("L}}}}\right)$

$P = \text{0.02754 atm}$

To convert this to mmHg, use the fact that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 atm = 760 mmHg}}}}$

You should end up with

$0.02754 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * "760 mmHg"/(1color(red)(cancel(color(black)("atm")))) = color(darkgreen)(ul(color(black)("21 mmHg}}}}$

The answer is rounded to two sig figs.