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

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At 65∘C, the half-life for the first-order decomposition of N2O5 is 2.38 min.

#N_2O_5(g)# →#2NO_2(g)+1/2O_2(g)#

1.50 g of N2O5 is introduced into an evacuated 14 −L flask at 65∘C.

At 65∘C, the half-life for the first-order decomposition of N2O5 is 2.38 min.

1.50 g of N2O5 is introduced into an evacuated 14 −L flask at 65∘C.

##### 1 Answer

#### 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**.

#color(blue)(ul(color(black)(PV = nRT)))#

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 theuniversal gas constant, equal to#0.0821("atm L")/("mol K")# #T# is theabsolute temperatureof 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

#PV = nRT implies P = (nRT)/V#

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

#P = (0.01389 color(red)(cancel(color(black)("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"))))#

#P = "0.02754 atm"#

To convert this to *mmHg*, use the fact that

#color(blue)(ul(color(black)("1 atm = 760 mmHg")))#

You should end up with

#0.02754 color(red)(cancel(color(black)("atm"))) * "760 mmHg"/(1color(red)(cancel(color(black)("atm")))) = color(darkgreen)(ul(color(black)("21 mmHg")))#

The answer is rounded to two **sig figs**.