Question

How much energy does the oxidation of 6.03 L of nitrogen absorb?

Answer 1

The reaction absorbs 2.47 kJ.

Explanation:

You don’t say what the oxidation product is, so I shall assume that it is dinitrogen tetroxide.

Then the reaction is

`"N"_2 + "2O"_2 → "N"_2"O"_4; Δ_text(f)H = "9.16 kJ/mol"`

The formula for the amount of heat `q` absorbed during the formation of `n` mol of a compound is

`color(blue)(barul|stackrel(" ")(q = nΔ_text(f)H)|)`

where `Δ_text(f)H` is the enthalpy of formation of the compound.

Step 1. Calculate the number of moles of `"N"_2`

We know that the volume of 1 mol of nitrogen at 1 atm and 273 K (the old definition of STP) is 22.4 L.

∴ `"Moles of N"_2 = 6.03 color(red)(cancel(color(black)("L N"_2))) × "1 mol N"_2/(22.4 color(red)(cancel(color(black)("L N"_2)))) = "0.2692 mol N"_2`

Step 2. Calculate the amount of heat absorbed

`q = 0.2692 color(red)(cancel(color(black)("mol N"_2"O"_4))) × "9.16 kJ"/(1 color(red)(cancel(color(black)("mol N"_2"O"_4)))) = "2.47 kJ"`