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

Aug 28, 2017

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

$\text{N"_2 + "2O"_2 → "N"_2"O"_4; Δ_text(f)H = "9.16 kJ/mol}$

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 ${\text{N}}_{2}$

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.

${\text{Moles of N"_2 = 6.03 color(red)(cancel(color(black)("L N"_2))) × "1 mol N"_2/(22.4 color(red)(cancel(color(black)("L N"_2)))) = "0.2692 mol N}}_{2}$

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"