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

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Answer 1 · 2017-08-28T20:17:15

The reaction absorbs 2.47 kJ.

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

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