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

Question

Natalie Anton · Accepted Answer

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"#

Jeremiah Adams · Answer

undefined The energy absorbed during the oxidation of 6.03 L of nitrogen depends on the specific reaction and conditions, so a precise answer cannot be given without additional information.