Question

The chemistry of nitrogen oxides is very versatile. Given the following numbered reactions and their standard enthalpy changes, calculate the standard enthalpy of reaction for

N­₂O₃(g) + N₂O₅(g) → 2N₂O₄(g)

NO(g) + NO₂(g) → N₂O₃(g)                         DH = -39.8kJ/mol

NO(g) + NO₂(g) + O₂(g) → N₂O₅(g) DH = -112.5kJ/mol

2NO₂(g) → N₂O₄(g) DH = -57.2kJ/mol

2NO(g) + O₂(g) → 2NO₂(g) DH = -114.2kJ/mol

N₂O₅(s) → N₂O₅(g) DH = 54.1kJ/mol

Answer 1

The equation should be N­2O3(g) + N2O5(s) → 2N2O4(g)

We have to arrange the equations in order to get the final equation from the statement when we add them, the first step to do is to put the compounds we already know are products and reactants on the correct side of the equation

N2O5(s) must appear as a reactant and N2O5 (g) should not appear in the equation

N2O3 (g) must appear on the left side

N2O4 must appear on the right side

Notice that NO, NO2 and O2 dont appear in the final equation, remember that if we reverse and equation the enthalpy changes from positive to negative and negative to positive

N2O5(g) → NO(g) + NO2(g) + O2(g) 112.5 KJ/mol

N2O5(s) → N2O5(g) 54.1

(2NO2(g) → N2O4(g)) * 2 -57.2 * 2

4NO₂ (g) → 2N₂O₄ -114.4

N2O3(g) →  NO(g) + NO2(g) 39.8

2NO(g) + O2(g)→ 2NO2(g) -114.2

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N2O5(s) + N2O5 (g) +4NO2 + 2 NO + O2 + N2O3(g)  → 4 NO2 + 2 NO + 2N2O4 + N2O5 (g) + O2

Now remove repeated terms on both sides

N₂O₅(s) + N₂O₃(g)  → 2N₂O₄

Now add all the enthalpies

-114.2 + 39.8 + 54.1 + 112.5 - 114.4 = -22.2 KJ / mol

That is the Standard Enthalpy of Rxn.

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