The equation should be N­2O3(g) + N2O5(s) 2N2O4(g) We have to arrange the equati

Question

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

---------------------------------------------------------

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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Explanation / Answer

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

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