Question

Compound

ΔHf (kJ/mol)


NO(g)


+90.4


NO2(g)


+33.9


Using the data given, what is the ΔHrxn (in kJ) for:




2 NO(g) + O2(g) → 2 NO2(g)




Hint: There is a reason no data is given for O 2(g

Answer 1

The reaction enthalpy ΔHrxn can be calculated using Hess’s law, which states that the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which a reaction can be divided.

The reaction is:

2 NO(g) + O2(g) → 2 NO2(g)

The formation reactions are:

1/2 N2(g) + 1/2 O2(g) → NO(g) ΔHf = +90.4 kJ/mol 1/2 N2(g) + O2(g) → NO2(g) ΔHf = +33.9 kJ/mol

We can write the enthalpy change of the reaction (ΔHrxn) as the sum of the enthalpy changes of formation (ΔHf) of the products minus the sum of the enthalpy changes of formation of the reactants, each multiplied by their stoichiometric coefficients.

ΔHrxn = [ 2 * ΔHf(NO2) ] - [ 2 * ΔHf(NO) + ΔHf(O2) ]

Since O2 is in its standard state, its enthalpy of formation is zero. So, we have:

ΔHrxn = [ 2 * (+33.9 kJ/mol) ] - [ 2 * (+90.4 kJ/mol) ] = -113 kJ

So, the enthalpy change for the reaction is -113 kJ.

Step-by-step explanation: