The standard enthalpy change for the reaction below is -905.4 kJ (for the reacti

Question

The standard enthalpy change for the reaction below is -905.4 kJ (for the reaction as written). What is the standard enthalpy of formation of NO(g)?
4 NH3(g) + 5 O2(g) --> 4 NO(g) + 6 H2O(g)

Give your answer in kilojoules per mole (kJ mol-1), accurate to one decimal place. Do not include units in your answer.
Data :

Substance and Standard enthalpy of formation (in kJ mol-1)
NH3(g) -46.1
H2O(g) -241.8

Explanation / Answer

The first equation is 4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g); ?H = -906 kJ The second equation is NO(g) + 3/2H2O(g) -> NH3(g) + 5/4O2(g); ?H = ? The second equation has been obtained by reversing the first equation and dividing each coefficient by four. Therefore, to obtain ?H for the second equation, the ?H for the first equation must be reversed in sign and divided by four: +906 ? 4 = 226.50 = 227 kJ.

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