Question

Ammonia will burn in oxygen and form nitric oxide (NO) and water vapor according to the following equation:

4 NH3 (g) + 5 O2 (g) → 4 NO(g) + 6 H2O(g); ΔHrxn= -906 kJ.

What is the enthalpy change for the following reaction:

NO(g) + 3/2 H2O(g) → NH3 (g) + 5/4 O2 (g) ; ΔHrxn= ???

Answer 1

You need the values/ each compound to calculate \(\Delta\)H\(\sf\color{}{_{rxn}}\)

Answer 2

doesnt give me that...gives me this hint: remember that enthalpy is extensive.)

Answer 3

You can see that equation TWO is just the reverse of equation ONE so the heat rxn will
reverse the sign of the heat of rxn

=906 kJ

notice though the equation on the bottom was divided by 4

so 906/4 kJ = your answer

that is extensive --enthalpy

Answer 4

thank you sooo much!!