Set up Hess\'s law cycle, and use the following information tp calculate deltaH

Question

Set up Hess's law cycle, and use the following information tp calculate deltaH degrees Fahrenheit for aqueous nitric acid, HNO3(aq). You will need to use fractional coefficients for some equations.

3NO2(g) + H2(l)--->2 HNO3(aq) +NO(g)

                                                        deltaHdegrees= -137.3kJ

2 NO(g) + O2(g)---->2 NO2(g)          deltaHdegrees= -116.2kJ

4 NH3g)                                            deltaHdegrees= -46.1kJ/mol

H2O(l)                                              deltaHdegrees= -285.8 kJ/mol

Explanation / Answer

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Eq.1) 4NH3(g)+5O2(g)------->4NO(g)+6H2O(l)... -908 kj/mol
Eq.2) 2NO(g)+O2(g)----------->2NO2(g) ............... -112 kj/mol
Eq.3) 3NO2(g)+H2O(l)-------->2HNO3(aq) + NO(g) -140 kj/mol

Since the product nitric acid (HNO3) appears in Eq.3, take this one as the starting point.

3NO2(g)+H2O(l)-------->2HNO3(aq) + NO(g) .... -140 kj/mol

Since the main reactant is NH3, take Eq.1 and add these equations.

4NH3(g)+5O2(g)------->4NO(g)+6H2O(l)..... -908 kj/mol
--------------------------------------...
4NH3(g)+5O2(g)+3NO2(g) -----> 2HNO3(aq) + 5NO(g) + 5H2O(l) ........ - 1048 kJ/mol

Since NO2is an intermediate substance it should not appear in the main equation. To cancel NO2, Multiply Eq.2 by 3/2 and add the last equation.

3NO(g)+3/2O2(g)----------->3NO2(g) ............... -168 kj/mol
4NH3(g)+5O2(g)+3NO2(g) -----> 2HNO3(aq) + 5NO(g) + 5H2O(l) ........ - 1048 kJ/mol
--------------------------------------...

4NH3(g) + 13/2O2 ---------> 2HNO3(aq) + 2NO(g) + 5H2O(l)
dH = - 1216 kJ/mol

Since the sign of delta H reaction is negative, the reaction is exothermic.

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