The following sequences of reactions occurs in the commercial production of aqueous nitric acid:
4nh3 + 5o2 -> 4no + 6h2o h= -907kj
2no + o2 -> 2no2 h = -113kj
3no2 + h2o -> 2hno2 + no h= -139kj
determine the total energy change for the production of onle mole of aqueous nitric acid by this process

H = -494.38 kJ

Explanation:

By Hess' Law, if there are stages to provide the production of some compound, it's possible to reduce these stages in only one reaction, by adding the reactions. To add the reactions, some elements may be simplified, so they will result in a simple reaction.

To use that, everything done in the reaction must be done is its enthalpy. It means that, if the reaction is inverted, then the enthalpy must have it's signal inverted (positive to negative, negative to positive). If the reaction is multiplied by some number, then the enthalpy must be multiplied by the same number.

After adding the reactions, the enthalpy of the new one will be the addition of the enthalpies of the reactions added. If there is the same compound in reagent in one reaction and product in the other one, so it must be subtracted.

Using the first and second reactions, and multiplying the second by 2, so it will be possible simplyfing the NO:

4NH3 + 5O2 -> 4NO + 6 H2O H = -907 kJ

4NO + 2O2 -> 4NO2 H = 2x(-113) = -226 kJ

Adding them:

4NH3 + 7O2 -> 6H2O + 4NO2 H = -907 + (-226) = -1133 kJ

Now, doing the same thing for this new reaction and the last one, to simplyfing NO2, its necessary to multiply these for 3, and the last one for 4:

12NH3 + 21O2 -> 18H2O + 12NO2 H = 3x(-1133) = -3399 kJ

12NO2 + 4H2O -> 8HNO2 + 4NO H = 4x(-139) = - 556 kJ

Adding them:

12NH3 + 21O2 -> 8HNO2 + 14H2O + 4NO H = -3399 + (-556) = -3955 kJ

So, to produce 8 moles of NO2 the energy change is -3955 kJ, so to produce 1 mol:

8 moles -3955 kJ

1 mol X

By direct three rule:

8X = -3955

X = -3955/8

X = -494.38 kJ

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