Question

Answer the enthalpy:

Answer 1

Answer:

ΔH = +140.9 kJ

Explanation:

This is an application of Hess's Law. Begin by rewriting the equations so that similar species NOT found in the final reaction will cancel each other out. Remember to flip the signs when you flip the equations.

3O₂ -> 2O₃   ΔH = +427 kJ

2O -> O₂   ΔH = +395.2 kJ

NO + O₃ -> NO₂ + O₂   ΔH = -199 kJ

Now, divide the coefficients of the reactions so that they will cancel each other out (remember to also divide the enthalpies).

O₂ -> 2/3 O₃   ΔH = +142.3 kJ

O -> 1/2 O₂   ΔH = +197.6 kJ

NO + O₃ -> NO₂ + O₂   ΔH = -199 kJ

The O₂ as a reactant in the first reaction will cancel out with the O₂ as a product in the last reaction.

The O₃ is a little tricky to think about, so bear with me here. We can think of O₃ also as 3/3 O₃. Two-thirds of this O₃ in the reactants of the final reaction will cancel out with the products of the first reaction. The final 1/3 O₃ (which is  just a single O!) will cancel out with 1/2 O₂ (also just a single O!).

This finally leaves us with:

NO + O -> NO₂

Adding the three reactions with the rewritten coefficients gives us:

142.3 + 197.6 + (-199) = +140.9 kJ = ΔH

Thus, NO + O -> NO₂   ΔH = +140.9 kJ