If the heat of the following reaction is -1125 kJ, what is the heat of the formation for H2S? Use the standard heat of formation: Is this endo or exo?

Question

2H2S(g) + 3O2(g) ➡ 2H2O(l) +2SO2(g)

Answer 1

From a table of standard heats of formation, the following data can be obtained.

∆Hf H2O(l) = -285.8 kJ/mole

∆Hf SO2(g) = -296.8 kJ/mole

∆Hf O2(g) = 0

 

∆Hrxn = -1125 kJ = ∑∆Hf products - ∑∆Hf reactants

-1125 kJ = (2x-285.8 + 2x-296.8) - 2x  where x = ∆HfH2S(g)

-1125 = -571.6 + -593.6 - 2x

-1125 = -1165.2 - 2x

2x = -40.2

x = - 20.1 kJ/mole = ∆Hf H2S(g)

This is exothermic as indicated by the negative sign