Calculate the enthalpy change, ∆h in kj, for the reaction h2o(s) → h2(g) + 1/2o2(g). use the following information: : +279.9 kj +291.9 kj –279.9 k –291.9 kj

+ 291.9 kJ

Solution:

The equation given is as;

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = ?

First, as we know the heat of formation of H₂O ₍l₎ is,

H₂ ₍g₎ + 1/2 O₂ ₍g₎ → H₂O ₍l₎ ΔH = - 285.9 kJ

Now, reversing the equation will reverse the sign of heat as,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

Also, we know that,

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

Now, adding last two equations,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 291.9 kJ

a. energy is absorbed in the reaction.

if in a given reaction the energy of products is greater than the energy of reactants it means that reaction is endothermic and during the reaction, the energy is absorbed by reactants. in an endothermic reaction, the products have more stored chemical energy than the reactants. and in this reaction the reactants are more stable than products before the compounds with less energy are always more stable.