Consider the production of liquid water,H 2 O(l), from hydrogen combustion: H 2

Question

Consider the production of liquid water,H2O(l), from hydrogen combustion: H2(g) + 1/2O2(g) ÆH2O(l) DH = -285.9 kJ
(a) What is the enthalpy change for the reverse reaction?

DHreverse = 1 kJ/mol.




(b) Balance the forward reaction with whole-number coefficients.What is DH for the reaction representedby this equation?

DH = 2 kJ/mol. H2(g) + 1/2O2(g) ÆH2O(l) DH = -285.9 kJ H2(g) + 1/2O2(g) ÆH2O(l) DH = -285.9 kJ

Explanation / Answer

a) the enthalpy change for the reverse reaction would be +285.9kj/mole this is because the reverse reaction would be H2O (l)-->H2(g)+1/2 O2(g) when you do the heat of formation of products -heat of formation ofreactants, you see that products =0, and reactants =-285.9 0- (-285.9)=285.9 kj/mole

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