What energy change is associated with the reaction to obtain one mole of h2 from one mole of water vapor? the balanced equation is 2 h2o(g) ® 2h2(g) + o2(g) and the relevant bond energies are: h — h = 436 kj/mol; h — o = 467 kj/mol; o — o = 146 kj/mol; o o = 498 kj/mol.

ΔH = 249 kJ/mol

Explanation:

The balanced reaction is:

2H₂O(g) → 2H₂(g) + O₂(g)    (1)

To calculate the energy change to obtain one mole of H₂(g) from one mole of H₂O(g), the coefficients of the reaction (1) must be halved:

H₂O(g) → H₂(g) + 1/2O₂(g)    (2)

The enthalpy of the reaction (2) is given by:

Where : is the bond enthalpy of reactants and : is the bond enthalpy of products.

For the reactants we have the next bond energies:

2 x (H-O) = 2 x (467)

And the bond energies for the products are:

H-H + (1/2) (O=O) =  436 + (1/2)(498)

So, the enthalpy of the reaction (2) is:

I hope it helps you!    

could you add the photo?

explanation: