Question

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.

Answer 1

Answer:

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:

$\Delta H = \Delta H_{r} - \Delta H_{p}$  

Where $\Delta H_{r}$: is the bond enthalpy of reactants and $\Delta H_{p}$: 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:

$\Delta H = 2 \cdot 467 kJ/mol - 436 kJ/mol - \frac{1}{2} \cdot 498 kJ/mol = 249 kJ/mol$  

I hope it helps you!