Question

1. given: h₂(g) + ½o₂ (g) → h₂o(l) δh°f = -286 kj what is the enthalpy change for the following reaction? 2 h₂o(l) → 2 h₂(g) + o₂(g) δh°rxn =? a. δh° = -286 kj b. δh° = +286 kj c. δh° = -572 kj d. δh° = +572 kj e. δh° = -143 kj

Explanation:

Step1: Reverse the given reaction

When a reaction is reversed, the sign of $\Delta H$ changes.
The given reaction is \(H_{2}(g)+\frac{1}{2}O_{2}(g)\to H_{2}O(l)\), \(\Delta H_{f}^{\circ}=- 286\space kJ\).
The reversed reaction is \(H_{2}O(l)\to H_{2}(g)+\frac{1}{2}O_{2}(g)\), and its \(\Delta H = + 286\space kJ\).

Step2: Multiply the reversed reaction by 2

When a reaction is multiplied by a factor \(n\), its \(\Delta H\) is also multiplied by \(n\).
Multiply \(H_{2}O(l)\to H_{2}(g)+\frac{1}{2}O_{2}(g)\) (\(\Delta H = + 286\space kJ\)) by 2.
We get \(2H_{2}O(l)\to 2H_{2}(g)+O_{2}(g)\).
\(\Delta H_{rxn}^{\circ}=2\times(+ 286\space kJ)=+572\space kJ\)

Answer:

D. \(\Delta H^{\circ}= + 572\space kJ\)