# Question #d203d

Aug 6, 2017

The $\text{heat of formation........}$

#### Explanation:

By definition, the $\text{heat of formation}$ of a compound is the enthalpy change associated with the formation of ONE MOLE of compound in its standard state under standard conditions, from its constituent elements in their standard states under standard conditions.

For $C {O}_{2} \left(g\right)$..........we thus get......

$C \left(s\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + \Delta {H}_{f}^{\circ}$, where $\Delta {H}_{f}^{\circ} = - 393.5 \cdot k J \cdot m o {l}^{-} 1$. In general $\Delta {H}_{f}^{\circ}$ must be established by measurement, tho energy summation laws such as $\text{Hess' Law}$ allows calculation of many $\Delta {H}_{f}^{\circ}$ values without doing the experiment. Anway this is the only value I am willing to quote....

But by definition, $\Delta {H}_{f}^{\circ}$ of an ELEMENT in its standard state under standard conditions is conceived to be $0 \cdot k J \cdot m o {l}^{-} 1$.

For $\text{cupric chloride}$....

$C u \left(s\right) + C {l}_{2} \left(g\right) \rightarrow C u C {l}_{2} \left(s\right) + \Delta {H}_{f}^{\circ}$

For $\text{cuprous chloride}$....

$C u \left(s\right) + \frac{1}{2} C {l}_{2} \left(g\right) \rightarrow C u C l \left(s\right) + \Delta {H}_{f}^{\circ}$

..........and I would expect that this $\Delta {H}_{f}^{\circ}$ is numerically LESS than the former. Why can I make this prediction?

For $\text{hydrazine}$....

${N}_{2} \left(g\right) + 2 {H}_{2} \left(g\right) \rightarrow {N}_{2} {H}_{4} \left(l\right) + \Delta {H}_{f}^{\circ}$

For $\text{ammonium chloride}$....

$\frac{1}{2} {N}_{2} \left(g\right) + 2 {H}_{2} \left(g\right) + \frac{1}{2} C {l}_{2} \left(g\right) \rightarrow N {H}_{4} C l \left(s\right) + \Delta {H}_{f}^{\circ}$

Anyway, you will have to find all these values in your inorganic/physical chem text......they should all be numerically negative.