# What is DeltaH_f^@ for ammonia, and how is it formulated?

Jun 26, 2017

#### Answer:

The $\text{standard enthalpy of formation}$ of ammonia is $- 46 \cdot k J \cdot m o {l}^{-} 1$.

#### Explanation:

What does this mean...? Well the given equation is associated with the RELEASE of a given quantity of heat to the surroundings........

$\frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {H}_{2} \left(g\right) \rightarrow N {H}_{3} \left(g\right) + 46 \cdot k J$

We write enthalpies (and other thermodynamic quantities) $\text{per mole of reaction}$ as written. And thus when we report $\Delta {H}_{\text{rxn}}^{\circ} = \Delta {H}_{f}^{\circ} \left(N {H}_{3}\right) = - 46 \cdot k J \cdot m o {l}^{-} 1$......The negative sign indicates that heat is RELEASED to the environment rather than absorbed. The enthalpy of the $3 \times N H$ formed is GREATER than the enthalpy of the $H - H$ and $N \equiv N$ bonds broken.

And so if there are $2 \cdot m o l$ ammonia formed, i.e. $34 \cdot g$ rather than $17 \cdot g$, the enthalpy is DOUBLED......... Are you clear on what I am saying?