# How do you calculate the molar enthalpy of condensation (DeltaH_(cond)) for ammonia when 100.0 g of NH_3 gas turn into a liquid at its boiling point?

## 68,500 J of energy are released in the process.

Jun 19, 2017

By calculating the quotient, $\text{energy released"/"moles of ammonia}$......and get $\Delta {H}_{\text{condensation}} = - 11.7 \cdot k J \cdot m o {l}^{-} 1$

#### Explanation:

We interrogate the process...........

$N {H}_{3} \left(g\right) \rightarrow N {H}_{3} \left(l\right) + {\Delta}_{\text{condensation}}$

We want $\Delta {H}_{\text{condensation}}$ with units of $k J \cdot m o {l}^{-} 1$.

We have parameters for a $100 \cdot g$ mass of ammonia.....

And thus, $\frac{- 68500 \cdot J}{\frac{100.0 \cdot g}{17.03 \cdot g \cdot m o {l}^{-} 1}} = - 11665.6 \cdot J \cdot m o {l}^{-} 1$, i.e. $- 11.7 \cdot k J \cdot m o {l}^{-} 1$.