Chapter 15.3, Problem 24PP

Interpretation Introduction

Interpretation:

The amount of heat released when 275g of ammonia gas condenses to a liquid at its boiling point is to be calculated.

Concept introduction:

Heat of vaporization: The amount of heat needed to turn a gram of liquid into vapor.

Answer to Problem 24PP

The heat of vaporisation, $\Delta H_{vap}$, for ammonia $N{H}_3=23.3\raisebox{1ex}{$kJ$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$ (from table 15.4)

Molar mass of ammonia $N{H}_3=14\times 1+1\times 3=17\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

$Q=N\times \Delta H_{\mathrm{vap}}$ (N in moles)

$N=\frac{\mathrm{mass}}{17\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}=\frac{275}{17}$

$\begin{array}{l}Q=\frac{275}{17}\mathrm{mol}\times 23.3\raisebox{1ex}{$kJ$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\\ Q=376.9kJ\end{array}$

Explanation of Solution

The following steps are needed to calculate the heat released:

• Calculate the molar mass of ammonia
• Calculate the number of moles contained in 275g of ammonia.
• Calculate the heat releasedwhen ammonia condenses using the formula,
$Q=N\times \Delta H_{vap}$

Conclusion

The heat required to melt 275g of ammonia is 376.9kJ.