Chapter 15, Problem 88A

A sample of ammonia $(\Delta H_{solid}=-5.66kJ/mol)$ liberates5.66 kJ of heat as it solidifies at its melting point.What is the mass of the sample?

Interpretation Introduction

Interpretation: Mass of ammonia to be calculated.

Concept introduction: Enthalpy: It is the state function whose changes indicates the amount of heat transferred from a system to its surrounding and vice versa.

Answer to Problem 88A

The mass of ammonia of the sample is $\text{17}\text{.0307g}{\text{NH}}_{\text{3}}^{}$ .

Explanation of Solution

Given:

5.66 kJ of heat is released, that is, heat flow at constant pressure, ${\text{q}}_{\text{p}}^{}$

  $\text{Δ}{H}_{\text{solid}}^{}\text{=-5}\text{.66 kJ/mol}$

Calculation:

Enthalpy of heat flow,

  $\Delta H_{frz}^{}=\frac{q_p^{}}{n_{compound}^{}}$

Where, $\Delta H_{frz}^{}\text{=molar}\text{enthalpy}\text{of}\text{freezing}$

     $q_{\text{p}}^{}\text{=heat}\text{flow}\text{at}\text{constatnt}\text{pressure}$

  $n_{\text{compound}}^{}\text{=moles}\text{in}\text{the}\text{compound}$

  $n_{\text{compound}}^{}=\frac{q_p^{}}{\Delta H_{frz}^{}}$

     $\text{=}\frac{\text{-5}\text{.66}\text{kJ}}{\text{-5}\text{.66}\text{kJ/mol}}$

     $\text{=1mol}$

Thus, the mass of ammonia is,

  $\text{=1mol×}\frac{\text{Molar}\text{mass}\text{of}{\text{NH}}_{\text{3}}^{}}{\text{1}\text{mol}{\text{NH}}_{\text{3}}^{}}$ (molar mass is the sum of all the atomic weights of atoms present in the element)

  $\begin{array}{l}\text{=1mol×}\frac{\text{(14}\text{.007)+3×1}{\text{.0079gNH}}_{\text{3}}^{}}{\text{1}\text{mol}{\text{NH}}_{\text{3}}^{}}\end{array}$

  $\begin{array}{c}\text{=17}\text{.0307g}{\text{NH}}_{\text{3}}^{}\end{array}$

Conclusion

It is important to realize that the 5.66kJ of heat is released which is q, the heat flow not the enthalpy of freezing, $\Delta H_{frz}^{}$ .