Chapter 15, Problem 15.1P

Interpretation Introduction

Interpretation:

The heat energy required to vaporize $5\text{kg}$ of ammonia molecule has to be determined.

Concept Introduction:

Molar enthalpy for a process is the quantity of heat energy needed by one mole of a substance to undergo change in the process. The heat energy absorbed by $n$ moles of substance to undergo change is as follows:

  $q=n\left(\Delta H\right)$

Here,

$q$ is the heat energy in $\text{kJ}$

$n$ is the number of moles of substance.

$\Delta H$ is the enthalpy for process in $\text{kJ/mol}$ .

The value of $\Delta H$ is positive always as energy is needed to overcome the interactions between the liquid molecules.

Answer to Problem 15.1P

The heat energy required to vaporize $5\text{kg}$ of ammonia molecule is $6849.2727\text{kJ}$.

Explanation of Solution

The formula to determine moles of ammonia is as follows:

  $\text{number}\text{of}\text{moles}\text{of}\text{ammonia}=\frac{\text{mass}\text{of}\text{ammonia}}{\text{molecular}\text{weight}\text{of}\text{ammonia}}$        (1)

Substitute $17.031\text{g/mol}$ for molecular weight of ammonia, $5\text{kg}$ for mass of ammonia in equation (1).

  $\begin{array}{c}\text{number}\text{of}\text{moles}\text{of}\text{ammonia}=\frac{5\text{kg}\left(\frac{1000\text{g}}{1\text{kg}}\right)}{17.031\text{g/mol}}\\ =293.5822\text{mol}\end{array}$

The formula for heat energy required to vaporize $5\text{kg}$ of ammonia molecule is as follows:

  $q_{\text{vap}}=n\left(\Delta H_{\text{vap}}\right)$        (2)

Substitute $23.33\text{kJ/mol}$ for $\Delta H_{\text{vap}}$ and $293.5822\text{mol}$ for $n$ in equation (2) to calculate the heat energy required to vaporize $5\text{kg}$ of ammonia molecule.

  $\begin{array}{c}{q}_{\text{vap}}=\left(293.5822\text{mol}\right)\left(23.33\text{kJ/mol}\right)\\ =6849.2727\text{kJ}\end{array}$

The heat energy needed to vaporize $5\text{kg}$ of ammonia molecule is $6849.2727\text{kJ}$.