Chapter 15.4, Problem 43SSC

Interpretation Introduction

Interpretation:

The heat of vaporization of water is to be calculated.

Concept introduction:

Heat of vaporization is calculated by subtracting enthalpy of formation of liquid water from enthalpy of formation of gaseous water.

Answer to Problem 43SSC

0…………………………………… $\Delta {{\rm H}}_f^0(kJ/mol)$

-241.8…………………………………… $H_{2}O(g)$

$\Delta {{\rm H}}_{vap}=+44kJ/mol$

-285.8……………………………………. $H_{2}O(l)$

Heat of vaporization of water at $298Kis44kJ/mol$

Explanation of Solution

Heat of vaporization= $\Delta {{\rm H}}_f^0(gaseouswater)-\Delta {{\rm H}}_f^0(liquidwater)$

$\begin{array}{l}=-241.8-(-285.8)\\ =44.0kJ/mol\end{array}$

Conclusion

Thus, heat of vaporization of water at $298Kis44kJ/mol$