Chapter 17, Problem 84A

Interpretation Introduction

Interpretation: The amount of heat that must be removed from liquid naphthalene at its freezing point to bring about solidification is to be found.

Concept Introduction: The enthalpy of solidification $\Delta H_{solid}$ , is the amount of energy that must be removed from a liquid substance to transform a quantity of that substance into a solid at constant pressure.

Answer to Problem 84A

The amount of heat that must be removed from $45.0g$ sample of liquid naphthalene at its freezing point to bring about solidification is $6.67\text{ kJ}$

Explanation of Solution

As fusion and solidification are opposite processes, the numerical value of the enthalpy of solidification is the same as the numerical value of the enthalpy of fusion but opposite in sign.

The enthalpy of solidification of naphthalene at its freezing point is $19.0\text{ kJ/mol}$

This means to convert $1mol$ of liquid naphthalene at its freezing point to solid, $19.0\text{ kJ}$ of heat must be removed.

Amount of naphthalene given = $45.0g$

Moles of naphthalene is found by dividing the given amount of naphthalene by the molecular mass of naphthalene.

Moles of naphthalene given = $\frac{45.0g}{128.17g/mol}=0.351mol$

For the formation of $1mol$ of solid naphthalene, heat removed = $19.0\text{ kJ}$

Thus, for the formation of $0.351mol$ of solid naphthalene, heat removed = $\frac{\text{19}\text{.0 kJ}}{\text{1}\text{.0 mol}}\text{×0}\text{.351mol=6}\text{.67 kJ}$

Conclusion

Thus, the amount of heat that must be removed from $45.0g$ sample of liquid naphthalene at its freezing point to bring about solidification is $\text{6}\text{.67 kJ}$