Heating Curve: Acetone

Overview

Acetone (C₃H₆O) is an organic compound and a common organic solvent that is miscible in water.

It is a well known household chemical and serves as an active ingredient in nail polish remover and paint thinner. Though a liquid at room temperature, it is a volatile substance and is readily noticeable as a “fruity” smell.

Properties of Acetone

\[\begin{align*} \mathrm{molar~mass} &= 58.08~\mathrm{g~mol^{-1}}\\[2ex] T_\mathrm{f} &= -95~^{\circ}\mathrm{C}\\ T_\mathrm{b} &= 56~^{\circ}\mathrm{C}\\[2ex] \Delta H_{\mathrm{fus}} &= 5.70~\mathrm{kJ~mol^{-1}}\\ \Delta H_{\mathrm{vap}} &= 31.30~\mathrm{kJ~mol^{-1}}\\[2ex] c_{\mathrm{s}} &= 1.653~\mathrm{J~g^{-1}~^{\circ}\mathrm{C}^{-1}}\\ c_{\mathrm{l}} &= 2.409~\mathrm{J~g^{-1}~^{\circ}\mathrm{C}^{-1}}\\ c_{\mathrm{g}} &= 1.283~\mathrm{J~g^{-1}~^{\circ}\mathrm{C}^{-1}} \end{align*}\]

Question 1

How much heat (in kJ to the nearest whole number) is required to transform 500.0 g of acetone at –100 °C to 100 °C?

Visualize this:

How much is 500.0 g of acetone? If the density of acetone at room temperature is 0.784 g mL^–1, we can get the volume.

\[\begin{align*} d &= \dfrac{m}{V} \\ V &= \dfrac{m}{d} = \dfrac{500.0~\mathrm{g}}{0.784~\mathrm{g~mL^{-1}}}= 632~\mathrm{mL} \end{align*}\]

632 mL is approximately 2.67 cups.

Rationalize:

Acetone is a solid at –100 °C and a gas at 100 °C based on the freezing and boiling points of acetone. Therefore, we will have to consider all five regions of a typical heating curve.

Solve:

\[\begin{align*} q_1 &= mc_{\mathrm{s}}\Delta T \\ &= 500.0~\mathrm{g} \left ( 1.653~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( -95~^{\circ}\mathrm{C} - -100~^{\circ}\mathrm{C} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}}\right)\\ &= 4.1325~\mathrm{kJ}\\[4ex] q_2 &= n\Delta H_{\mathrm{fus}}\\ &= 500.0~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{58.08~\mathrm{g}}\right ) \left ( \dfrac{5.70~\mathrm{kJ}}{\mathrm{mol}} \right )\\ &= 49.070~\mathrm{kJ}\\[4ex] q_3 &= mc_{\mathrm{l}}\Delta T \\ &= 500.0~\mathrm{g} \left ( 2.409~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 56~^{\circ}\mathrm{C} - -95~^{\circ}\mathrm{C} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}}\right)\\ &= 181.8795~\mathrm{kJ}\\[4ex] q_4 &= n\Delta H_{\mathrm{vap}}\\ &= 500.0~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{58.08~\mathrm{g}}\right ) \left ( \dfrac{31.30~\mathrm{kJ}}{\mathrm{mol}} \right )\\ &= 269.4559~\mathrm{kJ}\\[4ex] q_5 &= mc_{\mathrm{g}}\Delta T \\ &= 500.0~\mathrm{g} \left ( 1.283~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 100^{\circ}\mathrm{C} - 56~^{\circ}\mathrm{C} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}}\right)\\ &= 28.226~\mathrm{kJ}\\[4ex] q_{\mathrm{tot}} &= q_1 + q_2 + q_3 + q_4 + q_5 \\ &= 4.1325 + 49.070 + 181.8795 + 269.4559 + 28.226\\ &= 533~\mathrm{kJ} \end{align*}\]

Question 2

How much heat (in kJ to the nearest whole number) is required to transform 500.0 g of acetone at –150 °C to 80 °C?

Answer: 561 kJ

Question 3

How much heat (in kJ to the nearest whole number) is required to transform 500.0 g of acetone at –120 °C to 25 °C?

Answer: 214 kJ