Question

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) melts at \(-114{ }^{\circ} \mathrm{C}\) and boils at \(78{ }^{\circ} \mathrm{C}\). The enthalpy of fusion of ethanol is \(5.02 \mathrm{~kJ} / \mathrm{mol},\) and its enthalpy of vaporization is \(38.56 \mathrm{~kJ} / \mathrm{mol}\). The specific heats of solid and liquid ethanol are \(0.97 \mathrm{~J} / \mathrm{g}-\mathrm{K}\) and \(2.3 \mathrm{~J} / \mathrm{g}-\mathrm{K},\) respectively. (a) How much heat is required to convert \(42.0 \mathrm{~g}\) of ethanol at \(35^{\circ} \mathrm{C}\) to the vapor phase at \(78{ }^{\circ} \mathrm{C} ?\) (b) How much heat is required to convert the same amount of ethanol at \(-155^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C}\) ?

Short answer

(a) The total heat required to convert \(42.0 \mathrm{~g}\) of ethanol at \(35^{\circ} \mathrm{C}\) to the vapor phase at \(78{ }^{\circ} \mathrm{C}\) is the sum of the heat required to heat ethanol in the liquid phase and the heat required to vaporize ethanol: \(Q_{total} = Q_{liquid} + Q_{vaporization} = 42.0 \times 2.3 \times 43 \, \mathrm{J} + n \times 38.56 \, \mathrm{kJ/mol}\) (b) The total heat required to convert the same amount of ethanol at \(-155^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C}\) is the sum of the heat required to heat ethanol as a solid, the heat required to melt ethanol, the heat required to heat ethanol in the liquid phase, and the heat required to vaporize ethanol: \(Q_{total} = Q_{solid} + Q_{fusion} + Q_{liquid} + Q_{vaporization}\)

Step-by-step solution

Heat ethanol as a liquid

To find the heat required to heat liquid ethanol from \(35^{\circ} \mathrm{C}\) to \(78^{\circ} \mathrm{C}\), we can use the formula: \(Q = m \times c \times \Delta T\) Where: - \(Q\) is the heat required. - \(m\) is the mass of ethanol (\(42.0 \mathrm{~g}\)). - \(c\) is the specific heat of liquid ethanol (\(2.3 \mathrm {~J} / \mathrm{g} \cdot \mathrm{K}\)). - \(\Delta T\) is the change in temperature (\(78 - 35 = 43\)). So, the heat required to heat ethanol in the liquid phase will be : \(Q = 42.0 \times 2.3 \times 43\) J.

Vaporize ethanol

To find the amount of heat required to vaporize ethanol, we'll use the formula: \(Q = n \times \Delta H_{vap}\) But first, we need to convert the mass of ethanol to moles. We know that the molar mass of ethanol is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} = 2 \times 12.01 + 6 \times 1.01 + 1 \times 16.00 = 46.07 \, \mathrm{g/mol}\) So, the number of moles of ethanol is: \(n = 42.0 \, \mathrm{g} \times \frac{1 \, \mathrm{mol}}{46.07 \, \mathrm{g}}\) Next, we can calculate the amount of heat required to vaporize ethanol by multiplying the number of moles by the enthalpy of vaporization (\(38.56 \mathrm{~kJ/mol}\)): \(Q = n \times 38.56 \, \mathrm{kJ/mol}\) Now we can add the heat required in steps 1 and 2.

Total heat required

The total heat required to convert 42 g of ethanol at \(35^{\circ} \mathrm{C}\) to the vapor phase at \(78{ }^{\circ} \mathrm{C}\) is the sum of the heat required in steps 1 and 2: \(Q_{total} = Q_{liquid} + Q_{vaporization}\) (b) How much heat is required to convert the same amount of ethanol at \(-155^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C}\) ? We can break down this process into the following steps: 1. Heat ethanol from \(-155^{\circ} \mathrm{C}\) to \(-114^{\circ} \mathrm{C}\) as a solid 2. Melt ethanol at \(-114^{\circ} \mathrm{C}\) 3. Heat ethanol from \(-114^{\circ} \mathrm{C}\) to \(78^{\circ} \mathrm{C}\) as a liquid 4. Vaporize ethanol at \(78^{\circ} \mathrm{C}\) To find the heat required in each step, we can use the same procedure as before.

Steps 1 and 3: Heat ethanol as a solid and as a liquid

Calculate each separately using the specific heat capacity of each phase and the temperature difference. \(Q_{solid} = m \times c_{solid} \times \Delta T\) (from \(-155^{\circ} \mathrm{C}\) to \(-114^{\circ} \mathrm{C}\)) \(Q_{liquid} = m \times c_{liquid} \times \Delta T\) (from \(-114^{\circ} \mathrm{C}\) to \(78^{\circ} \mathrm{C}\))

Melt ethanol

Calculate the heat required to melt the ethanol using the enthalpy of fusion and the number of moles of ethanol. \(Q_{fusion} = n \times \Delta H_{fus}\)

Vaporize ethanol

Calculate the heat required to vaporize ethanol using the enthalpy of vaporization and the number of moles of ethanol. \(Q_{vaporization} = n \times \Delta H_{vap}\)

Total heat required

Finally, add the heat required for all four steps: \(Q_{total} = Q_{solid} + Q_{fusion} + Q_{liquid} + Q_{vaporization}\)

Key concepts

Enthalpy of Fusion

The enthalpy of fusion, often denoted as \( \Delta H_{fus} \), refers to the heat energy required to change a substance from solid to liquid at its melting point without changing its temperature. This phase change is also known as melting. For example, melting ice into water or, in the case of our problem, solid ethanol becoming liquid ethanol at \( -114{ }^{\circ} \mathrm{C} \).

The calculation to determine the heat needed for this process involves multiplying the number of moles of the substance by its molar enthalpy of fusion. In the context of the exercise, where the molar enthalpy of fusion for ethanol is given as \( 5.02 \mathrm{~kJ} / \mathrm{mol} \), we can understand the process as a crucial middle step in heating ethanol from a solid at sub-freezing temperatures to a vapor at its boiling point.

Enthalpy of Vaporization

On the other end of the temperature spectrum from fusion, we find the enthalpy of vaporization, represented as \( \Delta H_{vap} \). This is the amount of heat energy necessary to turn a liquid into a gas at its boiling point while maintaining constant temperature. For water, this occurs at \(100{ }^{\circ} \mathrm{C} \), and for ethanol, it's at \(78{ }^{\circ} \mathrm{C} \).

The formula to calculate the heat required for vaporization is similar to that for fusion: we multiply the number of moles of the liquid by its molar enthalpy of vaporization. In the problem's context, the enthalpy of vaporization for ethanol is \( 38.56 \mathrm{~kJ} / \mathrm{mol} \). This step is integral when determining the energy needed to convert liquid ethanol into vapor at its boiling point.

Specific Heat Capacity

The specific heat capacity, denoted as \( c \), is a property that defines how much heat energy is required to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin). Unlike enthalpy, which is concerned with phase changes, specific heat capacity involves heating a substance within a single phase - solid, liquid, or gas. The higher the specific heat capacity, the more energy is needed to achieve the same temperature change.

In the exercise, the specific heat capacity of liquid ethanol is given as \(2.3 \mathrm{~J} / \mathrm{g}-\mathrm{K} \), and for solid ethanol, it is \(0.97 \mathrm{~J} / \mathrm{g}-\mathrm{K} \). This means we need to account for different amounts of heat energy to raise the temperature of ethanol depending on whether it's in the solid or liquid state.

Heating Curve of Ethanol

A heating curve graphically represents the temperature of a substance as heat is continuously added. For a substance like ethanol, the curve would generally show a steady incline in temperature during heating in the solid phase, a plateau during melting (fusion), another incline when heated as a liquid, a second plateau during vaporization, and finally, an increase again as a gas.

In the provided problem, we are particularly interested in the temperature intervals and phase changes that ethanol undergoes on this curve. There is a stage of warming solid ethanol up to its melting point (\( -114{ }^{\circ} \mathrm{C} \)), another stage of heating the resulting liquid to its boiling point (\(78{ }^{\circ} \mathrm{C} \)), and the energy required for the phase changes themselves—melting and vaporizing. By following ethanol's heating curve, we find the total energy required to turn solid or liquid ethanol into vapor at various starting temperatures.