Question

The fluorocarbon compound \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) has a normal boiling point of \(47.6^{\circ} \mathrm{C}\). The specific heats of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(l)\) and \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(\mathrm{~g})\) are \(0.91 \mathrm{~J} / \mathrm{g}-\mathrm{K}\) and \(0.67 \mathrm{~J} / \mathrm{g}-\mathrm{K}\), respectively. The heat of vaporization for the compound is \(27.49 \mathrm{~kJ} / \mathrm{mol}\). Calculate the heat required to convert \(50.0 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) from a liquid at \(10.00^{\circ} \mathrm{C}\) to a gas at \(85.00^{\circ} \mathrm{C}\).

Short answer

The total heat required to convert \(50.0g\) of \(\mathrm{C}_{2}\mathrm{Cl}_{3}\mathrm{F}_{3}\) from a liquid at \(10.00^{\circ}C\) to a gas at \(85.00^{\circ}C\) is calculated by adding the heat required to increase the temperature of the liquid, the heat of vaporization, and the heat required to increase the temperature of the gas: $$q_{total} = 50.0g \times 0.91J/gK \times 37.6K + 50.0g \times 0.1466J/g + 50.0g \times 0.67J/gK \times 37.4K$$ After plugging the values and performing the calculations, we get: $$q_{total} = 1711.76J + 366.5J + 1252.45J = 3330.71J$$ Therefore, the heat required to convert \(50.0g\) of \(\mathrm{C}_{2}\mathrm{Cl}_{3}\mathrm{F}_{3}\) from a liquid at \(10.00^{\circ}C\) to a gas at \(85.00^{\circ}C\) is approximately \(3330.71J\).

Step-by-step solution

Calculate the heat needed to increase the temperature of the liquid

To find the heat required to raise the temperature of the liquid, we will use the formula: $$q = mc \Delta T$$ Where: - q is the heat required; - m is the mass of the substance; - c is the specific heat capacity of the substance (in this case, of the liquid); - ∆T is the change in temperature. Given: - mass (m) = \(50.0g\); - specific heat capacity of the liquid (c) = \(0.91 J/gK\); - initial temperature (T1) = \(10.00 ^{\circ}C\); - boiling point (T2) = \(47.6^{\circ}C\); First, let's find the change in temperature (∆T): $$\Delta T = T2 - T1 = 47.6^{\circ}C - 10.00^{\circ}C = 37.6^{\circ}C$$ Now we can calculate the heat required to increase the temperature of the liquid: $$q_{increasing\ temperature} = mc\Delta T = 50.0g \times 0.91J/gK \times 37.6K$$

Calculate the heat needed to convert the liquid to gas (heat of vaporization)

To find the heat required to convert the liquid to gas, we will multiply the given heat of vaporization per mole by the number of moles in the given mass: Given: - molar mass of \(\mathrm{C}_{2}\mathrm{Cl}_{3}\mathrm{F}_{3}\) = (2 × 12.01) + (3 × 35.45) + (3 × 18.998) = \(187.38 g/mol\); - heat of vaporization = \(27.49 kJ/mol\); Let's convert the heat of vaporization to J/g: $$\frac{27.49 kJ/mol}{187.38 g/mol} = 0.1466 J/g$$ Now we can calculate the heat required to convert the liquid to gas, using the same formula as in step 1: $$q_{vaporization} = mc\Delta T = 50.0g \times 0.1466J/g$$

Calculate the heat needed to increase the temperature of the gas

Now we need to find the heat required to raise the temperature of the gaseous substance to the final temperature: Given: - specific heat capacity of the gas (c) = \(0.67J/gK\); - final temperature (T3) = \(85.00^{\circ}C\); Let's find the change in temperature (∆T): $$\Delta T = T3 - T2 = 85.00^{\circ}C - 47.6^{\circ}C = 37.4^{\circ}C$$ Now we can calculate the heat required to increase the temperature of the gas: $$q_{increasing\ temperature\ gas} = mc\Delta T = 50.0g \times 0.67J/gK \times 37.4K$$

Calculate the total heat required

Finally, to get the total heat required, we will sum the heat required in all three steps: $$q_{total} = q_{increasing\ temperature} + q_{vaporization} + q_{increasing\ temperature\ gas}$$ Calculate each term from the previous steps and add them together to get the final answer.

Key concepts

Specific Heat Capacity

Imagine you're warming up a cup of water for your morning tea. You know it takes time and energy to get that water to the perfect drinking temperature. Specific heat capacity, the amount of heat needed to change the temperature of a substance by a certain amount, is like a measure of a substance's 'thermal stubbornness'. It tells us how much energy is needed to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin).

In our exercise, we had two different specific heat capacities to consider: one for the liquid state of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) at \(0.91 \mathrm{~J} / \mathrm{g}-\mathrm{K}\), and another for the gaseous state at \(0.67 \mathrm{~J} / \mathrm{g}-\mathrm{K}\). These values are crucial because they determine how much energy it will take to heat the substance in each phase before and after it vaporizes. A lower specific heat capacity means the substance heats up more quickly—like when your seat warms up fast on a hot day.

Phase Change

A phase change is like a personality shift for a substance—it's when a material goes from being a solid to a liquid, or from a liquid to a gas. We see this when ice melts into water, or when our hot tea releases steam. During a phase change, the substance absorbs or releases energy, but its temperature doesn't change until the transformation is complete. That's like running on a treadmill and not getting anywhere—energy is expended, but without a temperature change.

In the context of the exercise, we looked at the heat of vaporization, which is the energy required to turn the liquid \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) into a gas. It is a specific type of enthalpy change—an endothermic one, meaning it consumes heat. This phase change occurs without a change in temperature, despite the massive amount of energy involved.

Enthalpy Change

Enthalpy change is a term that's all about energy during chemical reactions or physical changes. Think of it as the energy budget for a substance's transformation. It's either absorbed from the surroundings (endothermic) or released into the surroundings (exothermic).

During the phase transition in our exercise, the heat of vaporization represents a significant enthalpy change, as energy is absorbed to convert the liquid \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) into a vapor. This enthalpy change of vaporization doesn't change the temperature but breaks the intermolecular forces that hold the liquid together, turning it into a gas. Additionally, when the substance heats up before and after this phase change, the enthalpy change occurs as the temperature rises—showing other forms of enthalpy changes beyond just the vaporization.