Problem 39 - Chemistry and Chemical Reactivity

Question

You are going to prepare a silicone polymer, and one of the starting materials is dichlorodimethylsilane, \(\mathrm{SiCl}_{2}\left(\mathrm{CH}_{3}\right)_{2} .\) You need its normal boiling point and to measure equilibrium vapor pressures at various temperatures. $$\begin{array}{cc} \text { Temperature }\left(^{\circ} \mathrm{C}\right) & \text { Vapor Pressure }(\mathrm{mm} \mathrm{Hg}) \\ \hline-0.4 & 40 . \\\ +17.5 & 100 \\ 51.9 & 400 \\ 70.3 & 760 \\ \hline \end{array}$$ (a) What is the normal boiling point of dichlorodimethylsilane? (b) Plot these data as In \(P\) versus \(1 / T\) so that you have a plot resembling the one in Figure \(12.18 .\) At what temperature does the liquid have an equilibrium vapor pressure of 250 mm Hg? At what temperature is it 650 mm Hg? (c) Calculate the molar enthalpy of vaporization for dichlorodimethylsilane using the Clausius-Clapeyron equation.

Step-by-Step Solution

Answer

(a) 70.3°C; (b) Use plot to find temperatures; (c) Use Clausius-Clapeyron equation.

Step 1: Determine Normal Boiling Point

The normal boiling point of a substance is the temperature at which its vapor pressure equals 760 mm Hg, which is the standard atmospheric pressure. From the data provided, the vapor pressure is 760 mm Hg at 70.3°C. Therefore, the normal boiling point of dichlorodimethylsilane is 70.3°C.

Step 2: Convert Temperatures to Kelvin

Before plotting, convert all temperatures from Celsius to Kelvin using the formula: \[ T(K) = T(°C) + 273.15 \]For example:- For -0.4°C: \(-0.4 + 273.15 = 272.75 \, K\)- For 17.5°C: \(17.5 + 273.15 = 290.65 \, K\)Repeat for the other temperatures.

Step 3: Calculate 1/T Values

Compute the reciprocal of each temperature in Kelvin calculated in the previous step. For instance, for 272.75 K: \[ \frac{1}{T} = \frac{1}{272.75} = 0.00367 \, \text{K}^{-1} \]Repeat this for all temperatures to get their corresponding \(1/T\) values.

Step 4: Calculate ln(P) Values

Calculate the natural logarithm (ln) of the vapor pressure for each given pressure value.For example, for a pressure of 40 mm Hg: \[ \ln(40) = 3.688 \]Repeat this for all vapor pressures to get their corresponding \(\ln(P)\) values.

Step 5: Create the ln(P) vs 1/T Plot

Plot the calculated values of \(\ln(P)\) on the vertical axis against \(1/T\) on the horizontal axis. The plot should be a straight line and the slope of this line can be used in the Clausius-Clapeyron equation to find the enthalpy of vaporization.

Step 6: Determine Temperatures for Specific Vapor Pressures

Use the plot from Step 5 to find the temperature corresponding to a vapor pressure of 250 mm Hg and 650 mm Hg:- For 250 mm Hg, find \(\ln(250)\), then find the corresponding \(1/T\) from the plot and convert it back to °C.- For 650 mm Hg, find \(\ln(650)\), then find the corresponding \(1/T\) from the plot and convert it back to °C.

Step 7: Use the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is:\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]Use two data points from the plot (for example, vapor pressures of 40 mm Hg and 400 mm Hg) to calculate the slope, which is equal to \(-\frac{\Delta H_{vap}}{R}\). Solve for the enthalpy of vaporization \(\Delta H_{vap}\), using the gas constant \(R = 8.314 \, \text{J/mol K}\).

Key Concepts

Boiling Point Determination

The boiling point of a substance is where its vapor pressure equals standard atmospheric pressure, 760 mm Hg. It's a vital property for understanding a compound's behavior during phase changes. In the given exercise, the normal boiling point of dichlorodimethylsilane is identified directly from provided data. When the vapor pressure reaches 760 mm Hg, corresponding to 70.3°C, this temperature is defined as the normal boiling point. Knowing the boiling point is crucial in processes where precise temperature control is needed, such as chemical synthesis and industrial manufacturing.

Molar Enthalpy of Vaporization

The molar enthalpy of vaporization (\(\Delta H_{vap}\)) is the heat required to convert one mole of a liquid into vapor without a temperature change. It reflects the strength of intermolecular forces in a substance. In this exercise, the \(\Delta H_{vap}\) for dichlorodimethylsilane can be found using the Clausius-Clapeyron equation:

• Utilize the natural logarithm of vapor pressures and the inverse of temperatures to determine the slope of a plot.
• The equation is \(\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\)

By calculating the slope and knowing the gas constant \(R\), \(\Delta H_{vap}\) can be extracted. This value helps in assessing the energy changes during vaporization, crucial for thermodynamic calculations.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid form at a given temperature. It's essential for predicting how a substance will behave when temperature changes. When analyzing the vapor pressures at different temperatures for dichlorodimethylsilane:

• Vapor pressures rise with temperature, as more molecules have enough energy to escape into the vapor phase.
• The provided dataset helps generate a \(\ln(P)\) vs. \(\frac{1}{T}\) plot for extrapolating unknown pressures or temperatures.

Understanding vapor pressure is key in determining boiling points, predicting evaporation rates, and designing equipment for chemical processes.

Temperature Conversion to Kelvin

Temperature in scientific settings is often converted to Kelvin to maintain consistency and accuracy in calculations, especially involving thermodynamic equations. The conversion formula is simple:\[ T(K) = T(°C) + 273.15 \]For the exercise data, converting temperatures from Celsius to Kelvin allows for a proper calculation of the inverse temperatures used in the Clausius-Clapeyron equation:

• For example, converting -0.4°C would result in 272.75 K.
• Similarly, 17.5°C becomes 290.65 K, and so on.

Using Kelvin is necessary for calculations because this scale starts at absolute zero, ensuring that mathematical manipulations are accurate and comply with physical laws.