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Chapter 10: Problem 126

Carbon tetrachloride, \(\mathrm{CCl}_{4},\) has a vapor pressure of 213 torr at \(40 .^{\circ} \mathrm{C}\) and 836 torr at \(80 .^{\circ} \mathrm{C}\) . What is the normal boiling point of \(\mathrm{CCl}_{4} ?\)

Short Answer

Expert verified
The normal boiling point of carbon tetrachloride (CCl4) is 77.8°C.

Step by step solution

01

Convert the given temperatures to Kelvin

To convert the given temperatures from Celsius to Kelvin, add 273.15 K to each temperature: T1 = 40°C + 273.15 = 313.15 K T2 = 80°C + 273.15 = 353.15 K
02

Convert the given vapor pressures to atmospheres

To convert the given vapor pressures from torr to atmospheres, divide each pressure by 760 torr/atm: P1 = 213 torr / 760 torr/atm = 0.2803 atm P2 = 836 torr / 760 torr/atm = 1.1000 atm
03

Set up the Clausius-Clapeyron equation

The Clausius-Clapeyron equation is as follows: \[\ln{\frac{P_2}{P_1}} = -\frac{\Delta H_\mathrm{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] Where \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\), \(\Delta H_\mathrm{vap}\) is the enthalpy of vaporization, and \(R\) is the ideal gas constant (8.314 J/mol·K).
04

Rearrange the Clausius-Clapeyron equation to solve for ΔHvap.

We will rearrange the equation to solve the enthalpy of vaporization, ΔHvap : \[\Delta H_\mathrm{vap} = -R\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \times \frac{\ln{(P_2/P_1)}}{\left(\frac{1}{T_2} - \frac{1}{T_1}\right)}\]
05

Calculate ΔHvap using the given values

Now that we have the appropriate equation, we can plug in the known values for T1, T2, P1, and P2 to find ΔHvap . \[\Delta H_\mathrm{vap} = -8.314\cdot\left(\frac{1}{353.15} - \frac{1}{313.15}\right)^{-1}\cdot \ln{\frac{1.1000~\text{atm}}{0.2803~\text{atm}}}\] ΔHvap ≈ 3.27 x 10^4 J/mol
06

Calculate the boiling point temperature at 1 atm pressure

Now that we have the enthalpy of vaporization, we can use the Clausius-Clapeyron equation to find the boiling point temperature (Tbp) at 1 atm pressure (Pbp). Rearranging the equation, we get: \[\frac{1}{T_\mathrm{bp}} = \frac{1}{T_1} + \frac{R}{\Delta H_\mathrm{vap}} \ln{\frac{P_\mathrm{bp}}{P_1}}\] Plugging in the known values, the equation becomes: \[\frac{1}{T_\mathrm{bp}} = \frac{1}{313.15} + \frac{8.314}{3.27 \times 10^4} \ln{\frac{1~\text{atm}}{0.2803~\text{atm}}}\]
07

Solve for the boiling point temperature and convert to Celsius

Now that we have the equation set up, we can solve for Tbp: \[T_\mathrm{bp} = \left(\frac{1}{313.15}+ \frac{8.314}{3.27 \times 10^4} \ln{\frac{1~\text{atm}}{0.2803~\text{atm}}}\right)^{-1} = 350.95~\text{K}\] Finally, convert the boiling point temperature from Kelvin to Celsius: Tbp = 350.95 K - 273.15 = 77.8°C So, the normal boiling point of carbon tetrachloride is 77.8°C.

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