Question

Dichloromethane, \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\), is widely used as a "degreaser" and paint stripper. Its vapor pressure is \(381.0 \mathrm{~mm} \mathrm{Hg}\) at \(21.9^{\circ} \mathrm{C}\) and \(465.8 \mathrm{~mm} \mathrm{Hg}\) at \(26.9^{\circ} \mathrm{C}\). Estimate (a) its heat of vaporization \(\left(\Delta H_{\text {vap }}\right)\). (b) its normal boiling point.

Short answer

a) The heat of vaporization of dichloromethane is approximately 3.1627 x 10^4 J/mol. b) The normal boiling point of dichloromethane is approximately 315.66 K, or 42.51°C.

Step-by-step solution

Rewrite Clausius-Clapeyron equation

We need to rewrite the Clausius-Clapeyron equation so that we can easily integrate it. The equation can be rewritten as: $$\frac{d\left(\ln P\right)}{\Delta H_{\text {vap }} / R} = \frac{dT}{T^2}$$

Integrate the equation

Now, we need to integrate both sides of the equation. Let's integrate with respect to temperature and vapor pressure: $$\int_{\ln P1}^{\ln P2}\frac{d\left(\ln P\right)}{\Delta H_{\text {vap }} / R} = \int_{T1}^{T2}\frac{dT}{T^2}$$

Solve the integral

Solving the integral, we get: $$\frac{\ln P2 - \ln P1}{\Delta H_{\text {vap }} / R} = -(\frac{1}{T2} - \frac{1}{T1})$$

Insert given values and solve for \(\Delta H_{\text {vap }}\)

Now, we insert the given values of vapor pressure \((P1 = 381.0 \text{ mm Hg}, P2 = 465.8 \text{ mm Hg})\) and temperature \((T1 = 21.9+273.15 \text{ K}, T2 = 26.9+273.15 \text{ K})\), and solve for the unknown \(\Delta H_{\text {vap }}\): $$\frac{\ln (465.8) - \ln (381.0)}{\Delta H_{\text {vap }} / R} = -(\frac{1}{26.9+273.15} - \frac{1}{21.9+273.15})$$ Rearrange and get: $$\Delta H_{\text {vap }} = \frac{R(\ln (465.8) - \ln (381.0))}{-\left(\frac{1}{26.9+273.15} - \frac{1}{21.9+273.15}\right)}$$ Using \(R = 8.314 \mathrm{~J} \mathrm{ K}^{-1} \mathrm{mol}^{-1}\), we find: $$\Delta H_{\text {vap }} \approx 3.1627414 \times10^4 \mathrm{~J} \mathrm{ mol}^{-1}$$ (a) So, the heat of vaporization is \(\Delta H_{\text {vap }} \approx 3.1627 \times10^4 \mathrm{~J} \mathrm{ mol}^{-1}\).

Determine the normal boiling point

To find the normal boiling point, we set the vapor pressure equal to 1 atmosphere (760 mm Hg) and solve for the corresponding temperature using the Clausius-Clapeyron equation: $$\ln\left(\frac{P2}{P1}\right) = \frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T1} - \frac{1}{T2}\right)$$ Where \(P1 = 381.0 \text{ mm Hg}\), \(T1 = 21.9+273.15 \text{ K}\), and \(P2 = 760 \text{ mm Hg}\). Rearrange and get: $$\frac{1}{T2} = \frac{1}{T1} - \frac{R\ln\left(\frac{P2}{P1}\right)}{\Delta H_{\text {vap }}}$$ Plugging in values, we get: $$\frac{1}{T2} = \frac{1}{21.9+273.15} - \frac{8.314\ln\left(\frac{760}{381.0}\right)}{3.1627 \times10^4}$$ Solving for \(T2\), we find: $$T2 \approx 315.66 \mathrm{~K}$$ (b) So, the normal boiling point of dichloromethane is approximately \(315.66 \mathrm{~K}\) or \(42.51^{\circ} \mathrm{C}\).